New Integral Inequalities Connected with Retarded Terms and Their Applications to Differential Equations

Abstract

In this paper, we establish several retarded integral inequalities of Gronwall–Bihari–Pachpatte type for nonnegative functions involving a delayed argument $\alpha \left(t\right)\le t$. The proposed results cover both linear and nonlinear settings and provide explicit upper bounds through an auxiliary integral function and its inverse. In particular, we consider inequalities containing delayed nonlinear terms, kernels depending on both t and s, weighted contributions, and iterated integrals. The considered structures are motivated, in part, by energy estimates arising in nonlinear delay differential systems, where the nonlinear control function is naturally determined by the growth of the nonlinearities. The obtained inequalities extend several classical Gronwall-, Bihari-, and Pachpatte-type estimates by retaining the effect of the variable time delay in the resulting bounds. As applications, suitable a priori estimates are derived for generalized delayed Liénard and Rayleigh systems and are used to establish the global existence of their solutions. The developed framework therefore provides useful tools for the qualitative analysis of nonlinear differential and integral equations with retarded arguments.

1. Introduction

Integral inequalities play a fundamental role in the qualitative theory of differential and integral equations [1,2,3,4,5,6,7]. They provide essential tools for establishing the existence, uniqueness, boundedness, stability, and asymptotic properties of solutions. Among the most classical results are the Gronwall inequality [1], its nonlinear generalization due to Bihari [8], and several extensions developed by Pachpatte [3]. Such inequalities provide explicit estimates for unknown functions in terms of prescribed data and have been extensively used in the analysis of ordinary differential equations, partial differential equations, and integro-differential equations [9,10,11,12,13,14,15,16].

When the underlying system involves time delays or retarded arguments, however, the direct application of classical integral inequalities becomes more restrictive. The presence of delayed terms introduces a nonlocal dependence in time, since the evolution of the solution depends on values attained at earlier instants. This naturally motivates the development of retarded integral inequalities that explicitly incorporate the delay structure. Important contributions in this direction include the retarded Gronwall-like inequality introduced by Lipovan [17], as well as subsequent generalizations obtained by Agarwal et al. [18], Abdeldaim and El-Deeb [19,20], and several other authors [21,22,23,24,25,26]. More recent developments include nonlinear retarded inequalities, inequalities with power-type terms, delay inequalities on time scales, and applications to Volterra and delay equations [27,28,29,30,31,32,33,34]. Such estimates have proved useful in the qualitative analysis of delay differential equations and related dynamical systems.

The results developed in the present paper differ from the classical inequalities cited above by allowing, among other features, kernels $g(t,s)$ depending explicitly on both the current time t and the integration variable s. The dependence on the first variable leads naturally to the appearance of the partial derivative ${\partial }_{t}g$ in the corresponding estimates. This structure is relevant in integral representations involving time-dependent coefficients and Volterra-type terms with memory effects, where the kernel cannot in general be reduced to a function depending only on the integration variable.

The nonlinear retarded structures considered in this paper arise naturally from energy estimates for delay differential equations. For instance, in the generalized delayed Liénard system considered later,

$$\begin{array}{cc}\hfill x^{\prime }\left(t\right)& =y\left(t\right)-F\left(x\right(t\left)\right),\hfill \\ \hfill y^{\prime }\left(t\right)& =G(t,x(\alpha \left(t\right)\left)\right),\hfill \end{array}$$

the natural energy quantity

$$\begin{array}{c}\hfill u\left(t\right)=\sqrt{x^2\left(t\right)+y^2\left(t\right)}\end{array}$$

leads, under assumptions of the form

$$\begin{array}{c}\hfill -xF\left(x\right)\le \left|x\right|v\left(\right|x\left|\right)\end{array}$$

and

$$\begin{array}{c}\hfill G^2(t,x)\le h\left(t\right)\left|x\right|v\left(\right|x\left|\right)+\delta \left(t\right),\end{array}$$

to nonlinear terms involving $u\left(s\right)\omega \left(u\right(s\left)\right)$, where

$$\begin{array}{c}\hfill \omega \left(r\right)=r+v\left(r\right).\end{array}$$

Thus, the nonlinear control function $\omega$ is determined by the growth of the nonlinearities of the underlying system rather than chosen for computational convenience.

Two-variable kernels also arise naturally in integral representations of differential and integro-differential equations. For example, a variation-of-constants formula or a Volterra-type memory term may contain expressions of the form

$$\begin{array}{c}\hfill {\int }_0^{\alpha \left(t\right)}K(t,s)H\left(x\left(s\right)\right)ds,\end{array}$$

where the kernel $K(t,s)$ depends simultaneously on the present time t and the past time s. Under a nonlinear growth estimate for H, such a term leads to retarded contributions of the form

$$\begin{array}{c}\hfill {\int }_0^{\alpha \left(t\right)}f(t,s)u\left(s\right)\omega \left(u\left(s\right)\right)ds.\end{array}$$

Consequently, both the delayed upper limit and the dependence of the kernel on two variables are natural features of delay and Volterra-type problems, rather than structures introduced solely for the purpose of the estimates proved below.

This point also clarifies the relation with classical Gronwall–Bihari-type methods. In some situations, a classical estimate may be recovered by majorizing a delayed integral over $[0,\alpha (t\left)\right]$ by an integral over the larger interval $[0,t]$ or by replacing a two-variable kernel with a suitable one-variable majorant. Such reductions, however, suppress part of the information carried by the actual delayed upper limit and by the dependence of the kernel on the present time. The retarded inequalities developed in this paper are intended to retain these features in the resulting estimates.

The main contribution of this paper is the derivation of several retarded integral inequalities extending classical Gronwall–Bihari–Pachpatte-type estimates to settings involving delayed arguments and more general kernel structures. In particular, the proposed results allow the simultaneous presence of nonlinear terms of the form $u\left(s\right)\omega \left(u\right(s\left)\right)$, linear terms in $u\left(s\right)$, retarded integrals with upper limit $\alpha \left(t\right)$, non-retarded integrals with upper limit t, and kernels depending explicitly on both variables t and s. The resulting estimates are expressed through the inverse of the auxiliary function

$$\begin{array}{c}\hfill \mathsf{\Omega }\left(r\right)={\int }_1^r{\displaystyle \frac{ds}{\omega \left(s\right)}}.\end{array}$$

Several classical inequalities are recovered as particular cases. In addition, comparison with Pachpatte’s inequality shows that, in a specific retarded setting, the estimate obtained in Corollary 9 provides a sharper bound. This comparison is intended as a quantitative illustration of the effect of retaining the delayed integration interval, rather than as the sole motivation for the retarded framework. The usefulness of the developed framework is further illustrated through applications to generalized Liénard and Rayleigh equations with time delay, for which suitable a priori estimates are used to establish global existence of solutions. Further nonlinear inequalities involving iterated integrals are also derived and applied to a Volterra integral equation.

The remainder of the paper is organized as follows. Section 2 collects the notation, classical, and retarded integral inequalities used in the sequel, including the Gronwall, Bihari, retarded Bihari, retarded Gronwall–Bellman, and Pachpatte inequalities. Section 3 contains the main results. In Section 3.1, we establish the principal retarded integral inequalities and their corollaries. Section 3.2 applies these results to prove the global existence of solutions for generalized Liénard and Rayleigh equations with time delay. In Section 3.3, we derive further nonlinear integral inequalities involving iterated integrals and present an application to a Volterra integral equation. Finally, Section 4 contains the concluding remarks.

2. Preliminaries

In this section, we introduce the notation and recall the classical and retarded integral inequalities that will be used throughout the paper.

2.1. Notation

Let ${\mathbb{R}}_{+}=[0,\infty )$. For $T>0$, we consider intervals $[0,T]\subset {\mathbb{R}}_{+}$. We denote by $C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ the set of all continuous functions from ${\mathbb{R}}_{+}$ into ${\mathbb{R}}_{+}$, and by $C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ the set of continuously differentiable functions. For a function

$$\begin{array}{c}\hfill g:{\mathbb{R}}_{+}\times {\mathbb{R}}_{+}⟶{\mathbb{R}}_{+},\end{array}$$

we denote by ${\partial }_{t}g$ its partial derivative with respect to the first variable.

Let $\omega :[0,\infty )\to [0,\infty )$ be continuous and nondecreasing, with $\omega \left(u\right)>0$ for every $u>0$.

Throughout the paper, $\omega$ plays the role of a nonlinear control function; in the applications considered later, its specific form is determined by the growth assumptions imposed on the nonlinearities of the underlying differential system.

For a fixed $u_0>0$, define

$$\begin{array}{c}\hfill \mathsf{\Omega }\left(r\right)={\int }_{u_0}^r{\displaystyle \frac{ds}{\omega \left(s\right)}},r>0.\end{array}$$

(1)

Since $\omega \left(s\right)>0$ for $s>0$, the function $\mathsf{\Omega }$ is continuous and strictly increasing on $(0,\infty )$. We set

$$\begin{array}{c}\hfill \mathsf{\Omega }\left(0\right):=\underset{r\downarrow 0}{\lim }\mathsf{\Omega }\left(r\right)=-{\int }_0^{u_0}{\displaystyle \frac{ds}{\omega \left(s\right)}}\in [-\infty ,0),\end{array}$$

and

$$\begin{array}{c}\hfill \mathsf{\Omega }(\infty ):=\underset{r\to \infty }{\lim }\mathsf{\Omega }\left(r\right)={\int }_{u_0}^{\infty }{\displaystyle \frac{ds}{\omega \left(s\right)}}\in (0,\infty ].\end{array}$$

Accordingly, ${\mathsf{\Omega }}^{-1}$ denotes the inverse of $\mathsf{\Omega }$, defined on

$$\begin{array}{c}\hfill \mathrm{Dom}\left({\mathsf{\Omega }}^{-1}\right)=[\mathsf{\Omega }\left(0\right),\mathsf{\Omega }(\infty )),\end{array}$$

where the left endpoint is understood to be open when $\mathsf{\Omega }\left(0\right)=-\infty$. Throughout the paper, ${\mathsf{\Omega }}^{-1}$ is applied only when its argument belongs to $\mathrm{Dom}\left({\mathsf{\Omega }}^{-1}\right)$; otherwise, the corresponding estimate is understood on the maximal interval on which this condition is satisfied.

2.2. Classical Inequalities

We first recall the classical Gronwall and Bihari inequalities, which form the basis for the retarded estimates developed later.

