Some retarded nonlinear integral inequalities and their applications in retarded differential equations

Abstract

In this article, we discuss some generalized retarded nonlinear integral inequalities, which not only include nonlinear compound function of unknown function but also include retard items, and give upper bound estimation of the unknown function by integral inequality technique. This estimation can be used as tool in the study of differential equations with the initial conditions.

2000 MSC: 26D10; 26D15; 26D20; 34A12; 34A40.

1 Introduction

Gronwall-Bellman inequalities [1, 2] and their various generalizations can be used important tools in the study of existence, uniqueness, boundedness, stability, and other qualitative properties of solutions of differential equations, integral equations, and integral-differential equations.

Lemma 1 (Gronwall [1]). Let u(t) be a continuous function defined on the interval [a, a + h], a, h are nonnegative constants and

$$0\le u\left(t\right)\le \underset{a}{\overset{t}{\int }}\left[bu\left(s\right)+a\right]ds,t\in \left[a,a+h\right].$$

(1.1)

Then, 0 ≤ u(t) ≤ ah exp(bh), ∀t ∈ [a,a + h].

Lemma 2 (Bellman [2]). Let f, u ∈ C([0, h], [0, ∞)), h, c are positive constants. If u satisfy the inequality

$$u\left(t\right)\le c+\underset{0}{\overset{t}{\int }}f\left(s\right)u\left(s\right)ds,t\in \left[0,h\right].$$

(1.2)

Then, $u\left(t\right)\le c\text{exp}\left({\int }_0^{t}f\left(s\right)ds\right),t\in \left[0,h\right]$.

Lemma 3 (Lipovan [3]). Let u, f ∈ C([t₀,T), R₊). Further, let α ∈ C¹([t₀,T),[t₀,T)) be nondecreasing with α(t) ≤ t on [t₀,T), and let c be a nonnegative constant. Then the inequality

$$u\left(t\right)\le c+\underset{\alpha \left(t_0\right)}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)u\left(s\right)ds,t\in \left[t_0,T\right)$$

(1.3)

implies that $u\left(t\right)\le c\text{exp}\left({\int }_{\alpha \left(t_0\right)}^{\alpha \left(t\right)}f\left(s\right)ds\right),t\in \left[t_0,T\right).$.

Lemma 4 (Abdeldaim and Yakout [4]). We assume that u(t) and f(t) are nonnegative real-valued continuous functions defined on I and satisfy the inequality

$$u\left(t\right)\le u_0+{\left(\underset{0}{\overset{t}{\int }}f\left(s\right)u\left(s\right)ds\right)}^2+\underset{0}{\overset{t}{\int }}f\left(s\right)u\left(s\right)\left(u\left(s\right)+2\underset{0}{\overset{s}{\int }}f\left(\tau \right)u\left(\tau \right)d\tau \right)ds,\forall t\in I,$$

(1.4)

where u ₀ be a positive constant. Then

$$u\left(t\right)\le u_0\text{exp}\left(\underset{0}{\overset{t}{\int }}f\left(s\right)B_1\left(s\right)ds\right),\forall t\in I,$$

(1.5)

where

$$B_1\left(t\right)=\frac{u_0\text{exp}\left(4{\int }_0^{t}f\left(s\right)ds\right)}{1-u_0{\int }_0^{t}f\left(s\right)\text{exp}\left(4{\int }_0^{s}f\left(\tau \right)d\tau \right)ds},\forall t\in I,$$

(1.6)

such that $u_0{\int }_0^{t}f\left(s\right)\text{exp}\left(4{\int }_0^{s}f\left(\tau \right)d\tau \right)ds<1$.

Lemma 5 (Abdeldaim and Yakout [4]). We assume that u(t) and f(t) are nonnegative real-valued continuous functions defined on I and satisfy the inequality

$$u^{p+1}\left(t\right)\le u_0+{\left(\underset{0}{\overset{t}{\int }}f\left(s\right)u^p\left(s\right)ds\right)}^2+2\underset{0}{\overset{t}{\int }}f\left(s\right)u^p\left(s\right)\left(u\left(s\right)+\underset{0}{\overset{s}{\int }}f\left(\tau \right)u^p\left(\tau \right)d\tau \right)ds,$$

