Proof. Let x be a solution of (1-1) and (1-2). Integrate (1-1) from t to $\infty $ ; according to the second condition of (1-2), we can get

(2-2) $$ \begin{align} x'(t)=-\phi^{-1}\bigg(\!\int_t^{\infty}f(s, x)\,ds\bigg). \end{align} $$

Integrate (2-2) from t to $\infty $ again; by the first condition of (1-2), it follows that (2-1) is valid.

Alternatively, let $x\in C^1[t_0, \infty )$ be a solution of the integral equation (2-1). Take the derivative on both sides of (2-1); then x satisfies (2-2). Since $\phi $ is odd, it follows from (2-2) that

$$ \begin{align*} -\phi(x'(t))=\phi(-x'(t))=\int_t^{\infty}f(s, x)\,ds. \end{align*} $$

Then, it is not difficult to prove that x satisfies (1-1). In addition, by the definition of $\phi ^{-1}$ , L’Hôpital’s rule, (1-10) and (1-14), we deduce that

$$ \begin{align*} \begin{aligned} \lim_{t\to\infty}e^t\phi^{-1}\bigg(\!\int_t^{\infty}f(s, x)\,ds\bigg) &=\lim_{t\to\infty}e^t\frac{\int_t^{\infty}f(s, x)\,ds}{\sqrt{1+(\int_t^{\infty}f(s, x)\,ds)^2}}\\ &=\lim_{t\to\infty}\frac{1}{\sqrt{1+(\int_t^{\infty}f(s, x)\,ds)^2}}\cdot\lim\limits_{t\to\infty}e^t\int_t^{\infty}f(s, x)\,ds\\ &=\lim\limits_{t\to\infty}e^t\int_t^{\infty}f(s, x)\,ds\\ &=\lim\limits_{t\to\infty}\frac{\int_t^{\infty}f(s, x)\,ds}{e^{-t}}\\ &=\lim_{t\to\infty}e^tf(t, x)\\ &=0. \end{aligned} \end{align*} $$

This, together with (2-2), implies that $\lim \limits _{t\to \infty } e^t x'(t)=0$ and, hence, the second boundary condition of (1-2) is valid. Moreover, by virtue of changing the order of integration and the definition of $\phi ^{-1}$ , we deduce that

$$ \begin{align*} |x(t)-\psi_0|&=\bigg|\!\int_t^{\infty}\phi^{-1}\bigg(\!\int_\tau^{\infty}f(s, x)\,ds\bigg)\,d\tau\bigg|\\&\leq\int_t^{\infty}\frac{\int_\tau^{\infty}|f(s, x)|\,ds}{\sqrt{1+(\int_\tau^{\infty}f(s, x)\,ds)^2}}\\&\leq\int_t^{\infty}\int_\tau^{\infty}|f(s, x)|\,ds\,d\tau\\&=\int_t^{\infty}\int_t^s|f(s, x)|\,d\tau \,ds\\&=\int_t^{\infty} (s-t)|f(s, x)|\,ds\\&\leq \int_t^{\infty} s|f(s, x)|\,ds. \end{align*} $$

Then, combining this with (1-11) in Condition (H1), we infer that $\lim _{t\to \infty }x(t)=\psi _0.$ Thus, (1-2) holds.

Proof of Theorem 1.1.

For any given $0<\delta <1$ , it follows from (1-11) and

$$ \begin{align*} \int_{t_0}^{\infty}t\ \bigg(\frac{\mu'(t)}{\mu(t)}\bigg)^2\,dt<\infty \end{align*} $$

that we can choose $T_0\geq \max \{t_0, 1\}$ large enough such that

(3-1) $$ \begin{align} w(|\psi_0|+\delta)\int_{T_0}^{\infty}t\ \bigg(\frac{\mu'(t)}{\mu(t)}\bigg)^2\,dt\leq\frac{\delta}{2},\quad \int_{T_0}^{\infty}tf(t, 0)\,dt\leq\frac{\delta}{2}. \end{align} $$

Set

(3-2) $$ \begin{align} \Omega=\{x\in X: \|x-\psi_0\|\leq\delta\}. \end{align} $$

Define the operator $F:\ \Omega \to X$ ,

$$ \begin{align*} [F(x)](t)=\psi_0+\int_t^{\infty}\phi^{-1}\bigg(\!\int_\tau^{\infty}f(s, x)\,ds\bigg)\,d\tau,\quad t\geq T_0. \end{align*} $$

According to the proof of Lemma 2.2, we have $\lim _{t\to \infty }[F(x)](t)=\psi _0$ and $\lim _{t\to \infty } e^t [F(x)]'(t)=0$ . Then, (1-12), (3-1), Remark 2.3 and the monotonicity property of w imply that for each $x\in \Omega $ and $t\ge T_0$ , we have that

(3-3) $$ \begin{align} |[F(x)](t)-\psi_0|&=\bigg|\!\int_t^{\infty}\phi^{-1}\bigg(\!\int_\tau^{\infty}f(s, x)\,ds\bigg)\,d\tau\bigg| \nonumber \\&\leq\int_t^{\infty}\frac{\int_\tau^{\infty}|f(s, x)|\,ds}{\sqrt{1+(\int_\tau^{\infty}f(s, x)\,ds)^2}} \nonumber \\&\leq\int_t^{\infty}\int_\tau^{\infty}|f(s, x)|\,ds\,d\tau \nonumber \\&=\int_t^{\infty}\int_t^s|f(s, x)|\,d\tau \,ds \nonumber \\&\leq\int_t^{\infty} (s-t)|f(s, x)|\,ds\\&\leq \int_t^{\infty} (s-t)|f(s, x)-f(s,\ 0)|\,ds+\int_t^{\infty} (s-t)|f(s,\ 0)|\,ds \nonumber \\&\leq \int_t^{\infty} s \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2w(|x(s)|)\,ds+\int_t^{\infty} s|f(s,\ 0)|\,ds \nonumber \\&\leq w(|\psi_0|+\delta)\int_{T_0}^{\infty} s \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2\,ds+\int_{T_0}^{\infty} s|f(s,\ 0)|\,ds \nonumber \\&\leq\delta. \nonumber \end{align} $$

Using similar processing methods to (3-3) and (1-14) in Remark 2.3 for any $x\in \Omega $ and $t\geq T_0$ , we can also get

(3-4) $$ \begin{align} \begin{aligned} |([F(x)](t)-\psi_0)'|&=\bigg|\phi^{-1}\bigg(\!\int_t^{\infty}f(s, x)\,ds\bigg)\bigg|\\ &\leq\int_t^{\infty} |f(s, x)|\,ds\\ &\leq\delta. \end{aligned} \end{align} $$

Then, $F:\ \Omega \to \Omega $ is well defined.