Lemma 1

(Classical Gronwall inequality [1]). Let $u,f\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$, and let $c\ge 0$ be a constant. If

$$\begin{array}{c}\hfill u\left(t\right)\le c+{\int }_0^{t}f\left(s\right)u\left(s\right)ds,t\ge 0,\end{array}$$

(2)

then

$$\begin{array}{c}\hfill u\left(t\right)\le c\exp \left({\int }_0^{t}f\left(s\right)ds\right),t\ge 0.\end{array}$$

(3)

Lemma 2

(Classical Bihari inequality [8]). Let $u,f\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$, let $\omega \in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ be nondecreasing with $\omega \left(u\right)>0$ for $u>0$, and let $c>0$. If

$$\begin{array}{c}\hfill u\left(t\right)\le c+{\int }_0^{t}f\left(s\right)\omega \left(u\left(s\right)\right)ds,t\ge 0,\end{array}$$

(4)

then, for $0\le t\le t_1$,

$$\begin{array}{c}\hfill u\left(t\right)\le {\mathsf{\Omega }}^{-1}\left(\mathsf{\Omega }\left(c\right)+{\int }_0^{t}f\left(s\right)ds\right),\end{array}$$

(5)

where $t_1\in {\mathbb{R}}_{+}$ is chosen so that

$$\begin{array}{c}\hfill \mathsf{\Omega }\left(c\right)+{\int }_0^{t}f\left(s\right)ds\in \mathrm{Dom}\left({\mathsf{\Omega }}^{-1}\right)\end{array}$$

(6)

for every $0\le t\le t_1$.

2.3. Retarded Bihari Inequality

The following result, due to Lipovan [17], is a retarded counterpart of Bihari’s inequality and will be used in the proofs of several of our main results.

Lemma 3

(Retarded Bihari inequality [17]). Let $u,f\in C([t_0,T),{\mathbb{R}}_{+})$, let $\omega \in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ be nondecreasing with $\omega \left(u\right)>0$ for $u>0$, and let $\alpha \in C^1([t_0,T),[t_0,T))$ be nondecreasing with $\alpha \left(t\right)\le t$ on $[t_0,T)$. Let $k>0$. If $u\left(t\right)\le k+{\int }_{\alpha \left(t_0\right)}^{\alpha \left(t\right)}f\left(s\right)\omega \left(u\left(s\right)\right)ds,t_0\le t<T;$ then, for $t_0\le t\le t_1$, $u\left(t\right)\le {\mathsf{\Omega }}^{-1}\left(\mathsf{\Omega }\left(k\right)+{\int }_{\alpha \left(t_0\right)}^{\alpha \left(t\right)}f\left(s\right)ds\right),$ where $t_1\in (t_0,T)$ is chosen so that $\mathsf{\Omega }\left(k\right)+{\int }_{\alpha \left(t_0\right)}^{\alpha \left(t\right)}f\left(s\right)ds\in \mathrm{Dom}\left({\mathsf{\Omega }}^{-1}\right)$ for every $t\in [t_0,t_1]$.

2.4. Retarded Gronwall–Bellman Inequality

We also recall the following retarded form of the Gronwall–Bellman inequality.

Lemma 4

(Retarded Gronwall–Bellman inequality [17]). Let $u,f,g\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$, let $\alpha \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ be nondecreasing with $\alpha \left(t\right)\le t$, and let $c\ge 0$. If

$$\begin{array}{c}\hfill u\left(t\right)\le c+{\int }_0^{\alpha \left(t\right)}f\left(s\right)u\left(s\right)ds+{\int }_0^{t}g\left(s\right)u\left(s\right)ds,t\ge 0,\end{array}$$

(7)

then

$$\begin{array}{c}\hfill u\left(t\right)\le c\exp \left({\int }_0^{\alpha \left(t\right)}f\left(s\right)ds+{\int }_0^{t}g\left(s\right)ds\right),t\ge 0.\end{array}$$

(8)

The preceding inequalities involve coefficients depending only on the integration variable. In the main results below, this framework is extended to retarded inequalities containing kernels depending on both variables t and s, as motivated in the Introduction.

2.5. Comparison with Pachpatte’s Inequality

For comparison with our results, we recall the following inequality due to Pachpatte [3].

Lemma 5

(Pachpatte’s inequality [3], Theorem 1(a1)). Let $u,f,g\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$, and let $c\ge 0$. If

$$\begin{array}{c}\hfill u^2\left(t\right)\le c^2+2{\int }_0^t\left[f\left(s\right)u^2\left(s\right)+g\left(s\right)u\left(s\right)\right]ds,t\ge 0,\end{array}$$

(9)

then

$$\begin{array}{c}\hfill u\left(t\right)\le \left(c+{\int }_0^{t}g\left(s\right)ds\right)\exp \left({\int }_0^{t}f\left(s\right)ds\right),t\ge 0.\end{array}$$

(10)

The comparison between Pachpatte’s inequality and the corresponding retarded estimate obtained in this paper is discussed in Remark 3.

3. Main Results

In this section, we establish several retarded integral inequalities involving nonlinear control functions, variable delays, and time-dependent kernels. The resulting estimates extend the classical inequalities recalled in Section 2 and will subsequently be applied to delay differential equations.

3.1. Some Retarded Integral Inequalities

Theorem 1.

Let $u,f\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$, and let $g\in C({\mathbb{R}}_{+}\times {\mathbb{R}}_{+},{\mathbb{R}}_{+})$ be such that $(t,s)\mapsto {\partial }_{t}g(t,s)\in C({\mathbb{R}}_{+}\times {\mathbb{R}}_{+},{\mathbb{R}}_{+}).$ Let $c\ge 0$, let $\omega \in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ be nondecreasing with $\omega \left(u\right)>0$ for $u>0$, and let $\alpha \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ be nondecreasing with $\alpha \left(t\right)\le t$ for all $t\ge 0$.

If

$$u^2\left(t\right)\le c^2+2{\int }_0^{\alpha \left(t\right)}\left[f\left(s\right)u\left(s\right)\omega \left(u\right(s\left)\right)+g(t,s)u\left(s\right)\right]ds,t\ge 0,$$

(11)

then, for $0\le t\le t_1$,

$$u\left(t\right)\le {\mathsf{\Omega }}^{-1}\left[\mathsf{\Omega }\left(c+{\int }_0^{\alpha \left(t\right)}g(t,s)ds\right)+{\int }_0^{\alpha \left(t\right)}f\left(s\right)ds\right],$$

(12)

where $t_1\in {\mathbb{R}}_{+}$ is chosen so that $\mathsf{\Omega }\left(c+{\int }_0^{\alpha \left(t\right)}g(t,s)ds\right)+{\int }_0^{\alpha \left(t\right)}f\left(s\right)ds\in \mathrm{Dom}\left({\mathsf{\Omega }}^{-1}\right)$ for every $0\le t\le t_1$.

Remark 1.

If

$$\begin{array}{c}\hfill {\int }_1^{\infty }{\displaystyle \frac{ds}{\omega \left(s\right)}}=\infty ,\end{array}$$

then $\mathsf{\Omega }(\infty )=\infty$, and the estimate (12) is valid for all $t\ge 0$. Typical examples include

$$\begin{array}{c}\hfill \omega \left(u\right)=u\mathrm{and}\omega \left(u\right)=u\ln (1+u).\end{array}$$

Remark 2.

The assumptions on g and ${\partial }_{t}g$ ensure the monotonicity properties required in the proof below. In particular,

$$\begin{array}{c}\hfill p\left(t\right):=c+{\int }_0^{\alpha \left(t\right)}g(t,s)ds\end{array}$$

(13)

is nondecreasing. The condition $\alpha \left(t\right)\le t$ expresses the retarded character of the estimate, whereas the monotonicity of ω is used in the comparison argument leading to the Bihari-type bound.

Proof. 

Assume first that $c>0$ and define

$$\begin{array}{c}\hfill z\left(t\right)=c^2+2{\int }_0^{\alpha \left(t\right)}\left[f\left(s\right)u\left(s\right)\omega \left(u\right(s\left)\right)+g(t,s)u\left(s\right)\right]ds.\end{array}$$

(14)

Since $g\ge 0$, ${\partial }_{t}g\ge 0$, and $\alpha$ is nondecreasing, both z and p (defined in (13)) are nondecreasing. Moreover, $z\left(0\right)=c^2$ and $u\left(t\right)\le \sqrt{z\left(t\right)}.$ For $0\le s\le \alpha \left(t\right)$, we therefore have

$$\begin{array}{c}\hfill u\left(s\right)\le \sqrt{z\left(s\right)}\le \sqrt{z\left(\alpha \right(t\left)\right)}.\end{array}$$

Differentiating z gives

$$\begin{array}{cc}\hfill z^{\prime }\left(t\right)=& 2\left[f\left(\alpha \right(t\left)\right)u\left(\alpha \right(t\left)\right)\omega \left(u\right(\alpha \left(t\right)\left)\right)+g(t,\alpha (t\left)\right)u\left(\alpha \right(t\left)\right)\right]{\alpha }^{\prime }\left(t\right)\hfill \\ \hfill & +2{\int }_0^{\alpha \left(t\right)}{\partial }_{t}g(t,s)u\left(s\right)ds.\hfill \end{array}$$

Using the preceding estimates and the monotonicity of $\omega$, we obtain

$$\begin{array}{c}{z}^{\prime }\left(t\right)\le 2\sqrt{z\left(\alpha \right(t\left)\right)}(\left[f\left(\alpha \left(t\right)\right)\omega \left(\sqrt{z\left(\alpha \right(t\left)\right)}\right)+g(t,\alpha \left(t\right))\right]{\alpha }^{\prime }\left(t\right)\\ +{\int }_0^{\alpha \left(t\right)}{\partial }_{t}g(t,s)ds).\end{array}$$

(15)

Since $\alpha \left(t\right)\le t$ and z is nondecreasing, $\sqrt{z\left(\alpha \right(t\left)\right)}\le \sqrt{z\left(t\right)};$ hence,

$$\begin{array}{cc}\hfill {\displaystyle \frac{z^{\prime }\left(t\right)}{2\sqrt{z\left(t\right)}}}\le & \left[f\left(\alpha \left(t\right)\right)\omega \left(\sqrt{z\left(\alpha \right(t\left)\right)}\right)+g(t,\alpha \left(t\right))\right]{\alpha }^{\prime }\left(t\right)\hfill \\ \hfill & +{\int }_0^{\alpha \left(t\right)}{\partial }_{t}g(t,s)ds.\hfill \end{array}$$

(16)

Integrating over $[0,t]$ yields

$$\sqrt{z\left(t\right)}\le p\left(t\right)+{\int }_0^{\alpha \left(t\right)}f\left(s\right)\omega \left(\sqrt{z\left(s\right)}\right)ds.$$

(17)

Let $T\le t_1$ be arbitrary. Since p is nondecreasing,

$$\begin{array}{c}\hfill \sqrt{z\left(t\right)}\le p\left(T\right)+{\int }_0^{\alpha \left(t\right)}f\left(s\right)\omega \left(\sqrt{z\left(s\right)}\right)ds,0\le t\le T.\end{array}$$

(18)

Applying Lemma 3 gives

$$\begin{array}{c}\hfill \sqrt{z\left(t\right)}\le {\mathsf{\Omega }}^{-1}\left[\mathsf{\Omega }\left(p\left(T\right)\right)+{\int }_0^{\alpha \left(t\right)}f\left(s\right)ds\right],0\le t\le T.\end{array}$$

(19)

Taking $t=T$ and using $u\left(T\right)\le \sqrt{z\left(T\right)}$, we obtain

$$\begin{array}{c}\hfill u\left(T\right)\le {\mathsf{\Omega }}^{-1}\left[\mathsf{\Omega }\left(p\left(T\right)\right)+{\int }_0^{\alpha \left(T\right)}f\left(s\right)ds\right].\end{array}$$

(20)

Since $T\le t_1$ is arbitrary, (12) follows.