(1.7)

for all t ∈ I, where u₀ > 0, p ∈ (0,1), are constants. Then

$$u\left(t\right)\le u_0^{\frac{1}{p+1}}+\frac{2}{p+1}\underset{0}{\overset{t}{\int }}f\left(s\right)B_2\left(s\right)ds,\forall t\in I,$$

(1.8)

where

$$B_2\left(t\right)=\text{exp}\left(\frac{2}{p+1}\underset{0}{\overset{t}{\int }}f\left(s\right)ds\right){\left(u_0^{\frac{1-p}{p+1}}+2\left(1-p\right)\underset{0}{\overset{t}{\int }}f\left(s\right)\text{exp}\left(-2\frac{1-p}{p+1}\underset{0}{\overset{s}{\int }}f\left(\tau \right)d\tau \right)ds\right)}^{\frac{1}{1-p}},$$

(1.9)

for all t ∈ I.

Lemma 6 (see [5]). Let φ ∈ C(R₊,R₊) be a increasing function, u,a,f ∈ C([t₀,T),R₊), a(t) be a increasing function, and α ∈ C¹([t₀,T), [t₀,T)) be nondecreasing with α(t) ≤ t on [t₀,T) where T ∈ (0,∞) is a constant. Then the inequality

$$u\left(t\right)\le a\left(t\right)+\underset{\alpha \left(0\right)}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)\phi \left(u\left(s\right)\right)ds,t\in \left[t_0,T\right)$$

(1.10)

implies that

$$u\left(t\right)\le W^{-1}\left(W\left(a\left(t\right)\right)+\underset{\alpha \left(0\right)}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds\right),t\in \left[t_0,T_1\right),$$

(1.11)

where

$$W\left(t\right)=\underset{1}{\overset{t}{\int }}\frac{dt}{\phi \left(t\right)}ds,t>0,$$

(1.12)

W ^-1 is the reverse function of W, T ₁ is the largest number such that

$$W\left(a\left(T_1\right)\right)+\underset{\alpha \left(0\right)}{\overset{\alpha \left(T_1\right)}{\int }}f\left(s\right)ds\le \underset{1}{\overset{\infty }{\int }}\frac{dt}{\phi \left(t\right)}ds.$$

There can be found a lot of generalizations of Gronwall-Bellman inequalities in various cases from literature (e.g., [3–13]).

In this article, we discuss some retarded nonlinear integral inequalities, where linear case u(t) in integral functions in [4] is changed into the nonlinear case ϕ(u(t)), and the non-retarded case t in [4] is changed into retarded case α(t), and give upper bound estimation of the unknown function by integral inequality technique.

2 Main result

In this section, we discuss some retarded integral inequalities of Gronwall-Bellman type. Throughout this article, let I = [0, ∞).

Theorem 1. Let φ,φ',α ∈ C¹(I, I) be increasing functions with φ'(t) ≤ k, α(t) ≤ t, α(0) = 0, ∀t ∈ I; k, u₀ be positive constants, we assume that u(t) and f(t) are nonnegative real-valued continuous functions defined on I and satisfy the inequality

$$u\left(t\right)\le u_0+{\left(\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)\phi \left(u\left(s\right)\right)ds\right)}^2+\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)\phi \left(u\left(s\right)\right)\left(\phi \left(u\left(s\right)\right)+2\underset{0}{\overset{s}{\int }}f\left(\tau \right)\phi \left(u\left(\tau \right)\right)d\tau \right)ds,$$

(2.1)

for all t ∈ I. If $u_0^{-1}-k{\int }_0^{\alpha \left(t\right)}f\left(s\right)\text{exp}\left(4{\int }_0^{s}f\left(\tau \right)d\tau \right)ds>0$, then

$$z\left(t\right)\le {\Phi }^{-1}\left(\Phi \left(u_0\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)B_3\left(s\right)ds\right),\forall t\in I,$$

(2.2)

where

$$\Phi \left(x\right):=\underset{1}{\overset{x}{\int }}\frac{ds}{\phi \left(s\right)},\forall x>0,$$

(2.3)

$$B_3\left(t\right):=\text{exp}\left(4\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds\right){\left({\left(\phi \left(u_0\right)\right)}^{-1}-k\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)\text{exp}\left(4\underset{0}{\overset{s}{\int }}f\left(\tau \right)d\tau \right)ds\right)}^{-1}.$$

(2.4)

Remark 1. If α(t) = t, φ(u(s)) = u(s), then Theorem 1 reduces Lemma 4.