Let $\{x_n\}\subset \Omega $ be an arbitrary sequence. By applying the Lagrange mean value theorem and (3-4), we obtain

(3-5) $$ \begin{align} |[{F}(x_{n})](t_{1})-[{F}(x_{n})](t_{2})|\leq\delta|t_{1}-t_{2}|,\quad t_{1}, t_{2}\geq T_{0},\ n\geq1. \end{align} $$

For any $t_{1}, t_{2}\geq T_{0}$ , without loss of generality, assuming that $t_1\le t_2$ , we have

(3-6) $$ \begin{align} \begin{aligned} |[{F}(x_{n})]'(t_{1})-[{F}(x_{n})]'(t_{2})|&\leq\bigg|\phi^{-1}\bigg(\!\int_{t_{1}}^{\infty}f(s, x_{n})\,ds\bigg)- \phi^{-1}\bigg(\!\int_{t_{2}}^{\infty}f(s, x_{n})\,ds\bigg)\bigg|\\ &\leq\frac{1}{(1+\xi^{2})^{{3}/{2}}}\bigg|\!\int_{t_{1}}^{\infty}f(s, x_{n})\,ds-\int_{t_{2}}^{\infty}f(s, x_{n})\,ds\bigg|\\ &\leq\int_{t_{1}}^{t_{2}}|f(s, x_{n})|\,ds\\ &\leq M(t_{2}-t_{1}), \end{aligned} \end{align} $$

where $\xi $ lives between $\int _{t_1}^{\infty }f(\tau , x_n)\,d\tau $ and $\int _{t_2}^{\infty }f(\tau , x_n)\,d\tau .$ Here, M is the supremum of f in $[t_0, \infty )\times \Omega $ in view of Condition (H1) and Remark 1.3. Thus, (3-5)–(3-6) show that $\{{F}(x_{n})\}$ is equicontinuous in $\Omega $ . Clearly, $\{{F}(x_{n})\}$ is uniformly bounded in $\Omega $ . Furthermore, it follows from the penultimate inequality of (3-3) that

(3-7) $$ \begin{align} |[{F}(x_{n})](t)-\psi_{0}|\leq w(|\psi_0|+\delta)\int_t^{\infty} s \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2\,ds+\int_t^{\infty} s|f(s, 0)|\,ds. \end{align} $$

Then, according to (1-8) and (1-11), for any $\varepsilon>0$ , there exists $T_1>T_{0}$ such that, for any $t\geq T_1$ ,

(3-8) $$ \begin{align} \int_t^{\infty} s \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2\,ds<\frac{\varepsilon}{2w(|\psi_0|+\delta)},\quad \int_t^{\infty} s|f(s, x)|\,ds<\frac{\varepsilon}{2}\ \text{uniformly in}\ x. \end{align} $$

Combining this with (3-7), it follows that

$$ \begin{align*} |[{F}(x_{n})](t)-\psi_{0}|<\varepsilon,\quad t\geq T_1, \ n\geq1. \end{align*} $$

Similar to (3-4), together with (1-11) and (3-8),

$$ \begin{align*} |[{F}(x_{n})]'(t)|\leq \int_t^{\infty} s|f(s, x_n)|\,ds<\varepsilon,\quad t\geq T_1, \ n\geq1. \end{align*} $$

Then, $\{{F}(x_{n})\}$ is equiconvergent in $\Omega $ . Thus, via the Arzela–Ascoli theorem [Reference Zeidler28], we obtain that $\{{F}(x_{n})\}$ is relatively compact in $\Omega $ .

Now, we just need to prove that ${F}:\ \Omega \rightarrow \Omega $ is continuous. From (1-8), for any fixed $\varepsilon>0$ , there exists $T_2\geq \max \{2, T_{0}\}$ such that

(3-9) $$ \begin{align} w(2\delta)\int_{T_2}^{\infty} s \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2\,ds<\frac{\varepsilon}{2}. \end{align} $$

It follows from Condition (H1) that $f:\ [t_{0}, T_2]\times [|\psi _{0}|-\delta , |\psi _{0}|+\delta ]\rightarrow \mathbb {R}$ is uniformly continuous. Then, there exists a constant $\gamma{\kern-1pt}>{\kern-1pt}0$ such that for any ${x, y{\kern-1pt}\in{\kern-1pt} [|\psi _{0}|{\kern-1pt}-{\kern-1pt}\delta , |\psi _{0}|{\kern-1pt}+{\kern-1pt}\delta ]}$ with $|x-y|<\gamma $ ,

(3-10) $$ \begin{align} |f(t, x)-f(t, y)|<\frac{\varepsilon}{T_2^{2}},\quad t\in[t_{0}, T_2]. \end{align} $$

Therefore, for any $t\geq T_0$ , $x_{1}, x_{2}\in \Omega $ with $\|x_{1}-x_{2}\|<\gamma $ , using a similar method to treat (3-6), we deduce from (3-9) and (3-10) that