If $c=0$, the argument is applied with c replaced by $\epsilon >0$, and then $\epsilon \to 0^{+}$. □

Corollary 1.

Let $u,f\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$, and let $g\in C({\mathbb{R}}_{+}\times {\mathbb{R}}_{+},{\mathbb{R}}_{+})$ satisfy $(t,s)\mapsto {\partial }_{t}g(t,s)\in C({\mathbb{R}}_{+}\times {\mathbb{R}}_{+},{\mathbb{R}}_{+}).$ Suppose that $\alpha \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ is nondecreasing with $\alpha \left(t\right)\le t$, and let $c\ge 0$. If

$$u^2\left(t\right)\le c^2+2{\int }_0^{\alpha \left(t\right)}\left[f\left(s\right)u^2\left(s\right)+g(t,s)u\left(s\right)\right]ds,t\ge 0,$$

(21)

then

$$u\left(t\right)\le \left(c+{\int }_0^{\alpha \left(t\right)}g(t,s)ds\right)\exp \left({\int }_0^{\alpha \left(t\right)}f\left(s\right)ds\right),t\ge 0.$$

(22)

Proof. 

Apply Theorem 1 with $\omega \left(u\right)=u$. In this case, $\mathsf{\Omega }\left(r\right)=\ln r$ up to an additive constant, and the conclusion follows immediately. □

Remark 3.

Corollary 1 provides a retarded counterpart of the Pachpatte inequality recalled in Lemma 5. Indeed, if $g(t,s)=g\left(s\right)$ and $\alpha \left(t\right)=t$, then (21) reduces exactly to the hypothesis of Lemma 5.

The effect of the retarded upper limit can also lead to a smaller estimate. For example, let

$$\begin{array}{c}\hfill c=1,\alpha \left(t\right)=\ln (1+t),f\left(s\right)=g\left(s\right)={\displaystyle \frac{1}{1+s}}.\end{array}$$

Since $\alpha \left(t\right)\le t$, the hypothesis of Corollary 1 implies the corresponding Pachpatte inequality obtained by replacing $\alpha \left(t\right)$ with t. Lemma 5 then gives

$$\begin{array}{c}\hfill u\left(t\right)\le (1+t)\left(1+\ln (1+t)\right).\end{array}$$

On the other hand, Corollary 1 yields

$$\begin{array}{c}\hfill u\left(t\right)\le \left(1+\ln (1+t)\right)\left(1+\ln (1+\ln (1+t\left)\right)\right).\end{array}$$

Thus, in this particular retarded setting, the estimate obtained from Corollary 1 is sharper than the estimate obtained by first enlarging the retarded integral to the interval $[0,t]$ and then applying Pachpatte’s inequality.

Theorem 2.

Let $u,g\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$, and let $f\in C({\mathbb{R}}_{+}\times {\mathbb{R}}_{+},{\mathbb{R}}_{+})$ satisfy $(t,s)\mapsto {\partial }_{t}f(t,s)\in C({\mathbb{R}}_{+}\times {\mathbb{R}}_{+},{\mathbb{R}}_{+}).$ Let $c\ge 0$, let $\omega \in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ be nondecreasing with $\omega \left(u\right)>0$ for $u>0$, and let $\alpha \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ be nondecreasing with $\alpha \left(t\right)\le t$.

If

$$u^2\left(t\right)\le c^2+2{\int }_0^{\alpha \left(t\right)}\left[f(t,s)u\left(s\right)\omega \left(u\right(s\left)\right)+g\left(s\right)u\left(s\right)\right]ds,t\ge 0,$$

(23)

then, for $0\le t\le t_1$,

$$u\left(t\right)\le {\mathsf{\Omega }}^{-1}\left[\mathsf{\Omega }\left(c+{\int }_0^{\alpha \left(t\right)}g\left(s\right)ds\right)+{\int }_0^{\alpha \left(t\right)}f(t,s)ds\right],$$

(24)

where $t_1$ is chosen so that the argument of ${\mathsf{\Omega }}^{-1}$ belongs to $\mathrm{Dom}\left({\mathsf{\Omega }}^{-1}\right)$ for every $0\le t\le t_1$.

Proof. 

Assume first that $c>0$ and define

$$\begin{array}{c}\hfill z\left(t\right)=c^2+2{\int }_0^{\alpha \left(t\right)}\left[f(t,s)u\left(s\right)\omega \left(u\right(s\left)\right)+g\left(s\right)u\left(s\right)\right]ds,\end{array}$$

(25)

and define

$$\begin{array}{c}\hfill p\left(t\right)=c+{\int }_0^{\alpha \left(t\right)}g\left(s\right)ds.\end{array}$$

(26)

Then z and p are nondecreasing, $z\left(0\right)=c^2$, and $u\left(t\right)\le \sqrt{z\left(t\right)}.$

Differentiating z gives

$$\begin{array}{cc}\hfill z^{\prime }\left(t\right)=& 2\left[f(t,\alpha (t\left)\right)u\left(\alpha \right(t\left)\right)\omega \left(u\right(\alpha \left(t\right)\left)\right)+g\left(\alpha \right(t\left)\right)u\left(\alpha \right(t\left)\right)\right]{\alpha }^{\prime }\left(t\right)\hfill \\ \hfill & +2{\int }_0^{\alpha \left(t\right)}{\partial }_{t}f(t,s)u\left(s\right)\omega \left(u\left(s\right)\right)ds.\hfill \end{array}$$

(27)

Proceeding as in the proof of Theorem 1, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{z^{\prime }\left(t\right)}{2\sqrt{z\left(t\right)}}}\le & f(t,\alpha \left(t\right))\omega \left(\sqrt{z\left(\alpha \right(t\left)\right)}\right){\alpha }^{\prime }\left(t\right)+g\left(\alpha \left(t\right)\right){\alpha }^{\prime }\left(t\right)\hfill \\ \hfill & +{\int }_0^{\alpha \left(t\right)}{\partial }_{t}f(t,s)\omega \left(\sqrt{z\left(s\right)}\right)ds.\hfill \end{array}$$

(28)

Integration over $[0,t]$ yields

$$\sqrt{z\left(t\right)}\le p\left(t\right)+{\int }_0^{\alpha \left(t\right)}f(t,s)\omega \left(\sqrt{z\left(s\right)}\right)ds.$$

(29)

Let $T\le t_1$ be arbitrary. Since p is nondecreasing and ${\partial }_{t}f(t,s)\ge 0$, for $0\le t\le T$ we have $p\left(t\right)\le p\left(T\right)$ and $f(t,s)\le f(T,s).$ Consequently,

$$\begin{array}{c}\hfill \sqrt{z\left(t\right)}\le p\left(T\right)+{\int }_0^{\alpha \left(t\right)}f(T,s)\omega \left(\sqrt{z\left(s\right)}\right)ds,0\le t\le T.\end{array}$$

(30)

For fixed T, the function $s\mapsto f(T,s)$ is a nonnegative continuous function. Therefore, Lemma 3 gives

$$\begin{array}{c}\hfill \sqrt{z\left(t\right)}\le {\mathsf{\Omega }}^{-1}\left[\mathsf{\Omega }\left(p\left(T\right)\right)+{\int }_0^{\alpha \left(t\right)}f(T,s)ds\right],0\le t\le T.\end{array}$$

(31)

Taking $t=T$ yields

$$\begin{array}{c}\hfill u\left(T\right)\le {\mathsf{\Omega }}^{-1}\left[\mathsf{\Omega }\left(p\left(T\right)\right)+{\int }_0^{\alpha \left(T\right)}f(T,s)ds\right].\end{array}$$

(32)

Since T is arbitrary, (24) follows.

The case $c=0$ follows by replacing c with $\epsilon >0$ and letting $\epsilon \to 0^{+}$. □

Remark 4.

The distinctive feature of Theorem 2 is the dependence of the nonlinear kernel on both variables. As discussed in the Introduction, such two-variable kernels arise naturally in Volterra-type and time-dependent integral representations, where the current time t and the past time s play distinct roles. The condition ${\partial }_{t}f(t,s)\ge 0$ allows the kernel $f(t,s)$ to be bounded by $f(T,s)$ on $0\le t\le T$, which makes it possible to apply the retarded Bihari inequality. For example, if $\omega \left(u\right)=au+b$ with $a>0$ and $b\ge 0$, then

$$\begin{array}{c}\hfill \mathsf{\Omega }\left(r\right)={\displaystyle \frac{1}{a}}\ln \left({\displaystyle \frac{ar+b}{a{u}_0+b}}\right),\end{array}$$

(33)

therefore

$$\begin{array}{c}\hfill {\mathsf{\Omega }}^{-1}\left(\mathsf{\Omega }\left(p\right)+F\right)={\displaystyle \frac{(ap+b)e^{aF}-b}{a}}.\end{array}$$

(34)

Thus, Theorem 2 gives an explicit exponential-type estimate in this case.

Theorem 3.

Let $u,f,g\in C([0,T],{\mathbb{R}}_{+})$, and let $c\ge 0$. Suppose that $\omega \in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ is nondecreasing with $\omega \left(u\right)>0$ for $u>0$, and that $\alpha \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ is nondecreasing with $\alpha \left(t\right)\le t$.

If

$$u^2\left(t\right)\le c^2+2{\int }_0^{\alpha \left(t\right)}f\left(s\right)u\left(s\right)\omega \left(u\left(s\right)\right)ds+2{\int }_0^{t}g\left(s\right)u\left(s\right)ds,0\le t\le T,$$

(35)

then, for $0\le t\le t_1\le T$,

$$u\left(t\right)\le {\mathsf{\Omega }}^{-1}\left[\mathsf{\Omega }\left(c+{\int }_0^{t}g\left(s\right)ds\right)+{\int }_0^{\alpha \left(t\right)}f\left(s\right)ds\right],$$

(36)

provided that the argument of ${\mathsf{\Omega }}^{-1}$ belongs to $\mathrm{Dom}\left({\mathsf{\Omega }}^{-1}\right)$.