Proof. Let z(t) denotes the function on the right-hand side of (2.1), which is a nonnegative and nondecreasing function on I with z(0) = u₀. Then (2.1) is equivalent to

$$u\left(t\right)\le z\left(t\right),u\left(\alpha \left(t\right)\right)\le z\left(\alpha \left(t\right)\right),\forall t\in I.$$

(2.5)

Differentiating z(t) with respect to t, we have

$$\begin{array}{ll}\hfill \frac{dz}{dt}& =2{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)\phi \left(u\left(\alpha \left(t\right)\right)\right)\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)\phi \left(u\left(s\right)\right)ds+{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)\phi \left(u\left(\alpha \left(t\right)\right)\right)\\ \times \left(\phi \left(u\left(\alpha \left(t\right)\right)\right)+2\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(\tau \right)\phi \left(u\left(\tau \right)\right)d\tau \right)ds,\forall t\in I.\end{array}$$

(2.6)

Using (2.5), we obtain

$$\frac{dz}{dt}\le {\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)\phi \left(z\left(\alpha \left(t\right)\right)\right)w\left(t\right),\forall t\in I,$$

(2.7)

where $w\left(t\right):=\phi \left(z\left(\alpha \left(t\right)\right)\right)+4{\int }_0^{\alpha \left(t\right)}f\left(s\right)\phi \left(z\left(s\right)\right)ds,w\left(0\right)=\phi \left(z\left(0\right)\right)=\phi \left(u_0\right)$, w is a nonnegative and nondecreasing function on I. By the monotonicity φ, φ',z, and α(t) ≤ t we have φ(z(α(t))) ≤ w(t), φ'(z(α(t))) ≤ k. Differentiating w(t) with respect to t, and using (2.7) we have

$$\begin{array}{ll}\hfill \frac{dw}{dt}& \le {\phi }^{\prime }\left(z\left(\alpha \left(t\right)\right)\right){\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)w^2\left(t\right)+4{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)w\left(t\right)\\ \le k{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)w^2\left(t\right)+4{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)w\left(t\right),\forall t\in I.\end{array}$$

(2.8)

By w(t) > 0, we have

$$w^{-2}\left(t\right)\frac{dw}{dt}\le k{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)+4{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)w^{-1}\left(t\right),\forall t\in I.$$

(2.9)

Let v(t) = w^-1(t), from (2.9) we have

$$\frac{dv}{dt}+4{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)v\left(t\right)\ge -k{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right),\forall t\in I.$$

(2.10)

Consider ordinary differential equation

$$\frac{dy}{dt}+4{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)y\left(t\right)=-k{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right),y\left(0\right)={\left(\phi \left(u_0\right)\right)}^{-1},\forall t\in I.$$

(2.11)

The solution of Equation (2.11) is

$$\begin{array}{ll}\hfill y\left(t\right)& ={\left(\phi \left(u_0\right)\right)}^{-1}\text{exp}\left(-4\underset{0}{\overset{t}{\int }}{\alpha }^{\prime }\left(s\right)f\left(\alpha \left(s\right)\right)ds\right)\\ -\text{exp}\left(-4\underset{0}{\overset{t}{\int }}{\alpha }^{\prime }\left(s\right)f\left(\alpha \left(s\right)\right)ds\right)\underset{0}{\overset{t}{\int }}k{\alpha }^{\prime }\left(s\right)f\left(\alpha \left(s\right)\right)\text{exp}\left(\underset{0}{\overset{s}{\int }}4{\alpha }^{\prime }\left(\tau \right)f\left(\alpha \left(\tau \right)\right)d\tau \right)ds\\ ={\left(\phi \left(u_0\right)\right)}^{-1}\text{exp}\left(-4\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds\right)-\text{exp}\left(-4\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds\right)\underset{0}{\overset{\alpha \left(t\right)}{\int }}kf\left(s\right)\text{exp}\left(4\underset{0}{\overset{s}{\int }}f\left(\tau \right)d\tau \right)ds.\\ =\text{exp}\left(-4\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds\right)\left({\left(\phi \left(u_0\right)\right)}^{-1}-k\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)\text{exp}\left(4\underset{0}{\overset{s}{\int }}f\left(\tau \right)d\tau \right)ds\right).\end{array}$$

(2.12)