$$ \begin{align*} &|[{F}(x_{1})](t)-[{F}(x_{2})](t)|\\& \quad =\bigg|\!\int_{t}^{\infty} \phi^{-1}\bigg(\!\int_{s}^{\infty}f(\tau, x_{1})\,d\tau\bigg)\,ds- \int_{t}^{\infty} \phi^{-1}\bigg(\!\int_{s}^{\infty}f(\tau, x_{2})\,d\tau\bigg)\,ds\bigg|\\& \quad \leq\int_{t}^{\infty} \bigg|\phi^{-1}\bigg(\!\int_{s}^{\infty}f(\tau, x_{1})\,d\tau\bigg)- \phi^{-1}\bigg(\!\int_{s}^{\infty}f(\tau, x_{2})\,d\tau\bigg)\bigg|ds\\& \quad \leq\int_{t}^{\infty}\int_{s}^{\infty}|f(\tau, x_{1})-f(\tau, x_{2})|\,d\tau\, ds\\& \quad =\int_t^{\infty}\int_t^\tau|f(\tau, x_{1})-f(\tau, x_{2})|\,ds \,d\tau \\& \quad =\int_{t}^{\infty}(\tau-t)|f(\tau, x_{1})-f(\tau, x_{2})|\,d\tau\\& \quad =\int_{t}^{\infty}(s-t)|f(s, x_{1})-f(s, x_{2})|\,ds\\& \quad \leq\int_{T_{0}}^{T_{2}}(s-T_0)|f(s, x_{1})-f(s, x_{2})|\,ds+\int_{T_2}^{\infty}(s-T_2)|f(s, x_{1})-f(s, x_{2})|\,ds\\& \quad \leq\frac{\varepsilon(T_{2}-T_{0})^{2}}{2T_2^{2}}+\int_{T_2}^{\infty} s \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2w(|x_1-x_2|)\,ds\\& \quad <\frac{\varepsilon}{2}+w(\gamma)\int_{T_2}^{\infty} s \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2\,ds\\& \quad <\varepsilon. \end{align*} $$

Then,

(3-11) $$ \begin{align} |[{F}(x_{1})](t)-[{F}(x_{2})](t)|<\varepsilon. \end{align} $$

Similarly, we can get

(3-12) $$ \begin{align} \begin{aligned} |[{F}(x_{1})]'(t)-[{F}(x_{2})]'(t)|&=\bigg|\phi^{-1}\bigg(\!\int_{t}^{\infty}f(s, x_{1})\,ds\bigg)-\phi^{-1}\bigg(\!\int_{t}^{\infty}f(s, x_{2})\,ds\bigg)\bigg|\\ &\leq\int_{t}^{\infty}|f(s, x_{1})-f(s, x_{2})|\,ds\\ &\leq\int_{T_{0}}^{T_{2}}|f(s, x_{1})-f(s, x_{2})|\,ds+\int_{T_{2}}^{\infty}s|f(s, x_{1})-f(s, x_{2})|\,ds\\ &\leq\frac{\varepsilon}{T_{2}^{2}}(T_{2}-T_{0})+\int_{T_2}^{\infty} s \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2w(|x_1-x_2|)\,ds\\ &\leq\frac{\varepsilon}{T_{2}}+\frac{\varepsilon}{2}\\ &<\varepsilon. \end{aligned} \end{align} $$

Thus, it follows from (3-11) and (3-12) that ${F}:\Omega \rightarrow \Omega $ is a continuous operator.

Consequently, it follows from Schauder fixed-point theorem that F admits a fixed point $x\in \Omega $ , so x is an abounded solution of (2-1). Indeed, by Lemma 2.2, x is also a solution of (1-1) and (1-2).

Next, we prove the uniqueness. Let $x, y\in \Omega $ be two solutions of (2-1). Then,

(3-13) $$ \begin{align} (x(t)-y(t))'=-\bigg[\phi^{-1}\bigg(\!\int_t^{\infty}f(s, x)\,ds\bigg)-\phi^{-1}\bigg(\!\int_t^{\infty}f(s, y)\,ds\bigg)\bigg], \end{align} $$

and, hence,

$$ \begin{align*} (x(t)-y(t))"=\bigg(1+\bigg(\!\int_t^{\infty}f(s, x)\,ds\bigg)^2\bigg)^{-{3}/{2}}f(t, x)-\bigg(1+\bigg(\!\int_t^{\infty}f(s, y)\,ds\bigg)^2\bigg)^{-{3}/{2}}f(t, y), \end{align*} $$

as well as

$$ \begin{align*} \lim_{t\to\infty}(x(t)-y(t))=0,\quad \lim_{t\to\infty}(x(t)-y(t))'e^t=0. \end{align*} $$

By virtue of the change of variables

$$ \begin{align*} \tau=\ln\mu(t),\quad t\in[t_0, \infty), \end{align*} $$

it follows that $\tau \in [\ln \mu (t_0), \infty )$ . Let $X(\tau )=x(t),\ Y(\tau )=y(t)$ . Then,

(3-14) $$ \begin{align} (X(\tau)-Y(\tau))'=\frac{\mu(t)}{\mu'(t)}(x(t)-y(t))'. \end{align} $$

Observe that ${\mu (t)}/{\mu '(t)}$ is a nondecreasing function since $\mu "(t)\leq 0$ . Combining this with (1-12), (3-13)–(3-14) and Remark 2.3, we obtain

(3-15) $$ \begin{align} \begin{aligned} |(X(\tau)-Y(\tau))'|&=\bigg|\frac{\mu(t)}{\mu'(t)}\bigg[\phi^{-1}\bigg(\!\int_t^{\infty}f(s, x)\,ds\bigg)-\phi^{-1}\bigg(\!\int_t^{\infty}f(s, y)\,ds\bigg)\bigg]\bigg|\\ &\leq |(\phi^{-1})'(\zeta)|\bigg(\frac{\mu(t)}{\mu'(t)}\bigg)\int_t^\infty|f(s, x)-f(s, y)|\,ds\\ &\leq\int_\tau^\infty\bigg(\frac{\mu(s)}{\mu'(s)}\bigg)^2|f(s, x)-f(s, y)|\,d\xi\ (\text{here}\ \xi=\ln\mu(s))\\ &\leq \int_\tau^\infty w(|X(\xi)- Y(\xi)|)\,d\xi, \end{aligned} \end{align} $$

where $\zeta $ is between $\int _t^{\infty }f(s, x)\,ds$ and $\int _t^{\infty }f(s, y)\,ds$ .