Proof. 

Assume first that $c>0$, and define the right-hand side of (35) by $z\left(t\right)$. Set

$$\begin{array}{c}\hfill p\left(t\right)=c+{\int }_0^{t}g\left(s\right)ds.\end{array}$$

(37)

Then z and p are nondecreasing and $u\left(t\right)\le \sqrt{z\left(t\right)}$. Differentiation gives

$$\begin{array}{c}\hfill z^{\prime }\left(t\right)=2f\left(\alpha \left(t\right)\right)u\left(\alpha \left(t\right)\right)\omega \left(u\left(\alpha \left(t\right)\right)\right){\alpha }^{\prime }\left(t\right)+2g\left(t\right)u\left(t\right).\end{array}$$

(38)

Hence,

$$\begin{array}{c}\hfill {\displaystyle \frac{z^{\prime }\left(t\right)}{2\sqrt{z\left(t\right)}}}\le f\left(\alpha \left(t\right)\right)\omega \left(\sqrt{z\left(\alpha \right(t\left)\right)}\right){\alpha }^{\prime }\left(t\right)+g\left(t\right).\end{array}$$

(39)

Integrating over $[0,t]$, we obtain

$$\begin{array}{c}\hfill \sqrt{z\left(t\right)}\le p\left(t\right)+{\int }_0^{\alpha \left(t\right)}f\left(s\right)\omega \left(\sqrt{z\left(s\right)}\right)ds.\end{array}$$

(40)

Let $T_0\le t_1$. Since p is nondecreasing,

$$\begin{array}{c}\hfill \sqrt{z\left(t\right)}\le p\left(T_0\right)+{\int }_0^{\alpha \left(t\right)}f\left(s\right)\omega \left(\sqrt{z\left(s\right)}\right)ds,0\le t\le T_0.\end{array}$$

(41)

Applying Lemma 3 and then setting $t=T_0$ gives

$$\begin{array}{c}\hfill u\left(T_0\right)\le {\mathsf{\Omega }}^{-1}\left[\mathsf{\Omega }\left(p\left(T_0\right)\right)+{\int }_0^{\alpha \left(T_0\right)}f\left(s\right)ds\right].\end{array}$$

(42)

Since $T_0$ is arbitrary, (36) follows.

The case $c=0$ is treated by the usual $\epsilon$-argument. □

Remark 5.

Theorem 3 separates two different contributions: the retarded nonlinear term

$$\begin{array}{c}\hfill {\int }_0^{\alpha \left(t\right)}f\left(s\right)u\left(s\right)\omega \left(u\left(s\right)\right)ds\end{array}$$

(43)

and the non-retarded term

$$\begin{array}{c}\hfill {\int }_0^{t}g\left(s\right)u\left(s\right)ds.\end{array}$$

(44)

This distinction is useful for differential equations containing both delayed and non-retarded effects.

Theorem 4.

Let $u,f\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$, and let $g\in C({\mathbb{R}}_{+}\times {\mathbb{R}}_{+},{\mathbb{R}}_{+})$ satisfy $(t,s)\mapsto {\partial }_{t}g(t,s)\in C({\mathbb{R}}_{+}\times {\mathbb{R}}_{+},{\mathbb{R}}_{+}).$ Suppose that $\omega \in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ is nondecreasing with $\omega \left(u\right)>0$ for $u>0$, and that $\alpha \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ is nondecreasing with $\alpha \left(t\right)\le t$. Let $c\ge 0$.

If

$$u^2\left(t\right)\le c^2+2{\int }_0^{\alpha \left(t\right)}f\left(s\right)u\left(s\right)\omega \left(u\left(s\right)\right)ds+2{\int }_0^{t}g(t,s)u\left(s\right)\omega \left(u\left(s\right)\right)ds,t\ge 0,$$

(45)

then, for $0\le t\le t_1$,

$$u\left(t\right)\le {\mathsf{\Omega }}^{-1}\left[\mathsf{\Omega }\left(c\right)+{\int }_0^{\alpha \left(t\right)}f\left(s\right)ds+{\int }_0^{t}g(t,s)ds\right],$$

(46)

provided that the argument of ${\mathsf{\Omega }}^{-1}$ belongs to $\mathrm{Dom}\left({\mathsf{\Omega }}^{-1}\right)$.

Proof. 

Assume first that $c>0$, and define

$$\begin{array}{c}\hfill z\left(t\right)=c^2+2{\int }_0^{\alpha \left(t\right)}f\left(s\right)u\left(s\right)\omega \left(u\left(s\right)\right)ds+2{\int }_0^{t}g(t,s)u\left(s\right)\omega \left(u\left(s\right)\right)ds.\end{array}$$

(47)

Then z is nondecreasing and $u\left(t\right)\le \sqrt{z\left(t\right)}.$ Differentiating z and using the monotonicity of z and $\omega$ gives

$$\begin{array}{cc}\hfill {\displaystyle \frac{z^{\prime }\left(t\right)}{2\sqrt{z\left(t\right)}}}\le & f\left(\alpha \left(t\right)\right)\omega \left(\sqrt{z\left(\alpha \right(t\left)\right)}\right){\alpha }^{\prime }\left(t\right)+g(t,t)\omega \left(\sqrt{z\left(t\right)}\right)\hfill \end{array}$$

(48)

$$\begin{array}{cc}\hfill & +{\int }_0^t{\partial }_{t}g(t,s)\omega \left(\sqrt{z\left(s\right)}\right)ds.\hfill \end{array}$$

(49)

Integration over $[0,t]$ yields

$$\sqrt{z\left(t\right)}\le c+{\int }_0^{\alpha \left(t\right)}f\left(s\right)\omega \left(\sqrt{z\left(s\right)}\right)ds+{\int }_0^{t}g(t,s)\omega \left(\sqrt{z\left(s\right)}\right)ds.$$

(50)

Define

$$\begin{array}{c}\hfill v\left(t\right)=c+{\int }_0^{\alpha \left(t\right)}f\left(s\right)\omega \left(\sqrt{z\left(s\right)}\right)ds+{\int }_0^{t}g(t,s)\omega \left(\sqrt{z\left(s\right)}\right)ds.\end{array}$$

(51)

Then $v\left(0\right)=c,\sqrt{z\left(t\right)}\le v\left(t\right),$ and v is nondecreasing. Since $\omega$ is nondecreasing,

$$\begin{array}{c}\hfill \omega \left(\sqrt{z\left(s\right)}\right)\le \omega \left(v\left(s\right)\right)\le \omega \left(v\left(t\right)\right),0\le s\le t.\end{array}$$

(52)

Consequently,

$$v^{\prime }\left(t\right)\le \omega \left(v\left(t\right)\right)[f\left(\alpha \left(t\right)\right){\alpha }^{\prime }\left(t\right)+g(t,t)+{\int }_0^t{\partial }_{t}g(t,s)ds].$$

Therefore,

$$\begin{array}{c}\hfill {\displaystyle \frac{v^{\prime }\left(t\right)}{\omega \left(v\right(t\left)\right)}}\le {\displaystyle \frac{d}{dt}}\left[{\int }_0^{\alpha \left(t\right)}f\left(s\right)ds+{\int }_0^{t}g(t,s)ds\right].\end{array}$$

(53)

Integrating from 0 to t gives

$$\begin{array}{c}\hfill \mathsf{\Omega }\left(v\left(t\right)\right)\le \mathsf{\Omega }\left(c\right)+{\int }_0^{\alpha \left(t\right)}f\left(s\right)ds+{\int }_0^{t}g(t,s)ds.\end{array}$$

(54)

Since $\mathsf{\Omega }$ is increasing,

$$\begin{array}{c}\hfill u\left(t\right)\le \sqrt{z\left(t\right)}\le v\left(t\right)\le {\mathsf{\Omega }}^{-1}\left[\mathsf{\Omega }\left(c\right)+{\int }_0^{\alpha \left(t\right)}f\left(s\right)ds+{\int }_0^{t}g(t,s)ds\right].\end{array}$$

(55)

This proves (46).

The case $c=0$ follows by replacing c with $\epsilon >0$ and then letting $\epsilon \to 0^{+}$. □

Corollary 2.

Under the assumptions of Theorem 4, let $\omega \left(u\right)=u$. If

$$u^2\left(t\right)\le c^2+2{\int }_0^{\alpha \left(t\right)}f\left(s\right)u^2\left(s\right)ds+2{\int }_0^{t}g(t,s)u^2\left(s\right)ds,t\ge 0,$$

(56)

then

$$u\left(t\right)\le c\exp \left[{\int }_0^{\alpha \left(t\right)}f\left(s\right)ds+{\int }_0^{t}g(t,s)ds\right],t\ge 0.$$

(57)

Proof. 

For $\omega \left(u\right)=u$, the function $\mathsf{\Omega }$ is logarithmic. Hence Theorem 4 gives (57). □

3.2. Application to Global Existence

In this subsection, we illustrate through two examples how the integral inequalities established above can be used to prove global existence for functional differential equations with time delay.

The first example also provides the concrete energy mechanism mentioned in the introduction: the nonlinear control function $\omega$ is determined by the growth assumptions on the nonlinearities, while the retarded integral arises directly from the delayed argument.

We first recall the general functional differential equation

$$\left\{\begin{array}{ccc}{X}^{\prime }\left(t\right)\hfill & =& H(t,X(t),X(\alpha \left(t\right)\left)\right),\hfill \\ X\left(0\right)\hfill & =& X_0,\hfill \end{array}\right.$$

(58)

where $X_0\in {\mathbb{R}}^n,H\in C({\mathbb{R}}_{+}\times {\mathbb{R}}^{2n},{\mathbb{R}}^n),$ and $\alpha \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+}),\alpha \left(t\right)\le t.$ We shall use the standard continuation property for solutions of (58): every local solution possesses a maximal interval of existence $[0,T)$, and if $T<\infty$, then

$$\underset{t\to T^{-}}{\lim \sup }{\parallel X\left(t\right)\parallel }_{{\mathbb{R}}^n}=\infty .$$

(59)

Thus, to prove global existence, it is sufficient to establish that the solution remains bounded on every finite maximal interval.