By (2.10), (2.11), and (2.13), we obtain

$$v\left(t\right)\ge \text{exp}\left(-4\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds\right)\left({\left(\phi \left(u_0\right)\right)}^{-1}-k\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)\text{exp}\left(4\underset{0}{\overset{s}{\int }}f\left(\tau \right)d\tau \right)ds\right).$$

(2.13)

By the definition of B₃(t) in (2.4) and the inequality (2.13), we have w(t) < B₃(t),∀t ∈ I. From (2.7), we get

$$\frac{dz}{dt}\le {\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)\phi \left(z\left(\alpha \left(t\right)\right)\right)B_3\left(\alpha \left(t\right)\right)\le {\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)B_3\left(\alpha \left(t\right)\right)\phi \left(z\left(t\right)\right),\forall t\in I.$$

(2.14)

By taking t = s in the inequality (2.14) and integrating (2.14) from 0 to t, by the definition (2.3) of Φ we obtain

$$z\left(t\right)\le {\Phi }^{-1}\left(\Phi \left(z\left(0\right)\right)+\underset{0}{\overset{t}{\int }}{\alpha }^{\prime }\left(s\right)f\left(\alpha \left(s\right)\right)B_3\left(\alpha \left(s\right)\right)ds\right)\le {\Phi }^{-1}\left(\Phi \left(z\left(0\right)\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)B_3\left(s\right)ds\right),$$

(2.15)

for all t ∈ I. This completes the proof of the Theorem 1.

Theorem 2. Let ψ(t),φ(t),φ(t)/t,α(t) ∈ C¹(I,I) be increasing functions with ψ'(t) = φ(t),α(t) ≤ t, α(0) = 0, ∀t ∈ I; k, u₀ be positive constants, we assume that u(t) and f(t) are nonnegative real-valued continuous functions defined on I and satisfy the inequality

$$\begin{array}{ll}\hfill \psi \left(u\left(s\right)\right)& \le u_0+{\left(\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)\phi \left(u\left(s\right)\right)ds\right)}^2\\ +\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)\phi \left(u\left(s\right)\right)\left(u\left(s\right)+2\underset{0}{\overset{s}{\int }}f\left(\tau \right)\phi \left(u\left(\tau \right)\right)d\tau \right)ds,\end{array}$$

(2.16)

for all t ∈ I. Then

$$u\left(t\right)\le \text{exp}\left({\Xi }^{-1}\left(\Xi \left(\text{ln}\left(1+{\psi }^{-1}\left(u_0\right)\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}4f\left(s\right)ds\right)\right),\forall t\in \left(0,T_2\right).$$

(2.17)

where

$$\Xi \left(t\right):=\underset{1}{\overset{t}{\int }}\frac{\text{exp}\left(s\right)ds}{\phi \left(\text{exp}\left(s\right)\right)},\forall t>0,$$

(2.18)

Ξ^-1,ψ^-1 are the reverse function of Ξ, ψ respectively, T₂ is the largest number such that

$$\Xi \left(\text{ln}\left(1+{\psi }^{-1}\left(u_0\right)\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}4f\left(s\right)ds\le \underset{1}{\overset{\infty }{\int }}\frac{\text{exp}\left(s\right)ds}{\psi \left(\text{exp}\left(s\right)\right)},\forall x\in {\mathbf{R}}_{+}.$$

Remark 2. If α(t) = t,φ(u(t)) = u^p(t),ψ(u(t)) = u^p+1(t)/(p + 1), by Theorem 2, we can obtain the result similar to Lemma 5.

Proof. Let ψ(z(t)) denotes the function on the right-hand side of (2.16), then z(t) is a nonnegative and nondecreasing function on I with z(0) = ψ^-1(u₀). Then (2.16) is equivalent to

$$u\left(t\right)\le z\left(t\right),u\left(\alpha \left(t\right)\right)\le z\left(\alpha \left(t\right)\right)\forall t\in I.$$

(2.19)

Differentiating ψ(z(t)) with respect to t, we have

$$\begin{array}{ll}\hfill {\psi }^{\prime }\left(z\left(t\right)\right)\frac{dz}{dt}& =2{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)\phi \left(u\left(\alpha \left(t\right)\right)\right)\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)\phi \left(u\left(s\right)\right)ds+{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)\phi \left(u\left(\alpha \left(t\right)\right)\right)\\ \times \left(u\left(\alpha \left(t\right)\right)+2\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(\tau \right)\phi \left(u\left(\tau \right)\right)d\tau \right)ds,\forall t\in I.\end{array}$$