However, it follows from the continuity of f, L’Hôpital’s rule, (1-13), (3-13) and Remark 1.3 that

$$ \begin{align*} &\lim_{\tau\to\infty}(e^\tau(X(\tau)-Y(\tau)))\\ & \quad =\lim_{t\to\infty}\frac{x(t)-y(t)}{{1}/{\mu(t)}}\\ & \quad =\lim_{t\to\infty}\frac{-\bigg[\phi^{-1}\bigg(\!\int_t^{\infty}f(s, x)\,ds\bigg)-\phi^{-1}\bigg(\!\int_t^{\infty}f(s, y)\,ds\bigg)\bigg]}{({1}/{\mu(t)})'(t)}\\ & \quad =\lim_{t\to\infty}-(\phi^{-1})'(\zeta)\frac{\int_t^{\infty}f(s, x)\,ds-\int_t^{\infty}f(s, y)\,ds}{({1}/{\mu(t)})'(t)}\\& \quad =\lim_{t\to\infty}-(\phi^{-1})'(\zeta)\lim_{t\to\infty}\frac{\int_t^{\infty}f(s, x)\,ds-\int_t^{\infty}f(s, y)\,ds}{({1}/{\mu(t)})'(t)}\\ & \quad =\lim_{t\to\infty}\frac{f(t, x(t))-f(t, y(t))}{({1}/{\mu(t)})"(t)}=0, \end{align*} $$

where $\zeta $ is introduced in (3-15), so

(3-16) $$ \begin{align} \lim_{\tau\to\infty}e^\tau|X(\tau)-Y(\tau)|=0, \end{align} $$

which implies that

$$ \begin{align*} \lim\limits_{\tau\to\infty}|X(\tau)-Y(\tau)|=0. \end{align*} $$

Set $\Phi (\tau )=|X(\tau )-Y(\tau )|$ . In view of (3-16), define the normed space

$$ \begin{align*} Y=\bigg\{\Phi\in C[\tau_0, \infty): \Phi(\tau)\geq0\ \text{on}\ \tau\geq\tau_0,\ \sup_{\tau\geq\tau_0}\{\Phi(\tau)e^\tau\}<\infty\bigg\} \end{align*} $$

endowed with the norm

$$ \begin{align*} \|\Phi\|_\infty=\sup_{\tau\geq\tau_0}\{\Phi(\tau)e^\tau\}, \end{align*} $$

where $\tau _0=\ln \mu (t_0)$ . Then, it follows from (3-16) that there exists $\tau _1\geq \tau _0$ such that

(3-17) $$ \begin{align} \|\Phi\|_\infty=\Phi(\tau_1)e^{\tau_1}>\Phi(\tau)e^{\tau},\quad \Phi\in Y,\ \tau>\tau_1. \end{align} $$

Assume that $\|\Phi \|_\infty>0$ (if $\|\Phi \|_\infty =0$ , then the uniqueness is proved). Together with (3-15),

(3-18) $$ \begin{align} \begin{aligned} \Phi(\tau)&=|X(\tau)-Y(\tau)|\leq\int_\tau^\infty|X'(\xi)-Y'(\xi)|\,d\xi\\ &\leq\int_\tau^\infty\int_\xi^\infty w(|X(\zeta)-Y(\zeta)|)\,d\zeta\, d\xi\\ &=\int_\tau^\infty\int_\xi^\infty w(\Phi(\zeta))\,d\zeta\, d\xi. \end{aligned} \end{align} $$

By (3-17), (3-18), the definition of $\mathcal {F}$ and $\|\Phi \|_\infty $ , we deduce that

$$ \begin{align*} \|\Phi\|_\infty e^{-\tau_1}=\Phi(\tau_1)&\leq\int_{\tau_1}^\infty\int_\xi^\infty w(\Phi(\zeta))\,d\zeta \,d\xi\\&=\int_{\tau_1}^\infty\int_\xi^\infty w(\Phi(\zeta)e^{-\zeta}e^{\zeta})\,d\zeta \,d\xi\\&<\int_{\tau_1}^\infty\int_\xi^\infty w(\|\Phi\|e^{-\zeta})\,d\zeta \,d\xi\\&\leq\int_{\tau_1}^\infty\int_0^{\|\Phi\|_\infty e^{-\xi}} \frac{w(r)}{r}\,dr \,d\xi\\&\leq\int_{\tau_1}^\infty\|\Phi\|_\infty e^{-\xi}\,d\xi\\&=\|\Phi\|_\infty e^{-\tau_1}, \end{align*} $$

which is a contradiction. Thus, $\|\Phi \|=0$ . Thus, there exists a unique fixed point x of F in $\Omega $ . Consequently, (1-1) and (1-2) admit a unique bounded solution x.

Proof of Theorem 1.2.

Let x and $\tilde {x}$ be two solutions of the integral equation (2-1) with

$$ \begin{align*} \lim_{t\to\infty}x(t)=\psi_0, \quad \lim_{t\to\infty} \tilde{x}(t)=\tilde{\psi}_0. \end{align*} $$

Since $\mu \in \mathcal {L}$ , we may choose $T_1\geq 1$ large enough such that

$$ \begin{align*} \alpha_1=\int_{T_1}^\infty s\bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2<\dfrac{1}{e}. \end{align*} $$

For any $t\geq T_1$ , it follows from (1-12), Lemma 2.1 and the mean value theorem that

$$ \begin{align*} |x(t)-\tilde{x}(t)|&\leq|\psi_0-\tilde{\psi}_0|+ \bigg|\!\int_t^{\infty}\phi^{-1}\bigg(\!\int_\tau^{\infty}f(s, x)\,ds\bigg)\,d\tau-\int_t^{\infty}\phi^{-1}\bigg(\!\int_\tau^{\infty}f(s, \tilde{x})\,ds\bigg)\,d\tau\bigg|\\ &\leq|\psi_0-\tilde{\psi}_0|+\int_t^{\infty}|(\phi^{-1})'(\xi)|\int_\tau^{\infty}|f(s, x)-f(s, \tilde{x})|\,ds\,d\tau\\ &\leq|\psi_0-\tilde{\psi}_0|+\int_t^{\infty}\int_\tau^{\infty}|f(s, x)-f(s, \tilde{x})|\,ds\,d\tau\\ &\leq|\psi_0-\tilde{\psi}_0|+\int_t^{\infty}\int_t^{s}|f(s, x)-f(s, \tilde{x})|\,d\tau \,ds\\ &=|\psi_0-\tilde{\psi}_0|+\int_t^{\infty}(s-t)|f(s, x)-f(s, \tilde{x})|\,ds\\ &\leq |\psi_0-\tilde{\psi}_0|+\int_t^{\infty}s\bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2w(|x(s)-\tilde{x}(s)|)\,ds\\ &\leq |\psi_0-\tilde{\psi}_0|+e\|x-\tilde{x}\|_\infty\int_t^{\infty}s\bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2\,ds\\ &\leq |\psi_0-\tilde{\psi}_0|+e\alpha_1\|x-\tilde{x}\|_\infty \end{align*} $$