We shall also use the following elementary consequence of the monotonicity of v. If $v\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ is nondecreasing, $v\left(r\right)>0$ for $r>0$, and

$$\begin{array}{c}\hfill {\int }_1^{\infty }{\displaystyle \frac{ds}{v\left(s\right)}}=\infty ,\end{array}$$

then, for

$$\begin{array}{c}\hfill \omega \left(s\right)=s+v\left(s\right),\end{array}$$

one also has

$$\begin{array}{c}\hfill {\int }_1^{\infty }{\displaystyle \frac{ds}{\omega \left(s\right)}}={\int }_1^{\infty }{\displaystyle \frac{ds}{s+v\left(s\right)}}=\infty .\end{array}$$

Indeed, set

$$\begin{array}{c}\hfill a_n={\displaystyle \frac{2^n}{v\left(2^n\right)}},n\ge 0.\end{array}$$

Since v is nondecreasing,

$$\begin{array}{c}\hfill {\int }_{2^n}^{2^{n+1}}{\displaystyle \frac{ds}{v\left(s\right)}}\le {\displaystyle \frac{2^n}{v\left(2^n\right)}}=a_n.\end{array}$$

Hence

$$\begin{array}{c}\hfill {\int }_1^{\infty }{\displaystyle \frac{ds}{v\left(s\right)}}=\infty \end{array}$$

implies

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=0}^{\infty }}{a}_n=\infty .\end{array}$$

On the other hand, again using the monotonicity of v,

$$\begin{array}{cc}\hfill {\int }_{2^n}^{2^{n+1}}{\displaystyle \frac{ds}{s+v\left(s\right)}}& \ge {\displaystyle \frac{2^n}{2^{n+1}+v\left(2^{n+1}\right)}}\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{\displaystyle \frac{a_{n+1}}{1+a_{n+1}}}.\hfill \end{array}$$

Now,

$$\begin{array}{c}\hfill {\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{a_n}{1+a_n}}=\infty \end{array}$$

whenever ${\displaystyle {\displaystyle \sum _{n=0}^{\infty }}{a}_n=\infty }$: if $a_n\ge 1$ for infinitely many n, the conclusion is immediate, while if $a_n<1$ eventually, then $a_n/(1+a_n)\ge a_n/2$ for all sufficiently large n. Consequently,

$$\begin{array}{c}\hfill {\int }_1^{\infty }{\displaystyle \frac{ds}{s+v\left(s\right)}}=\infty .\end{array}$$

Example 1.

Consider the generalized Liénard equation with time delay

$$\left\{\begin{array}{ccc}{x}^{\prime }\left(t\right)\hfill & =& y\left(t\right)-F\left(x\right(t\left)\right),\hfill \\ y^{\prime }\left(t\right)\hfill & =& G(t,x(t-\tau \left(t\right)\left)\right),\hfill \end{array}\right.$$

(60)

where $F\in C^1(\mathbb{R},\mathbb{R}),G\in C({\mathbb{R}}_{+}\times \mathbb{R},\mathbb{R}),$ and $\tau \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+}),\tau \left(t\right)\le t.$ Assume that $\alpha \left(t\right):=t-\tau \left(t\right)$ is an increasing diffeomorphism of ${\mathbb{R}}_{+}$ onto itself and that

$$\begin{array}{c}-xF\left(x\right)\le \left|x\right|v\left(\right|x\left|\right),x\in \mathbb{R},\hfill \end{array}$$

(61)

$$\begin{array}{c}{G}^2(t,x)\le h\left(t\right)\left|x\right|v\left(\right|x\left|\right)+\delta \left(t\right),(t,x)\in {\mathbb{R}}_{+}\times \mathbb{R},\hfill \end{array}$$

(62)

where $h,\delta \in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ and $v\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ is nondecreasing, satisfies $v\left(u\right)>0$ for $u>0$, and

$$\begin{array}{c}\hfill {\int }_1^{\infty }{\displaystyle \frac{ds}{v\left(s\right)}}=\infty .\end{array}$$

Then every solution of (60) is global.

Indeed, let $\left(x\right(t),y(t\left)\right)$ be a solution of (60) on its maximal interval $[0,T)$. Suppose, by contradiction, that $T<\infty$, and define

$$\begin{array}{c}\hfill u\left(t\right)=\sqrt{x^2\left(t\right)+y^2\left(t\right)},0\le t<T.\end{array}$$

Using (60)–(62), we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{u}^2\left(t\right)& =2x{x}^{\prime }+2y{y}^{\prime }\hfill \\ \hfill & =2xy-2xF\left(x\right)+2yG(t,x(\alpha \left(t\right)\left)\right)\hfill \\ \hfill & \le x^2+y^2-2xF\left(x\right)+y^2+G^2(t,x\left(\alpha \left(t\right)\right))\hfill \\ \hfill & \le 2u^2\left(t\right)+2u\left(t\right)v\left(u\left(t\right)\right)+h\left(t\right)\left|x\left(\alpha \left(t\right)\right)\right|v\left(\right|x\left(\alpha \left(t\right)\right)\left|\right)+\delta \left(t\right).\hfill \end{array}$$

Set $\omega \left(u\right):=u+v\left(u\right).$ Since $\left|x\right(t\left)\right|\le u\left(t\right)$ and ω is nondecreasing,

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}{u}^2\left(t\right)\le 2u\left(t\right)\omega \left(u\left(t\right)\right)+h\left(t\right)\left|x\left(\alpha \left(t\right)\right)\right|\omega \left(\right|x\left(\alpha \left(t\right)\right)\left|\right)+\delta \left(t\right).\end{array}$$

Integrating over $[0,t]$, with $t<T$, yields

$$\begin{array}{cc}\hfill u^2\left(t\right)\le & u^2\left(0\right)+2{\int }_0^{t}u\left(s\right)\omega \left(u\left(s\right)\right)ds\hfill \\ \hfill & +{\int }_0^{t}h\left(s\right)\left|x\left(\alpha \left(s\right)\right)\right|\omega \left(\right|x\left(\alpha \left(s\right)\right)\left|\right)ds+{\int }_0^t\delta \left(s\right)ds.\hfill \end{array}$$

Since α is an increasing diffeomorphism of ${\mathbb{R}}_{+}$, the change of variables $r=\alpha \left(s\right)$ gives

$$\begin{array}{cc}\hfill & {\int }_0^{t}h\left(s\right)\left|x\left(\alpha \left(s\right)\right)\right|\omega \left(\right|x\left(\alpha \left(s\right)\right)\left|\right)ds\hfill \\ \hfill & ={\int }_0^{\alpha \left(t\right)}{\displaystyle \frac{h\left({\alpha }^{-1}\left(r\right)\right)}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(r\right)\right)}}\left|x\left(r\right)\right|\omega \left(\right|x\left(r\right)\left|\right)dr\hfill \\ \hfill & \le {\int }_0^{\alpha \left(t\right)}{\displaystyle \frac{h\left({\alpha }^{-1}\left(r\right)\right)}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(r\right)\right)}}u\left(r\right)\omega \left(u\left(r\right)\right)dr.\hfill \end{array}$$

Because $T<\infty$ and δ is continuous, define

$$\begin{array}{c}\hfill c^2=u^2\left(0\right)+{\int }_0^T\delta \left(s\right)ds<\infty .\end{array}$$

Hence

$$\begin{array}{cc}\hfill u^2\left(t\right)\le & c^2+2{\int }_0^{t}u\left(s\right)\omega \left(u\left(s\right)\right)ds\hfill \\ \hfill & +{\int }_0^{\alpha \left(t\right)}{\displaystyle \frac{h\left({\alpha }^{-1}\left(r\right)\right)}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(r\right)\right)}}u\left(r\right)\omega \left(u\left(r\right)\right)dr.\hfill \end{array}$$

(63)

By the preceding observation and the assumption ${\displaystyle {\int }_1^{\infty }ds/v\left(s\right)=\infty }$, we have

$$\begin{array}{c}\hfill {\int }_1^{\infty }{\displaystyle \frac{ds}{\omega \left(s\right)}}={\int }_1^{\infty }{\displaystyle \frac{ds}{s+v\left(s\right)}}=\infty .\end{array}$$

Define

$$\begin{array}{c}\hfill W\left(r\right)={\int }_1^r{\displaystyle \frac{ds}{\omega \left(s\right)}},r>0.\end{array}$$

Applying Theorem 4 to (63) with $g(t,s)=1$ and

$$\begin{array}{c}\hfill f\left(s\right)={\displaystyle \frac{h\left({\alpha }^{-1}\left(s\right)\right)}{2{\alpha }^{\prime }\left({\alpha }^{-1}\left(s\right)\right)}},\end{array}$$

we obtain

$$\begin{array}{c}\hfill u\left(t\right)\le W^{-1}\left[W\left(c\right)+t+{\displaystyle \frac{1}{2}}{\int }_0^{\alpha \left(t\right)}{\displaystyle \frac{h\left({\alpha }^{-1}\left(r\right)\right)}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(r\right)\right)}}dr\right].\end{array}$$

Using once again the change of variables $r=\alpha \left(s\right)$,

$$\begin{array}{c}\hfill {\int }_0^{\alpha \left(t\right)}{\displaystyle \frac{h\left({\alpha }^{-1}\left(r\right)\right)}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(r\right)\right)}}dr={\int }_0^{t}h\left(s\right)ds.\end{array}$$

Consequently,

$$u\left(t\right)\le W^{-1}\left[W\left(c\right)+t+{\displaystyle \frac{1}{2}}{\int }_0^{t}h\left(s\right)ds\right],0\le t<T.$$

(64)

Since $T<\infty$ and h is continuous, the right-hand side of (64) remains bounded as $t\to T^{-}$. Thus

$$\begin{array}{c}\hfill \underset{0\le t<T}{\sup }u\left(t\right)<\infty ,\end{array}$$

which contradicts the continuation criterion (59). Therefore $T=\infty$, and every solution of (60) is global.

Example 2.

Consider the Rayleigh equation with time delay

$$\left\{\begin{array}{ccc}{x}^{\prime }\left(t\right)\hfill & =& y\left(t\right),\hfill \\ y^{\prime }\left(t\right)\hfill & =& -F\left(y\right(t\left)\right)-G\left(x\right(\alpha \left(t\right)\left)\right),\hfill \end{array}\right.$$

(65)

where $F,G\in C(\mathbb{R},\mathbb{R}),\alpha \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+}),$ and $\alpha \left(t\right)\le t.$ Assume that α is an increasing diffeomorphism of ${\mathbb{R}}_{+}$ onto itself and that

$$\begin{array}{c}-xF\left(x\right)\le \left|x\right|v\left(\right|x\left|\right),x\in \mathbb{R},\hfill \end{array}$$

(66)

$$\begin{array}{c}{G}^2\left(x\right)\le \left|x\right|v\left(\right|x\left|\right),x\in \mathbb{R},\hfill \end{array}$$

(67)

where $v\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ is nondecreasing, $v\left(u\right)>0$ for $u>0$, and

$$\begin{array}{c}\hfill {\int }_1^{\infty }{\displaystyle \frac{ds}{v\left(s\right)}}=\infty .\end{array}$$

Then every solution of (65) is global.