(2.20)

Using (2.19) and the relation ψ'(z(t)) = φ(z(t)), from (2.20) we obtain

$$\frac{dz}{dt}\le {\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)\left(z\left(t\right)+4\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)\phi \left(z\left(s\right)\right)ds\right),\forall t\in I.$$

(2.21)

Let $w\left(t\right):=z\left(t\right)+4{\int }_0^{\alpha \left(t\right)}f\left(s\right)\phi \left(z\left(s\right)\right)ds$, then w(0) = z(0) = ψ^-1(u₀), z(t) ≤ w(t), w is a nonnegative and nondecreasing function on I. Differentiating w(t) with respect to t, and using (2.21) we have

$$\begin{array}{ll}\hfill \frac{dw}{dt}& \le {\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)w\left(t\right)+4{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)\phi \left(z\left(\alpha \left(t\right)\right)\right)\\ \le {\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)w\left(t\right)+4{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)\phi \left(w\left(\alpha \left(t\right)\right)\right),\forall t\in I.\end{array}$$

(2.22)

By w(t) > 0, we have

$$\frac{dw}{w\left(t\right)dt}\le {\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)+4{\alpha }^{\prime }\left(t\right)f\left(\alpha \left(t\right)\right)\phi \left(w\left(\alpha \left(t\right)\right)\right)/w\left(\alpha \left(t\right)\right),\forall t\in I.$$

(2.23)

Integrating (2.23) from 0 to t, we have

$$\begin{array}{ll}\hfill \text{ln}w\left(t\right)& \le \text{ln}\left(1+w\left(0\right)\right)+\underset{0}{\overset{t}{\int }}{\alpha }^{\prime }\left(s\right)f\left(\alpha \left(s\right)\right)ds+\underset{0}{\overset{t}{\int }}4{\alpha }^{\prime }\left(s\right)f\left(\alpha \left(s\right)\right)\phi \left(w\left(\alpha \left(s\right)\right)\right){\left(w\left(\alpha \left(s\right)\right)\right)}^{-1}ds\\ \le \text{ln}\left(1+w\left(0\right)\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds+\underset{0}{\overset{\alpha \left(t\right)}{\int }}4f\left(s\right)\phi \left(w\left(s\right)\right){\left(w\left(s\right)\right)}^{-1}ds\\ \le \text{ln}\left(1+w\left(0\right)\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds+\underset{0}{\overset{\alpha \left(t\right)}{\int }}4f\left(s\right)\phi \left(\text{exp}\left(\text{ln}w\left(s\right)\right)\right){\left(\text{exp}\left(\text{ln}w\left(s\right)\right)\right)}^{-1}ds,\end{array}$$

(2.24)

for all t ∈ I. Using Lemma 6 and the Definition (2.18) of Ξ, we obtain

$$\begin{array}{ll}\hfill \text{ln}w\left(t\right)& \le {\Xi }^{-1}\left(\Xi \left(\text{ln}\left(1+w\left(0\right)\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}4f\left(s\right)ds\right)\\ ={\Xi }^{-1}\left(\Xi \left(\text{ln}\left(1+{\psi }^{-1}\left(u_0\right)\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}f\left(s\right)ds\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}4f\left(s\right)ds\right),\forall t\in \left(0,T_2\right).\end{array}$$

(2.25)

Using the relation u(t) ≤ z(t) ≤ w(t), we can obtain the estimation (2.17) of (2.16).

3 Application

In this section, we apply our result to the following nonlinear differential equation [4]

$$\left\{\begin{array}{c}\frac{dx\left(t\right)}{dt}=F\left(t,x\left(\alpha \left(t\right)\right)\right)+H\left(t,x\left(\alpha \left(t\right)\right),K\left(t,x\left(\alpha \left(t\right)\right)\right)\right),\forall t\in I,\hfill \\ x\left(0\right)=x_0,\hfill \end{array}\right.$$

(3.26)

where x₀ is a constant, F, K ∈ C(I × I, R), H ∈ C(I³, R), satisfy the following conditions

$$\left|F\left(t,x\left(\alpha \left(t\right)\right)\right)\right|\le f^2\left(\alpha \left(t\right)\right){\left|\phi \left(x\left(\alpha \left(t\right)\right)\right)\right|}^2,\left|K\left(t,x\left(\alpha \left(t\right)\right)\right)\right|\le f\left(\alpha \left(t\right)\right)\left|\phi \left(x\left(\alpha \left(t\right)\right)\right)\right|,$$

(3.27)

$$\left|H\left(t,x,y\right)\right|\le \left|y\right|\left(\phi \left(\left|x\right|\right)+2\underset{0}{\overset{t}{\int }}\left|y\right|ds\right),$$

(3.28)

where f(t) is nonnegative real-valued continuous function defined on I.