(here, $\xi $ is between $\int _\tau ^{\infty }f(s, x)\,ds$ and $\int _\tau ^{\infty }f(s, \tilde {x})\,ds$ ). Then,

(4-1) $$ \begin{align} |x(t)-\tilde{x}(t)|\leq\|x-\tilde{x}\|_\infty\leq\frac{|\psi_0-\tilde{\psi}_0|}{1-e\alpha_1},\quad t\geq T_1, \end{align} $$

since

$$ \begin{align*} \alpha_1=\int_{T_1}^\infty s\bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2\geq\int_{T_1}^\infty\bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2. \end{align*} $$

Then, by (2-1) and (4-1), for any $t\geq T_1$ , it follows that

(4-2) $$ \begin{align} \begin{aligned} |x'(t)-\tilde{x}'(t)|&=\bigg|\bigg[\phi^{-1}\bigg(\!\int_t^{\infty}f(s, x(s))\,ds\bigg)-\phi^{-1}\bigg(\!\int_t^{\infty}f(s, \tilde{x}(s))\,ds\bigg)\bigg]\bigg|\\ &\leq |(\phi^{-1})'(\zeta)|\int_t^\infty|f(s, x)-f(s, \tilde{x})|\,ds\\ &\leq \int_t^\infty \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2w(|x(s)- \tilde{x}(s)|)\,ds\\ &\leq \int_t^\infty e\bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2|x(s)- \tilde{x}(s)|\,ds\\ &\leq \frac{e\alpha_1}{1-e\alpha_1} |\psi_0-\tilde{\psi}_0|; \end{aligned} \end{align} $$

here, $\zeta $ is between $\int _t^{\infty }f(s, x)\,ds$ and $\int _t^{\infty }f(s, \tilde {x})\,ds$ . Thus, we can also get

(4-3) $$ \begin{align} |x(T_1)-\tilde{x}(T_1)|+|x'(T_1)-\tilde{x}'(T_1)|\leq\frac{1+e\alpha_1}{1-e\alpha_1} |\psi_0-\tilde{\psi}_0|. \end{align} $$

If $t_0=T_1$ , then the proof is complete via (4-1) and (4-2). If $T_1>t_0$ , set

$$ \begin{align*} s=T_{1}-t,~y(s)=x(T_{1}-s),~z(s)=y'(s),\quad 0\leq s\leq T_{1}-t_{0}. \end{align*} $$

Observe that a solution of (1-1) satisfies

$$ \begin{align*} x"=(1-(x')^{2})^{{3}/{2}}f(t, x),\quad t\geq t_0; \end{align*} $$

then y satisfies

(4-4) $$ \begin{align} \begin{aligned} y"(s)=x"(T_{1}-s)&=(1-(x'(T_{1}-s))^{2})^{{3}/{2}}f(T_{1}-s, x(T_{1}-s))\\ &=(1-(z(s))^{2})^{{3}/{2}}f(T_{1}-s, y(s)) \end{aligned} \end{align} $$

with the initial condition

(4-5) $$ \begin{align} y(0)=x(T_{1}), \quad y'(0)=-x'(T_{1}). \end{align} $$

Consider the linear system

(4-6) $$ \begin{align} \begin{pmatrix} y'(s)\\z'(s) \end{pmatrix} = \begin{pmatrix} 0&1\\ -\bigg(\dfrac{\mu'(T_1-s)}{\mu(T_1-s)}\bigg)^2&0 \end{pmatrix} \begin{pmatrix} y(s)\\z(s) \end{pmatrix}, \quad 0\leq s\leq T_{1}-t_{0}. \end{align} $$

Denote by

$$ \begin{align*} A(s)= \begin{pmatrix} a_{11}(s)&a_{12}(s)\\ a_{21}(s)&a_{22}(s) \end{pmatrix} \end{align*} $$

the fundamental matrix solution of the system (4-6). Then, (4-4)–(4-5) can be written as

$$ \begin{align*} \begin{pmatrix} y'(s)\\ z'(s) \end{pmatrix} & = \begin{pmatrix} 0&1\\ -\bigg(\dfrac{\mu'(T_1-s)}{\mu(T_1-s)}\bigg)^2&0 \end{pmatrix} \begin{pmatrix} y(s)\\ z(s) \end{pmatrix}\\ &\quad + \begin{pmatrix} 0\\ (1-(z(s))^{2})^{{3}/{2}}f(T_{1}-s, y(s))+\bigg(\dfrac{\mu'(T_1-s)}{\mu(T_1-s)}\bigg)^2y(s) \end{pmatrix} \end{align*} $$

with

$$ \begin{align*} \begin{pmatrix} y(0)\\ z(0) \end{pmatrix}= \begin{pmatrix} x(T_{1})\\ -x'(T_{1}) \end{pmatrix}.\\ \end{align*} $$

It follows from the variation of constants formula that

$$ \begin{align*} \begin{pmatrix} y(s)\\ z(s) \end{pmatrix} &=A(s) \begin{pmatrix} x(T_{1})\\ -x'(T_{1}) \end{pmatrix}\\ &\quad +\int_{0}^{s}A(s-\tau) \begin{pmatrix} 0\\ (1-(z(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, y(\tau))+\bigg(\frac{\mu'(T_1-\tau)}{\mu(T_1-\tau)}\bigg)^2y(\tau)\end{pmatrix} \,d\tau. \end{align*} $$

Then, for any $0\leq s\leq T_{1}-t_{0}$ ,

$$ \begin{align*} \begin{aligned} y(s)&=a_{11}(s)x(T_{1})-a_{12}(s)x'(T_{1})\\ &\quad+\int_{0}^{s}a_{12}(s-\tau)(1-(z(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, y(\tau))\,d\tau.\\ &\quad+\int_{0}^{s}a_{12}(s-\tau)\bigg(\frac{\mu'(T_1-\tau)}{\mu(T_1-\tau)}\bigg)^2y(\tau)\,d\tau.\\ \end{aligned} \end{align*} $$