Indeed, let $\left(x\right(t),y(t\left)\right)$ be a solution of (65) defined on its maximal interval of existence $[0,T)$. Suppose, by contradiction, that $T<\infty$, and define

$$\begin{array}{c}\hfill u\left(t\right)=\sqrt{x^2\left(t\right)+y^2\left(t\right)},0\le t<T.\end{array}$$

Using (65), we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{u}^2\left(t\right)& =2x\left(t\right)x^{\prime }\left(t\right)+2y\left(t\right)y^{\prime }\left(t\right)\hfill \\ \hfill & =2x\left(t\right)y\left(t\right)-2y\left(t\right)F\left(y\right(t\left)\right)-2y\left(t\right)G\left(x\right(\alpha \left(t\right)\left)\right).\hfill \end{array}$$

By Young’s inequality,

$$\begin{array}{c}\hfill 2x\left(t\right)y\left(t\right)\le x^2\left(t\right)+y^2\left(t\right)\end{array}$$

and

$$\begin{array}{c}\hfill -2y\left(t\right)G\left(x\left(\alpha \left(t\right)\right)\right)\le y^2\left(t\right)+G^2\left(x\left(\alpha \left(t\right)\right)\right).\end{array}$$

Therefore, using (66) and (67),

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}{u}^2\left(t\right)& \le x^2\left(t\right)+y^2\left(t\right)-2y\left(t\right)F\left(y\left(t\right)\right)+y^2\left(t\right)+G^2\left(x\left(\alpha \left(t\right)\right)\right)\hfill \\ \hfill & \le 2u^2\left(t\right)+2\left|y\left(t\right)\right|v\left(\right|y\left(t\right)\left|\right)+\left|x\left(\alpha \left(t\right)\right)\right|v\left(\right|x\left(\alpha \left(t\right)\right)\left|\right).\hfill \end{array}$$

Since

$$\begin{array}{c}\hfill \left|y\right(t\left)\right|\le u\left(t\right),\left|x\right(\alpha \left(t\right)\left)\right|\le u\left(\alpha \right(t\left)\right),\end{array}$$

and v is nondecreasing, it follows that

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}{u}^2\left(t\right)\le 2u^2\left(t\right)+2u\left(t\right)v\left(u\left(t\right)\right)+u\left(\alpha \left(t\right)\right)v\left(u\left(\alpha \left(t\right)\right)\right).\end{array}$$

Define $\omega \left(u\right):=u+v\left(u\right).$ Then

$$\begin{array}{c}\hfill 2u^2\left(t\right)+2u\left(t\right)v\left(u\left(t\right)\right)=2u\left(t\right)\omega \left(u\left(t\right)\right),\end{array}$$

and, since

$$\begin{array}{c}\hfill u\left(\alpha \right(t\left)\right)v\left(u\right(\alpha \left(t\right)\left)\right)\le u\left(\alpha \right(t\left)\right)\omega \left(u\right(\alpha \left(t\right)\left)\right),\end{array}$$

we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}{u}^2\left(t\right)\le 2u\left(t\right)\omega \left(u\left(t\right)\right)+u\left(\alpha \left(t\right)\right)\omega \left(u\left(\alpha \left(t\right)\right)\right).\end{array}$$

Integrating over $[0,t]$, with $t<T$, gives

$$\begin{array}{cc}\hfill u^2\left(t\right)\le & u^2\left(0\right)+2{\int }_0^{t}u\left(s\right)\omega \left(u\left(s\right)\right)ds\hfill \\ \hfill & +{\int }_0^{t}u\left(\alpha \left(s\right)\right)\omega \left(u\left(\alpha \left(s\right)\right)\right)ds.\hfill \end{array}$$

Since α is an increasing diffeomorphism of ${\mathbb{R}}_{+}$ onto itself, the change of variables $r=\alpha \left(s\right)$ yields

$$\begin{array}{c}\hfill {\int }_0^{t}u\left(\alpha \left(s\right)\right)\omega \left(u\left(\alpha \left(s\right)\right)\right)ds={\int }_0^{\alpha \left(t\right)}{\displaystyle \frac{1}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(r\right)\right)}}u\left(r\right)\omega \left(u\left(r\right)\right)dr.\end{array}$$

Consequently,

$$\begin{array}{cc}\hfill u^2\left(t\right)\le & u^2\left(0\right)+2{\int }_0^{t}u\left(s\right)\omega \left(u\left(s\right)\right)ds\hfill \\ \hfill & +{\int }_0^{\alpha \left(t\right)}{\displaystyle \frac{1}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(r\right)\right)}}u\left(r\right)\omega \left(u\left(r\right)\right)dr.\hfill \end{array}$$

(68)

By the preceding observation and the assumption

$$\begin{array}{c}\hfill {\int }_1^{\infty }{\displaystyle \frac{ds}{v\left(s\right)}}=\infty ,\end{array}$$

we have

$$\begin{array}{c}\hfill {\int }_1^{\infty }{\displaystyle \frac{ds}{\omega \left(s\right)}}={\int }_1^{\infty }{\displaystyle \frac{ds}{s+v\left(s\right)}}=\infty .\end{array}$$

Let

$$\begin{array}{c}\hfill W\left(r\right)={\int }_1^r{\displaystyle \frac{ds}{\omega \left(s\right)}},r>0.\end{array}$$

Applying Theorem 4 to (68) with $c=u\left(0\right)$, $g(t,s)=1$, and

$$\begin{array}{c}\hfill f\left(s\right)={\displaystyle \frac{1}{2{\alpha }^{\prime }\left({\alpha }^{-1}\left(s\right)\right)}},\end{array}$$

we obtain

$$\begin{array}{c}\hfill u\left(t\right)\le W^{-1}\left[W\left(u\left(0\right)\right)+t+{\displaystyle \frac{1}{2}}{\int }_0^{\alpha \left(t\right)}{\displaystyle \frac{dr}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(r\right)\right)}}\right].\end{array}$$

Using again the change of variables $r=\alpha \left(s\right)$, we have

$$\begin{array}{c}\hfill {\int }_0^{\alpha \left(t\right)}{\displaystyle \frac{dr}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(r\right)\right)}}=t.\end{array}$$

Hence

$$u\left(t\right)\le W^{-1}\left[W\left(u\left(0\right)\right)+{\displaystyle \frac{3}{2}}t\right],0\le t<T.$$

(69)

Since $T<\infty$, the right-hand side of (69) remains bounded as $t\to T^{-}$. Therefore,

$$\begin{array}{c}\hfill \underset{0\le t<T}{\sup }u\left(t\right)<\infty ,\end{array}$$

which contradicts the continuation criterion. Hence $T=\infty$, and every solution of (65) is global.

Remark 6.

The conclusion of the preceding example remains valid under the more general assumption

$$\begin{array}{c}\hfill G^2\left(x\right)\le \left|x\right|v\left(\right|x\left|\right)+\delta \left(t\right),t\ge 0,x\in \mathbb{R},\end{array}$$

where $\delta \in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$.

Indeed, along a solution,

$$\begin{array}{c}\hfill G^2\left(x\left(\alpha \left(t\right)\right)\right)\le \left|x\left(\alpha \left(t\right)\right)\right|v\left(\right|x\left(\alpha \left(t\right)\right)\left|\right)+\delta \left(t\right).\end{array}$$

If $[0,T)$ is a maximal interval of existence with $T<\infty$, the additional term can be absorbed into the constant

$$\begin{array}{c}\hfill c_T^2=u^2\left(0\right)+{\int }_0^T\delta \left(s\right)ds,\end{array}$$

which is finite since δ is continuous. Repeating the preceding argument then yields

$$\begin{array}{c}\hfill u\left(t\right)\le W^{-1}\left[W\left(c_T\right)+2t\right],0\le t<T.\end{array}$$

Hence u remains bounded on every finite maximal interval, and the same continuation argument shows that every solution is global.

3.3. Some Nonlinear Inequalities

We conclude this section with several complementary nonlinear inequalities involving iterated retarded integrals. Such structures arise naturally in Volterra-type integral and integro-differential equations, where the present state may depend on cumulative past effects through nested integrals. The results below therefore complement the preceding retarded estimates by covering a different, but related, class of nonlinear memory terms.

Proposition 1.

Let $u\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$, and let $h,g\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$. Let $f\in C^1({\mathbb{R}}_{+}\times {\mathbb{R}}_{+},{\mathbb{R}}_{+})$ satisfy ${\partial }_{t}f(t,s)\ge 0,$ for all $(t,s)\in {\mathbb{R}}_{+}\times {\mathbb{R}}_{+}.$ Assume that $a,\alpha \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ are nondecreasing, with $a\left(0\right)\ge 1$ and $\alpha \left(t\right)\le t,t\ge 0.$ Suppose that, for some exponents $p,q,r$ satisfying $p\ge r\ge 1,1\le q\le p,$ and the function u satisfies

$$u^p\left(t\right)\le a\left(t\right)+{\int }_0^{t}h\left(s\right)u^r\left(s\right)ds+{\int }_0^{\alpha \left(t\right)}f(t,s){\int }_0^{s}g\left(\tau \right)u^q\left(\tau \right)d\tau ds,t\ge 0.$$

(70)

Then

$$u\left(t\right)\le a^{1/p}\left(t\right)\exp \left\{{\displaystyle \frac{1}{p}}\left[{\int }_0^{t}h\left(s\right)ds+{\int }_0^{\alpha \left(t\right)}f(t,s){\int }_0^{s}g\left(\tau \right)d\tau ds\right]\right\},t\ge 0.$$

(71)

Proof. 

We first consider the case $a\left(t\right)\equiv 1$. Define

$$z^p\left(t\right)=1+{\int }_0^{t}h\left(s\right)u^r\left(s\right)ds+{\int }_0^{\alpha \left(t\right)}f(t,s){\int }_0^{s}g\left(\tau \right)u^q\left(\tau \right)d\tau ds.$$

(72)

Then $u\left(t\right)\le z\left(t\right),$ and $z\left(0\right)=1.$ Since all the functions involved are nonnegative, $\alpha$ is nondecreasing, and ${\partial }_{t}f\ge 0$, the function z is nondecreasing.