Corollary 1. Consider nonlinear system (3.26) and suppose that F,K, H satisfy the conditions (3.27) and (3.28). Let φ,φ', α ∈ C¹(I, I) be increasing functions with φ'(t) ≤ k,α(t) ≤ t, α(0) = 0,∀t ∈ I, k be positive constants; then all solutions of Equation (3.26) exist on I and satisfy the following estimation

$$\left|x\left(t\right)\right|\le {\Phi }^{-1}\left(\Phi \left(\left|x_0\right|\right)+\underset{0}{\overset{\alpha \left(t\right)}{\int }}\frac{f\left(s\right)}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(s\right)\right)}B\left(s\right)ds\right),\forall t\in I,$$

(3.29)

where

$$\Phi \left(x\right):=\underset{1}{\overset{x}{\int }}\frac{ds}{\phi \left(s\right)},\forall x>0,$$

(3.30)

$$\begin{array}{ll}\hfill B\left(t\right)& :=\text{exp}\left(4\underset{0}{\overset{\alpha \left(t\right)}{\int }}\frac{f\left(s\right)}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(s\right)\right)}ds\right)\\ \times {\left({\left(\phi \left(\left|x_0\right|\right)\right)}^{-1}-k\underset{0}{\overset{\alpha \left(t\right)}{\int }}\frac{f\left(s\right)}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(s\right)\right)}\text{exp}\left(4\underset{0}{\overset{s}{\int }}\frac{f\left(\tau \right)}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(\tau \right)\right)}d\tau \right)ds\right)}^{-1}.\end{array}$$

(3.31)

Proof. Integrating both sides of the Equation (3.26) from 0 to t, we get

$$x\left(t\right)=x_0+\underset{0}{\overset{t}{\int }}F\left(s,x\left(\alpha \left(s\right)\right)\right)ds+\underset{0}{\overset{t}{\int }}H\left(s,x\left(\alpha \left(s\right)\right),K\left(s,x\left(\alpha \left(s\right)\right)\right)\right)ds,\forall t\in I.$$

(3.32)

From (3.27), (3.28), and (3.32) we obtain

$$\begin{array}{ll}\hfill \left|x\left(t\right)\right|& \le \left|x_0\right|+\underset{0}{\overset{t}{\int }}{f}^2\left(\alpha \left(s\right)\right){\left|\phi \left(x\left(\alpha \left(t\right)\right)\right)\right|}^{2}ds\\ +\underset{0}{\overset{t}{\int }}f\left(\alpha \left(s\right)\right)\left|\phi \left(x\left(\alpha \left(t\right)\right)\right)\right|\left(\phi \left(\left|x\left(\alpha \left(s\right)\right)\right|\right)+2\underset{0}{\overset{s}{\int }}f\left(\alpha \left(\tau \right)\right)\left|\phi \left(x\left(\alpha \left(\tau \right)\right)\right)\right|d\tau \right)ds\\ \le \left|x_0\right|+{\left(\underset{0}{\overset{\alpha \left(t\right)}{\int }}\frac{f\left(s\right)\left|\phi \left(x\left(s\right)\right)\right|}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(s\right)\right)}ds\right)}^2\\ +\underset{0}{\overset{\alpha \left(t\right)}{\int }}\frac{f\left(s\right)\left|\phi \left(x\left(s\right)\right)\right|}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(s\right)\right)}\left(\phi \left(\left|x\left(\alpha \left(s\right)\right)\right|\right)+2\underset{0}{\overset{s}{\int }}\frac{f\left(\tau \right)\left|\phi \left(x\left(\tau \right)\right)\right|}{{\alpha }^{\prime }\left({\alpha }^{-1}\left(\tau \right)\right)}d\tau \right)ds,\forall t\in I.\end{array}$$

(3.33)

Applying Theorem 1 to (3.33), we get the estimation (3.29). This completes the proof of the Corollary 1.