Similarly, set

$$ \begin{align*} \tilde{y}(s)=\tilde{x}(T_{1}-s), \quad \tilde{z}(s)=\tilde{y}'(s),\quad 0\leq s\leq T_{1}-t_{0}. \end{align*} $$

Thus, for any $0\leq s\leq T_{1}-t_{0}$ ,

(4-7) $$ \begin{align} &y(s)-\tilde{y}(s) =a_{11}(s)[x(T_{1})-\tilde{x}(T_{1})]-a_{12}(s)[x'(T_{1})-\tilde{x}'(T_1)] \nonumber\\& \quad +\int_{0}^{s}\!\!a_{12}(s-\tau)[(1-(z(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, y(\tau))- (1-(\tilde{z}(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, \tilde{y}(\tau))]\,d\tau\nonumber\\& \quad + \int_{0}^{s}a_{12}(s-\tau)[y(\tau)-\tilde{y}(\tau)]\bigg(\frac{\mu'(T_1-\tau)}{\mu(T_1-\tau)}\bigg)^2\,d\tau. \end{align} $$

Set

$$ \begin{align*} a_{11}^{*}=\max\limits_{s\in[0,T_{1}-t_{0}]}|a_{11}(s)|,\quad a_{12}^{*}=\max\limits_{s\in[0,T_{1}-t_{0}]}|a_{12}(s)|,\quad a_{1}^{*}=\max\{a_{11}^{*}, a_{12}^{*}\}. \end{align*} $$

Then, (4-7) becomes

(4-8) $$ \begin{align} &|y(s)-\tilde{y}(s)|\nonumber\\ &\quad\leq a_{11}^{*}|x(T_{1})-\tilde{x}(T_{1})|+a_{12}^{*}|x'(T_{1})-\tilde{x}'(T_{1})|\nonumber\\ & \qquad+\int_{0}^{s}a_{12}^{*}|(1-(z(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, y(\tau)) -(1-(\tilde{z}(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, \tilde{y}(\tau))|\,d\tau \nonumber\\ & \qquad+\int_{0}^{s}a_{12}^{*} |y(\tau)-\tilde{y}(\tau)|\bigg(\frac{\mu'(T_1-\tau)}{\mu(T_1-\tau)}\bigg)^2\,d\tau. \end{align} $$

For any fixed $\tau \in [0,\ T_{1}-t_{0}]$ , note that it follows from the mean value theorem that there exists $z^\ast $ between $z(\tau )$ and $\widetilde {z}(\tau )$ such that

(4-9) $$ \begin{align} \begin{aligned} |(1-(z(\tau))^{2})^{{3}/{2}}-(1-(\tilde{z}(\tau))^{2})^{{3}/{2}}|&=|-3z^\ast(1-z^{\ast2})^{1}/{2}(z(\tau)-\tilde{z}(\tau))|\\ & \leq 3|z(\tau)-\tilde{z}(\tau)|. \end{aligned} \end{align} $$

Then, from (1-12), (4-9) and Lemma 2.1,

(4-10) $$ \begin{align} &|(1-(z(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, y(\tau))-(1-(\tilde{z}(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, \tilde{y}(\tau))| \nonumber \\ & \quad =|(1-(z(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, y(\tau))-(1-(z(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, \tilde{y}(\tau)) \nonumber \\ & \qquad +(1-(z(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, \tilde{y}(\tau))-(1-(\tilde{z}(\tau))^{2})^{{3}/{2}}f(T_{1}-\tau, \tilde{y}(\tau))| \nonumber \\ & \quad \leq(1-(z(\tau))^{2})^{{3}/{2}}|f(T_{1}-\tau, y(\tau))-f(T_{1}-\tau, \tilde{y}(\tau))| \\ & \qquad +|(1-(z(\tau))^{2})^{{3}/{2}}-(1-(\tilde{z}(\tau))^{2})^{{3}/{2}}|\cdot|f(T_{1}-\tau, \tilde{y}(\tau))| \nonumber \\ & \quad \leq|f(T_{1}-\tau, y(\tau))-f(T_{1}-\tau, \tilde{y}(\tau))|+3|f(T_{1}-\tau, \tilde{y}(\tau))|\cdot|z(\tau)-\tilde{z}(\tau)| \nonumber \\ & \quad \leq e\bigg(\frac{\mu'(T_1-\tau)}{\mu(T_1-\tau)}\bigg)^2|y(\tau)-\tilde{y}(\tau)|+3M|z(\tau)-\tilde{z}(\tau)|. \nonumber \end{align} $$

In fact, by Condition (H1), for any $\tau \in [0, T_{1}-t_{0}]$ , there exists some constant $M>0$ such that

$$ \begin{align*} |f(T_{1}-\tau, \tilde{y}(\tau))|\leq M. \end{align*} $$

Then, by (4-2), (4-8) and (4-10),

(4-11) $$ \begin{align} \begin{aligned} |y(s)-\tilde{y}(s)|&\leq a_{11}^{*}|x(T_{1})-\tilde{x}(T_{1})|+a_{12}^{*}|x'(T_{1})-\tilde{x}'(T_{1})|\\ &\quad +\int_{0}^{s}(1+e)a_{12}^{*}\bigg(\frac{\mu'(T_1-\tau)}{\mu(T_1-\tau)}\bigg)^2|y(\tau)-\tilde{y}(\tau)|\,d\tau\\ &\quad +(T_{1}-t_{0})3a_{12}^{*}M\frac{e\alpha_1}{1-e\alpha_1} |\psi_0-\tilde{\psi}_0|. \end{aligned} \end{align} $$