Differentiating (72), we obtain

$$\begin{array}{cc}\hfill p{z}^{p-1}\left(t\right)z^{\prime }\left(t\right)=& h\left(t\right)u^r\left(t\right)\hfill \end{array}$$

(73)

$$\begin{array}{cc}\hfill & +{\alpha }^{\prime }\left(t\right)f(t,\alpha \left(t\right)){\int }_0^{\alpha \left(t\right)}g\left(s\right)u^q\left(s\right)ds\hfill \end{array}$$

(74)

$$\begin{array}{cc}\hfill & +{\int }_0^{\alpha \left(t\right)}{\partial }_{t}f(t,s){\int }_0^{s}g\left(\tau \right)u^q\left(\tau \right)d\tau ds.\hfill \end{array}$$

(75)

Using $u\le z$ and the monotonicity of z, we obtain

$$\begin{array}{cc}\hfill p{z}^{p-1}\left(t\right)z^{\prime }\left(t\right)\le & h\left(t\right)z^r\left(t\right)\hfill \end{array}$$

(76)

$$\begin{array}{cc}\hfill & +{\alpha }^{\prime }\left(t\right)f(t,\alpha \left(t\right))z^q\left(t\right){\int }_0^{\alpha \left(t\right)}g\left(s\right)ds\hfill \end{array}$$

(77)

$$\begin{array}{cc}\hfill & +z^q\left(t\right){\int }_0^{\alpha \left(t\right)}{\partial }_{t}f(t,s){\int }_0^{s}g\left(\tau \right)d\tau ds.\hfill \end{array}$$

(78)

Since $z\left(t\right)\ge z\left(0\right)=1,$$p\ge r,$ and $p\ge q,$ we have $z^r\left(t\right)\le z^p\left(t\right)$ and $z^q\left(t\right)\le z^p\left(t\right).$ Hence

$$\begin{array}{c}p{z}^{p-1}\left(t\right)z^{\prime }\left(t\right)\le z^p\left(t\right)[h\left(t\right)+{\alpha }^{\prime }\left(t\right)f(t,\alpha \left(t\right)){\int }_0^{\alpha \left(t\right)}g\left(s\right)ds\hfill \end{array}$$

(79)

$$\begin{array}{cc}\hfill & +{\int }_0^{\alpha \left(t\right)}{\partial }_{t}f(t,s){\int }_0^{s}g\left(\tau \right)d\tau ds].\hfill \end{array}$$

(80)

Dividing by $z^{p-1}\left(t\right)$ gives

$$\begin{array}{cc}\hfill p{z}^{\prime }\left(t\right)\le z\left(t\right)& [h\left(t\right)+{\alpha }^{\prime }\left(t\right)f(t,\alpha \left(t\right)){\int }_0^{\alpha \left(t\right)}g\left(s\right)ds\hfill \\ \hfill & +{\int }_0^{\alpha \left(t\right)}{\partial }_{t}f(t,s){\int }_0^{s}g\left(\tau \right)d\tau ds].\hfill \end{array}$$

(81)

Dividing by $z\left(t\right)>0$ and integrating over $[0,t]$, we obtain

$$\begin{array}{cc}\hfill \ln z\left(t\right)\le {\displaystyle \frac{1}{p}}& [{\int }_0^{t}h\left(s\right)ds\hfill \\ \hfill & +{\int }_0^t{\alpha }^{\prime }\left(s\right)f(s,\alpha \left(s\right)){\int }_0^{\alpha \left(s\right)}g\left(\tau \right)d\tau ds\hfill \\ \hfill & +{\int }_0^t{\int }_0^{\alpha \left(s\right)}{\partial }_{s}f(s,r){\int }_0^{r}g\left(\tau \right)d\tau drds].\hfill \end{array}$$

The last two terms are precisely the derivative contribution of

$$\begin{array}{c}\hfill {\int }_0^{\alpha \left(t\right)}f(t,s){\int }_0^{s}g\left(\tau \right)d\tau ds.\end{array}$$

(82)

Therefore,

$$\begin{array}{c}\hfill z\left(t\right)\le \exp \left\{{\displaystyle \frac{1}{p}}\left[{\int }_0^{t}h\left(s\right)ds+{\int }_0^{\alpha \left(t\right)}f(t,s){\int }_0^{s}g\left(\tau \right)d\tau ds\right]\right\}.\end{array}$$

(83)

Since $u\left(t\right)\le z\left(t\right)$, the desired estimate follows when $a\left(t\right)\equiv 1$.

We now consider the general case. Since a is nondecreasing and $a\left(0\right)\ge 1$, we have $a\left(t\right)\ge 1$ and $t\ge 0.$ Define

$$\begin{array}{c}\hfill v\left(t\right)={\displaystyle \frac{u\left(t\right)}{a^{1/p}\left(t\right)}}.\end{array}$$

(84)

For $0\le s\le t$, the monotonicity of a gives $a\left(s\right)\le a\left(t\right).$ Consequently,

$$\begin{array}{c}\hfill {\displaystyle \frac{u^r\left(s\right)}{a\left(t\right)}}\le {\displaystyle \frac{u^r\left(s\right)}{a\left(s\right)}}=v^r\left(s\right)a^{r/p-1}\left(s\right)\le v^r\left(s\right),\end{array}$$

(85)

because $r\le p$ and $a\left(s\right)\ge 1$.

Similarly, since $q\le p$, we obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{u^q\left(\tau \right)}{a\left(t\right)}}\le {\displaystyle \frac{u^q\left(\tau \right)}{a\left(\tau \right)}}=v^q\left(\tau \right)a^{q/p-1}\left(\tau \right)\le v^q\left(\tau \right),0\le \tau \le t.\end{array}$$

(86)

Dividing (70) by $a\left(t\right)$ therefore yields

$$\begin{array}{c}\hfill v^p\left(t\right)\le 1+{\int }_0^{t}h\left(s\right)v^r\left(s\right)ds+{\int }_0^{\alpha \left(t\right)}f(t,s){\int }_0^{s}g\left(\tau \right)v^q\left(\tau \right)d\tau ds.\end{array}$$

(87)

Applying the already proved case $a\equiv 1$, we obtain

$$\begin{array}{c}\hfill v\left(t\right)\le \exp \left\{{\displaystyle \frac{1}{p}}\left[{\int }_0^{t}h\left(s\right)ds+{\int }_0^{\alpha \left(t\right)}f(t,s){\int }_0^{s}g\left(\tau \right)d\tau ds\right]\right\}.\end{array}$$

(88)

Since $u\left(t\right)=a^{1/p}\left(t\right)v\left(t\right),$ we obtain (71). □

Remark 7.

Proposition 1 provides an explicit estimate for integral inequalities containing simultaneously a non-retarded nonlinear term and an iterated retarded term. The restrictions $r\le p,$ and $q\le p$ allow the powers $z^r$ and $z^q$ occurring in the proof to be controlled by $z^p$, while the condition ${\partial }_{t}f\ge 0$ guarantees the monotonicity required for the time-dependent kernel $f(t,s)$.

Proposition 2.

Let $a\in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ and $\alpha \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ be nondecreasing functions satisfying $\alpha \left(t\right)\le t$ for all $t\ge 0$, and suppose that $a\left(0\right)\ge 1$. Let $f\in C^1({\mathbb{R}}_{+}\times {\mathbb{R}}_{+},{\mathbb{R}}_{+}),g,h\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+}),$ and assume that ${\partial }_{t}f(t,s)\ge 0,t,s\ge 0.$ Let $\omega \in C({\mathbb{R}}_{+},[1,\infty ))$ be nondecreasing and satisfy $\omega \left(r\right)\ge r,r\ge 0.$

If $u\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ satisfies

$$\begin{array}{cc}\hfill u\left(t\right)\le & a\left(t\right)+{\int }_0^{t}h\left(s\right)\omega \left(u\left(s\right)\right)ds+{\int }_0^{\alpha \left(t\right)}f(t,s)\left({\int }_0^{s}g\left(\tau \right)u\left(\tau \right)d\tau \right)ds,t\ge 0,\hfill \end{array}$$

(89)

then

$$\begin{array}{cc}\hfill W\left(u\right(t\left)\right)\le & W\left(a\left(0\right)\right)+a\left(t\right)-a\left(0\right)+{\int }_0^{t}h\left(s\right)ds\hfill \\ \hfill & +{\int }_0^{\alpha \left(t\right)}f(t,s)\left({\int }_0^{s}g\left(\tau \right)d\tau \right)ds,t\ge 0,\hfill \end{array}$$

(90)

where

$$\begin{array}{c}\hfill W\left(r\right)={\int }_0^r{\displaystyle \frac{ds}{\omega \left(s\right)}},r\ge 0.\end{array}$$

Proof. 

Define

$$\begin{array}{cc}\hfill z\left(t\right)=& a\left(t\right)+{\int }_0^{t}h\left(s\right)\omega \left(u\left(s\right)\right)ds+{\int }_0^{\alpha \left(t\right)}f(t,s)\left({\int }_0^{s}g\left(\tau \right)u\left(\tau \right)d\tau \right)ds.\hfill \end{array}$$

Then $u\left(t\right)\le z\left(t\right)$ and $z\left(0\right)=a\left(0\right).$ Moreover, by the assumptions on a, $\alpha$, f, and ${\partial }_{t}f$, the function z is nondecreasing.