By virtue of (4-3), Inequality (4-11) becomes

$$ \begin{align*} \begin{aligned} |y(s)-\tilde{y}(s)|&\leq\frac{a_{1}^{*}(1+e\alpha_1)}{1-e\alpha_1}|\psi_0-\tilde{\psi}_0|+\frac{3(T_{1}-t_{0})a_{12}^{*}Me\alpha_1}{1-e\alpha_1} |\psi_0-\tilde{\psi}_0|\\ &\quad +\int_{0}^{s}(1+e)a_{12}^{*}\bigg(\frac{\mu'(T_1-\tau)}{\mu(T_1-\tau)}\bigg)^2 |y(\tau)-\tilde{y}(\tau)|\,d\tau, \end{aligned} \end{align*} $$

which, on account of the Bellman–Gronwall inequality, implies that

$$ \begin{align*} \begin{aligned} |y(s)-\tilde{y}(s)| &\leq\bigg[\frac{a_{1}^{*}(1+e\alpha_1)}{1-e\alpha_1}+\frac{3a_{12}^{*}M(T_{1}-t_{0})e\alpha_1}{1-e\alpha_1}\bigg]|\psi_{0}-\tilde{\psi_{0}}| e^{(\int_{0}^{T_1-t_0}(1+e)a_{12}^{*}({\mu'(T_1-\tau)}/{\mu(T_1-\tau)})^2\,d\tau)}\\ &\leq\bigg[\frac{a_{1}^{*}(1+e\alpha_1)}{1-e\alpha_1}+\frac{3a_{12}^{*}M(T_{1}-t_{0})e\alpha_1}{1-e\alpha_1}\bigg]|\psi_{0}-\tilde{\psi_{0}}|e^{(\int_{t_{0}}^{T_{1}}(1+e)a_{12}^{*} ({\mu'(s)}/{\mu(s)})^2\,ds)}. \end{aligned} \end{align*} $$

Then, we equivalently have

(4-12) $$ \begin{align} |x(t)-\tilde{x}(t)|\leq\bigg[\frac{a_{1}^{*}(1+e\alpha_1)}{1-e\alpha_1}+\frac{3a_{12}^{*}M(T_{1}-t_{0})e\alpha_1}{1-e\alpha_1}\bigg] |\psi_{0}-\tilde{\psi_{0}}|e^{(\int_{t_{0}}^{T_{1}}(1+e)a_{12}^{*}({\mu'(s)}/{\mu(s)})^2\,ds)}. \end{align} $$

Similarly, we can also deduce that

(4-13) $$ \begin{align} |x'(t)-\tilde{x}'(t)|\leq\bigg[\frac{a_{2}^{*}(1+e\alpha_1)}{1-e\alpha_1}+\frac{3a_{22}^{*}M(T_{1}-t_{0})e\alpha_1}{1-e\alpha_1}\bigg] |\psi_{0}-\tilde{\psi_{0}}|e^{(\int_{t_{0}}^{T_{1}} (1+e)a_{22}^{*}({\mu'(s)}/{\mu(s)})^2\,ds)}, \end{align} $$

where

$$ \begin{align*} a_{21}^{*}=\max\limits_{s\in[0,T_{1}-t_{0}]}|a_{21}(s)|,\quad a_{22}^{*}=\max\limits_{s\in[0,T_{1}-t_{0}]}|a_{22}(s)|, \quad a_{2}^{*}=\max\{a_{21}^{*}, a_{22}^{*}\}. \end{align*} $$

Consequently, it follows from (4-1), (4-2), (4-12) and (4-13) that the solution obtained in Theorem 1.1 is stable in the norm (1-9) with respect to the values of $\psi _{0}$ .

Proof. We first show that $\{x_n(t)\}\subset \Omega $ , where $\Omega $ is defined as in (3-2). It is not difficult to see that $\{x_n\}\subset X.$ Since $T_0$ satisfies (3-1), now, we proceed as in (3-3). It follows that

(5-2) $$ \begin{align} \begin{aligned} |x_1(t)-\psi_0|&=\bigg|\!\int_t^{\infty}\phi^{-1}\bigg(\!\int_\tau^{\infty}f(s, x_0)\,ds\bigg)\,d\tau\bigg|\\ &\leq\int_t^{\infty} (s-t)|f(s, x_0)|\,ds\\ &\leq \int_t^{\infty} (s-t)|f(s, x_0)-f(s,\ 0)|\,ds+\int_t^{\infty} (s-t)|f(s,\ 0)|\,ds\\ &\leq \int_t^{\infty} s \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2w(|x_0(s)|)\,ds+\int_t^{\infty} s|f(s,\ 0)|\,ds\\ &\leq w(|\psi_0|+\delta)\int_{T_0}^{\infty} s \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2\,ds+\int_{T_0}^{\infty} s|f(s,\ 0)|\,ds\\ &\leq\delta. \end{aligned} \end{align} $$

For any $n\geq 1$ , now suppose that $x_{n-1}(t)\in \Omega $ . Repeating the arguments in (5-2), we can also get

(5-3) $$ \begin{align} |x_{n}(t)-\psi_0|\leq\delta,\quad t\geq T_0. \end{align} $$

Thus, $\{x_n(t)\}\subset \Omega $ .

In addition, it follows from (5-3) that $\{x_n(t)\}$ is uniformly bounded. However, by virtue of (5-1), for any $t\geq T_0$ , using methods similar to (3-3),

$$ \begin{align*} |x_{n+1}'(t)|&=\bigg|\phi^{-1}\bigg(\!\int_t^{\infty}f(s, x_n)\,ds\bigg)\bigg|\\ &\leq\int_t^{\infty} |f(s, x_n)|\,ds\\ &\leq \int_t^{\infty} s|f(s, x_n)|\,ds\\ &\leq\delta. \end{align*} $$

Combining the preceding inequality with the mean value theorem, we obtain that

(5-4) $$ \begin{align} |x_{n+1}(t_1)-x_{n+1}(t_2)|\leq\delta|t_1-t_2|,\quad t_1, t_2\geq T_0,\ n\geq0. \end{align} $$

Let $\zeta $ be between $\int _{t_1}^{\infty }f(s, x_n)\,ds$ and $\int _{t_2}^{\infty }f(s, x_{n})\,ds$ . Then, it follows from (5-1) that