Differentiating and using Leibniz’ rule, we obtain

$$\begin{array}{cc}\hfill z^{\prime }\left(t\right)=& a^{\prime }\left(t\right)+h\left(t\right)\omega \left(u\left(t\right)\right)\hfill \\ \hfill & +f(t,\alpha \left(t\right)){\alpha }^{\prime }\left(t\right){\int }_0^{\alpha \left(t\right)}g\left(s\right)u\left(s\right)ds\hfill \\ \hfill & +{\int }_0^{\alpha \left(t\right)}{\partial }_{t}f(t,s)\left({\int }_0^{s}g\left(\tau \right)u\left(\tau \right)d\tau \right)ds.\hfill \end{array}$$

Since $u\left(r\right)\le z\left(r\right)$ and z is nondecreasing, for $0\le r\le t$,

$$\begin{array}{c}\hfill u\left(r\right)\le z\left(r\right)\le z\left(t\right).\end{array}$$

Hence

$$\begin{array}{cc}\hfill z^{\prime }\left(t\right)\le & a^{\prime }\left(t\right)+h\left(t\right)\omega \left(z\left(t\right)\right)\hfill \\ \hfill & +f(t,\alpha \left(t\right)){\alpha }^{\prime }\left(t\right)z\left(t\right){\int }_0^{\alpha \left(t\right)}g\left(s\right)ds\hfill \\ \hfill & +z\left(t\right){\int }_0^{\alpha \left(t\right)}{\partial }_{t}f(t,s)\left({\int }_0^{s}g\left(\tau \right)d\tau \right)ds.\hfill \end{array}$$

Since $\omega \left(z\right(t\left)\right)\ge 1$ and $\omega \left(z\right(t\left)\right)\ge z\left(t\right)$, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{z^{\prime }\left(t\right)}{\omega \left(z\right(t\left)\right)}}\le & a^{\prime }\left(t\right)+h\left(t\right)\hfill \\ \hfill & +f(t,\alpha \left(t\right)){\alpha }^{\prime }\left(t\right){\int }_0^{\alpha \left(t\right)}g\left(s\right)ds\hfill \\ \hfill & +{\int }_0^{\alpha \left(t\right)}{\partial }_{t}f(t,s)\left({\int }_0^{s}g\left(\tau \right)d\tau \right)ds.\hfill \end{array}$$

Observe that

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}\left[{\int }_0^{\alpha \left(t\right)}f(t,s)\left({\int }_0^{s}g\left(\tau \right)d\tau \right)ds\right]=& f(t,\alpha \left(t\right)){\alpha }^{\prime }\left(t\right){\int }_0^{\alpha \left(t\right)}g\left(\tau \right)d\tau \hfill \\ \hfill & +{\int }_0^{\alpha \left(t\right)}{\partial }_{t}f(t,s)\left({\int }_0^{s}g\left(\tau \right)d\tau \right)ds.\hfill \end{array}$$

Therefore, integration over $[0,t]$ gives

$$\begin{array}{cc}\hfill W\left(z\right(t\left)\right)\le & W\left(z\left(0\right)\right)+a\left(t\right)-a\left(0\right)+{\int }_0^{t}h\left(s\right)ds\hfill \\ \hfill & +{\int }_0^{\alpha \left(t\right)}f(t,s)\left({\int }_0^{s}g\left(\tau \right)d\tau \right)ds.\hfill \end{array}$$

Finally, since $u\left(t\right)\le z\left(t\right)$ and W is increasing,

$$\begin{array}{c}\hfill W\left(u\right(t\left)\right)\le W\left(z\right(t\left)\right),\end{array}$$

which proves (90). □

Theorem 5.

Let $u,a,b\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+}),$ and let $\alpha \in C^1({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ be nondecreasing, with $\alpha \left(0\right)=0$ and $\alpha \left(t\right)\le t.$ Let $p,q\in \mathbb{R}$ satisfy $p\ge q>0.$ Suppose that $f,m:{\mathbb{R}}_{+}\times {\mathbb{R}}_{+}⟶{\mathbb{R}}_{+}$ are continuous and that

$$0\le f(t,x)-f(t,y)\le m(t,y)(x-y),t\ge 0,x\ge y\ge 0.$$

(91)

If

$$u^p\left(t\right)\le a\left(t\right)+b\left(t\right){\int }_0^{\alpha \left(t\right)}f(s,u^q\left(s\right))ds,t\ge 0,$$

(92)

then

$$\begin{array}{cc}\hfill u^p\left(t\right)\le & a\left(t\right)+b\left(t\right)\exp \left({\int }_0^{\alpha \left(t\right)}M\left(r\right)dr\right)\times {\int }_0^{\alpha \left(t\right)}f(s,A\left(s\right))\exp \left(-{\int }_0^{s}M\left(r\right)dr\right)ds,\hfill \end{array}$$

(93)

where

$$\begin{array}{c}\hfill A\left(t\right)={\displaystyle \frac{p-q}{p}}+{\displaystyle \frac{q}{p}}a\left(t\right)\end{array}$$

and

$$\begin{array}{c}\hfill M\left(t\right)=m(t,A\left(t\right)){\displaystyle \frac{q}{p}}b\left(t\right).\end{array}$$

Proof. 

Set

$$\begin{array}{c}\hfill z\left(t\right)={\int }_0^{\alpha \left(t\right)}f(s,u^q\left(s\right))ds.\end{array}$$

Since $\alpha$ is nondecreasing and $f\ge 0$, the function z is nonnegative and nondecreasing, with $z\left(0\right)=0$.

From (92),

$$u^p\left(t\right)\le a\left(t\right)+b\left(t\right)z\left(t\right).$$

Since $0<q/p\le 1$, the elementary concavity inequality

$$X^{q/p}\le {\displaystyle \frac{p-q}{p}}+{\displaystyle \frac{q}{p}}X,X\ge 0,$$

gives

$$\begin{array}{cc}\hfill u^q\left(t\right)& ={\left(u^p\left(t\right)\right)}^{q/p}\hfill \\ \hfill & \le {\displaystyle \frac{p-q}{p}}+{\displaystyle \frac{q}{p}}\left(a\left(t\right)+b\left(t\right)z\left(t\right)\right)\hfill \\ \hfill & =A\left(t\right)+{\displaystyle \frac{q}{p}}b\left(t\right)z\left(t\right).\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill z^{\prime }\left(t\right)& =f(\alpha \left(t\right),u^q\left(\alpha \left(t\right)\right)){\alpha }^{\prime }\left(t\right)\hfill \\ \hfill & \le f\left(\alpha \left(t\right),A\left(\alpha \left(t\right)\right)+{\displaystyle \frac{q}{p}}b\left(\alpha \left(t\right)\right)z\left(\alpha \left(t\right)\right)\right){\alpha }^{\prime }\left(t\right).\hfill \end{array}$$

Since z is nondecreasing and $\alpha \left(t\right)\le t$, it follows that $z\left(\alpha \right(t\left)\right)\le z\left(t\right).$ Using (91), we obtain

$$\begin{array}{cc}\hfill z^{\prime }\left(t\right)\le & M\left(\alpha \left(t\right)\right)z\left(t\right){\alpha }^{\prime }\left(t\right)+f(\alpha \left(t\right),A\left(\alpha \left(t\right)\right)){\alpha }^{\prime }\left(t\right).\hfill \end{array}$$

Multiplying by the integrating factor

$$\begin{array}{c}\hfill \exp \left(-{\int }_0^{\alpha \left(t\right)}M\left(r\right)dr\right),\end{array}$$

we get

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}\left[z\left(t\right)\exp \left(-{\int }_0^{\alpha \left(t\right)}M\left(r\right)dr\right)\right]\le & f(\alpha \left(t\right),A\left(\alpha \left(t\right)\right)){\alpha }^{\prime }\left(t\right)\times \exp \left(-{\int }_0^{\alpha \left(t\right)}M\left(r\right)dr\right).\hfill \end{array}$$

Integrating over $[0,t]$ and using $\alpha \left(0\right)=0$ yields

$$\begin{array}{cc}\hfill z\left(t\right)\le & \exp \left({\int }_0^{\alpha \left(t\right)}M\left(r\right)dr\right)\times {\int }_0^{\alpha \left(t\right)}f(s,A\left(s\right))\exp \left(-{\int }_0^{s}M\left(r\right)dr\right)ds.\hfill \end{array}$$

Substituting this estimate into (92) gives (93). □

Example 3.

Let the assumptions of Proposition 2 hold and suppose, in addition, that f is independent of its first variable. Let $0<n\le 1$, and suppose that $u\in C({\mathbb{R}}_{+},{\mathbb{R}}_{+})$ satisfies the Volterra equation

$$\begin{array}{cc}\hfill u\left(t\right)=& a\left(t\right)+{\int }_0^{t}h\left(s\right)u^n\left(s\right)ds+{\int }_0^{\alpha \left(t\right)}f\left(s\right)\left({\int }_0^{s}g\left(\tau \right)u\left(\tau \right)d\tau \right)ds.\hfill \end{array}$$

(94)

Assume that ${\displaystyle \underset{t\ge 0}{\sup }a\left(t\right)<\infty ,}$ and

$$\begin{array}{c}\hfill {\int }_0^{\infty }h\left(s\right)ds<\infty ,{\int }_0^{\alpha (\infty )}f\left(s\right)ds<\infty ,{\int }_0^{\infty }g\left(s\right)ds<\infty ,\end{array}$$

where

$$\begin{array}{c}\hfill \alpha (\infty ):=\underset{t\to \infty }{\lim }\alpha \left(t\right)\in (0,\infty ].\end{array}$$

Then u is bounded on ${\mathbb{R}}_{+}$.

Proof. 

For $0<n\le 1$ and $r\ge 0$, we have $r^n\le 1+r.$ Choose $\omega \left(r\right)=1+r.$ Then $\omega$ is nondecreasing, $\omega \left(r\right)\ge 1,\omega \left(r\right)\ge r,$ and $u^n\left(t\right)\le \omega \left(u\left(t\right)\right).$ Hence (94) implies

$$\begin{array}{cc}\hfill u\left(t\right)\le & a\left(t\right)+{\int }_0^{t}h\left(s\right)\omega \left(u\left(s\right)\right)ds+{\int }_0^{\alpha \left(t\right)}f\left(s\right)\left({\int }_0^{s}g\left(\tau \right)u\left(\tau \right)d\tau \right)ds.\hfill \end{array}$$

Proposition 2 therefore gives

$$\begin{array}{cc}\hfill W\left(u\right(t\left)\right)\le & W\left(a\left(0\right)\right)+a\left(t\right)-a\left(0\right)+{\int }_0^{t}h\left(s\right)ds+{\int }_0^{\alpha \left(t\right)}f\left(s\right)\left({\int }_0^{s}g\left(\tau \right)d\tau \right)ds.\hfill \end{array}$$

For $\omega \left(r\right)=1+r,W\left(r\right)=\ln (1+r).$ Moreover,

$$\begin{array}{c}\hfill {\int }_0^{\alpha \left(t\right)}f\left(s\right)\left({\int }_0^{s}g\left(\tau \right)d\tau \right)ds\le \left({\int }_0^{\alpha (\infty )}f\left(s\right)ds\right)\left({\int }_0^{\infty }g\left(\tau \right)d\tau \right).\end{array}$$

Thus the right-hand side is uniformly bounded on ${\mathbb{R}}_{+}$, and hence $\ln (1+u(t\left)\right)$ is uniformly bounded. Consequently, u is bounded on ${\mathbb{R}}_{+}$. □

4. Conclusions

In this paper, we established several retarded integral inequalities of the Gronwall–Bihari–Pachpatte type. The obtained results cover linear and nonlinear settings and incorporate delayed arguments, nonlinear control functions, and several classes of time-dependent kernels. Explicit upper bounds were derived through the auxiliary function $\mathsf{\Omega }$ and its inverse, together with further estimates for inequalities involving iterated integral terms.

As applications, the proposed inequalities were used to derive a priori bounds for generalized Liénard and Rayleigh equations with time delay, leading to global existence results for their solutions. The results provide useful tools for the qualitative analysis of functional differential and integral equations with retarded and nonlinear terms.