(5-5) $$ \begin{align} \begin{aligned} |x_{n+1}'(t_1)-x_{n+1}'(t_2)|&=\bigg|\phi^{-1}\bigg(\!\int_{t_1}^{\infty}f(s, x_n)\,ds\bigg)-\phi^{-1}\bigg(\!\int_{t_2}^{\infty}f(s, x_{n})\,ds\bigg)\bigg|\\ &\leq |(\phi^{-1})'(\zeta)|\cdot\bigg|\!\int_{t_1}^{t_2}|f(s, x_{n}(s))|\,ds\bigg|\\ &\leq M|t_1-t_2|, \end{aligned} \end{align} $$

where $|f(s, x)|\leq M$ for some constant $M>0$ when s belongs to either $[t_1,\ t_2]$ or $[t_2,\ t_1]$ . Then, it follows from (5-4) and (5-5) that $\{x_n(t)\}$ is equicontinuous in X. Moreover, similar to (3-3), it is not difficult to prove that $\{x_n(t)\}$ is equiconvergent. By applying the Arzela–Ascoli theorem, we obtain that there exists a subsequence $\{x_{n_k}(t)\}$ that converges uniformly to a continuous function $h(t)$ on $[T_0, \infty )$ as $k\to \infty $ . Since

$$ \begin{align*} x_{n_{k}+1}(t)=\psi_0+\int_t^{\infty}\phi^{-1}\bigg(\!\int_\tau^{\infty}f(s, x_{n_k})\,ds\bigg)\,d\tau, \end{align*} $$

by the continuity of f and $\phi ^{-1}$ , the sequence $\{x_{n_k+1}(t)\}$ converges uniformly to

$$ \begin{align*} \tilde{h}(t)=\psi_0+\int_t^{\infty}\phi^{-1}\bigg(\!\int_\tau^{\infty}f(s,\ h(s))\,ds\bigg)\,d\tau. \end{align*} $$

In the following, we prove that $\tilde {h}(t)=h(t)$ on $[T_0, \infty )$ . Now, we proceed as in [Reference Chu and Wang11], repeating the arguments for completeness. We only need to show that

$$ \begin{align*} \lim_{n\to\infty}\|x_{n+1}(t)-x_{n}(t)\|_\infty=0 \quad \text{uniformly on}\ [T_0, \infty). \end{align*} $$

By (5-1),

(5-6) $$ \begin{align} \begin{aligned} |x_{n+1}'(t)-x_{n}'(t)|&=\bigg|\phi^{-1}\bigg(\!\int_t^{\infty}f(s, x_n)\,ds\bigg)-\phi^{-1}\bigg(\!\int_t^{\infty}f(s, x_{n-1})\,ds\bigg)\bigg|\\ &\leq |(\phi^{-1})'(\zeta)|\int_t^\infty|f(s, x_{n}(s))-f(s, x_{n-1}(s))|\,ds\\ &\leq \int_t^\infty \bigg(\frac{\mu'(s)}{\mu(s)}\bigg)^2w(|x_{n}(s)- x_{n-1}(s)|)\,ds, \end{aligned} \end{align} $$

where $\zeta $ is between $\int _t^{\infty }f(s, x_{n})\,ds$ and $\int _t^{\infty }f(s, x_{n-1})\,ds$ . Set

$$ \begin{align*} \tau=\ln\mu(t),\quad X_n(\tau)=x_n(t). \end{align*} $$

Similar to (3-15), according to (5-6), it becomes

$$ \begin{align*} |X_{n+1}'(\tau)-X_{n}'(\tau)|\leq\int_\tau^\infty w(|X_{n}(s)- X_{n-1}(s)|)\,ds,\quad \tau\geq\ln\mu(T_0), \end{align*} $$

and, thus,

$$ \begin{align*} \begin{aligned} |X_{n+1}(\tau)-X_{n}(\tau)|&\leq \int_\tau^\infty|X_{n}'(s)-X_{n-1}'(s)|\,ds\\ &\leq\int_\tau^\infty\int_\xi^\infty w(|X_{n}(s)- X_{n-1}(s)|)\,ds\,d\xi. \end{aligned} \end{align*} $$

Let $y_n(\tau )=|X_{n+1}(\tau )-X_{n}(\tau )|$ . Then,

(5-7) $$ \begin{align} y_n(\tau)\leq\int_\tau^\infty\int_\xi^\infty w(y_{n-1}(s))\,ds\,d\xi. \end{align} $$

Set

$$ \begin{align*} Y_n(\tau)=\sup_{k\geq n}\{y_k(\tau)\}. \end{align*} $$

It is easy to see that the sequence $\{Y_n(\tau )\}$ is nonincreasing. From (5-7),

$$ \begin{align*} Y_{n+1}(\tau)\leq \int_\tau^\infty\int_\xi^\infty w(Y_n(s))\,ds\,d\xi. \end{align*} $$

The same type of argument shows that the sequence $\{Y_n\}$ is equicontinuous and equiconvergent on $[\ln \mu (T_0), \infty )$ , as well as that

$$ \begin{align*} Y(\tau)=\sup_{n\geq1}\{Y_n(\tau)\} \end{align*} $$

is continuous on $[\ln \mu (T_0), \infty )$ .

Since the pointwise convergence of the function sequence $\{Y_n(\tau )\}$ to the continuous $Y(\tau )$ is monotonic on $[\ln \mu (T_0), \infty )$ , Dini’s theorem (see [Reference Hewitt and Stromberg22]) shows that this convergence is uniform on $[\ln \mu (T_0), \infty )$ . It follows from the Lebesgue dominated convergence theorem that

$$ \begin{align*} Y(\tau)\leq \int_\tau^\infty\int_\xi^\infty w(Y(s))\,ds\,d\xi,\quad \tau\geq\ln\mu(T_0), \end{align*} $$

together with $Y(0)=0$ as $Y_n(0)=0$ . Reasoning in a manner analogous to the uniqueness part of the proof shows $Y(\tau )\equiv 0$ for all $\tau \in [\ln \mu (T_0), \infty )$ . Thus,

$$ \begin{align*} \lim_{n\to\infty}Y_n(\tau)=0\quad \text{uniformly on}\ [\ln\mu(T_0), \infty), \end{align*} $$

and hence,

$$ \begin{align*} \lim_{n\to\infty}\|x_{n+1}(\tau)-x_n(\tau)\|_\infty=0\quad \text{uniformly on}\ [T_0, \infty). \end{align*} $$

The proof is complete.