To prove Theorem 2.1, we shall use the fixed point technique by applying the Schauder fixed point theorem [14], whereas Theorem 2.2 is established by the Banach contraction principle [15]. We state our main tools below.

**Schauder fixed point theorem** [14]

*Let* *E* *be a Banach space and* Ω *be any nonempty convex and closed subset of* *E*. *If* *M* *is a continuous mapping of* Ω *into itself and* *M* Ω *is relatively compact*, *then the mapping* *M* *has at least one fixed point*, *i*.*e*., *there exists an* $x\in \mathrm{\Omega}$ *such that* $x=Mx$.

Let

$BC([0,\mathrm{\infty}),\mathbb{R})$ be the Banach space of all bounded continuous real-valued functions on

$[0,\mathrm{\infty})$, endowed with the sup-norm

$\parallel \cdot \parallel $ given by

$\parallel \psi \parallel =\underset{t\ge 0}{sup}|\psi (t)|\phantom{\rule{1em}{0ex}}\text{for}\psi \in BC([0,\mathrm{\infty}),\mathbb{R}).$

We need the following compactness criterion for a subset of $BC([0,\mathrm{\infty}),\mathbb{R})$, which is a consequence of the well-known Arzela-Ascoli theorem. This compactness criterion is an adaptation of a lemma due to Avramescu [16]. In order to formulate this criterion, we note that a set *U* of real-valued functions defined on $[0,\mathrm{\infty})$ is said to be *equiconvergent at* ∞ if all the functions in *U* are convergent in ℝ at the point ∞ and, in addition, for each $\u03f5>0$, there exists $T\equiv T(\u03f5)>0$ such that, for any function $\psi \in U$, we have $|\psi (t)-{lim}_{s\to \mathrm{\infty}}\psi (s)|<\u03f5$ for $t\ge T$.

**Compactness criterion** [16]

*Let* *U* *be an equicontinuous and uniformly bounded subset of the Banach space* $BC([0,\mathrm{\infty}),\mathbb{R})$. *If* *U* *is equiconvergent at* ∞, *it is also relatively compact*.

**Banach contraction principle** [15]

*Let* *E* *be a Banach space and* Ω *be any nonempty closed subset of* *E*. *If* *M* *is a contraction of* Ω *into itself*, *then the mapping* *M* *has a unique fixed point*, *i*.*e*., *there exists a unique* $x\in \mathrm{\Omega}$ *such that* $x=Mx$.

Throughout this section, let

*E* denote the set of all functions

*x* in

$C([-r,\mathrm{\infty}),\mathbb{R})$, which are continuously differentiable on the interval

$[0,\mathrm{\infty})$ and have bounded continuous derivatives on

$[0,\mathrm{\infty})$. The set

*E* is a Banach space endowed with the norm

${\parallel \cdot \parallel}_{E}$ given by

${\parallel x\parallel}_{E}=max\{\underset{-r\le t\le 0}{max}|x(t)|,\underset{t\ge 0}{sup}|{x}^{\prime}(t)|\}\phantom{\rule{1em}{0ex}}\text{for}x\in E.$

We shall first establish a lemma which will be needed to prove the main results.

**Lemma 3.1** *Suppose that* (2.7)

*holds*,

*where* *F* *is a nonnegative real*-

*valued function defined on* $[0,\mathrm{\infty})\times C([-r,0],[0,\mathrm{\infty}))\times [0,\mathrm{\infty})$,

*which satisfies the continuity condition* (C).

*Assume that* (A)

*is satisfied*.

*Let* Ω

*be the subset of the Banach space* *E* *defined by* $\mathrm{\Omega}=\{x\in E:x(t)=\xi (t)\phantom{\rule{0.1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.1em}{0ex}}-r\le t\le 0,-c\le {x}^{\prime}(t)\le c\phantom{\rule{0.1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.1em}{0ex}}t\ge 0\}.$

*For* $x\in \mathrm{\Omega}$,

*define a mapping* *M* *on* Ω

*by* $(Mx)(t)=\{\begin{array}{cc}\xi (t)\hfill & \mathit{\text{for}}\phantom{\rule{0.1em}{0ex}}-r\le t\le 0,\hfill \\ {\int}_{0}^{t}{\varphi}_{q}({\int}_{\theta}^{\mathrm{\infty}}f(s,{x}_{s},{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds)\phantom{\rule{0.2em}{0ex}}d\theta \hfill & \mathit{\text{for}}\phantom{\rule{0.1em}{0ex}}t\ge 0.\hfill \end{array}$

(3.1)

*Then* *M* *maps* Ω *into* *E*. *Moreover*, *M* Ω *is relatively compact and the mapping* $M:\mathrm{\Omega}\to E$ *is continuous*.

*Proof* Since the condition (A) is satisfied, from (2.8) we have

${\int}_{0}^{\mathrm{\infty}}F(t,{\eta}_{t},c)\phantom{\rule{0.2em}{0ex}}dt\le {\varphi}_{p}(c)={c}^{p-1}<\mathrm{\infty}.$

(3.2)

First, we shall show that

*M* is a mapping from Ω into

*E*,

*i.e.*,

$M\mathrm{\Omega}\subseteq E$. Let

*x* be an arbitrary function in Ω. By the definition of Ω, the function

*x* satisfies (1.2) and (2.11). Since

$\xi (0)=0$, it follows from (1.2) that

$x(0)=0$. By taking into account this fact and using (2.11), we can easily obtain

$|x(t)|\le ct\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

(3.3)

By virtue of (1.2), (3.3) and (2.9), it follows that

$|x(t)|\le \eta (t)$ for all

$t\ge -r$, which ensures that

$|{x}_{t}|\le {\eta}_{t}\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

(3.4)

In view of (3.4), (2.11) and the condition (A), we get

$F(t,|{x}_{t}|,|{x}^{\prime}(t)|)\le F(t,{\eta}_{t},c)\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

On the other hand, (2.7) guarantees that

$\left|f(t,{x}_{t},{x}^{\prime}(t))\right|\le F(t,|{x}_{t}|,|{x}^{\prime}(t)|)\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

Thus, we have

$\left|f(t,{x}_{t},{x}^{\prime}(t))\right|\le F(t,{\eta}_{t},c)\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

(3.5)

From (3.2) and (3.5) it follows that

${\int}_{0}^{\mathrm{\infty}}\left|f(t,{x}_{t},{x}^{\prime}(t))\right|\phantom{\rule{0.2em}{0ex}}dt<\mathrm{\infty}$

(3.6)

and consequently,

${\int}_{0}^{\mathrm{\infty}}f(t,{x}_{t},{x}^{\prime}(t))\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\text{exists in}\mathbb{R}.$

(3.7)

Furthermore, we can conclude that

${\int}_{0}^{t}{\varphi}_{q}\left({\int}_{\theta}^{\mathrm{\infty}}f(s,{x}_{s},{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\right)\phantom{\rule{0.2em}{0ex}}d\theta \phantom{\rule{1em}{0ex}}\text{exists in}\mathbb{R}.$

(3.8)

Since (3.8) holds for any function $x\in \mathrm{\Omega}$, we immediately see that the formula (3.1) makes sense for any $x\in \mathrm{\Omega}$, and this formula defines a mapping *M* from Ω into $C([-r,\mathrm{\infty}),\mathbb{R})$.

Next, using (3.5) and (2.8), from (3.1) we obtain for

$t\ge 0$,

$\begin{array}{rcl}|{(Mx)}^{\prime}(t)|& =& \left|{\varphi}_{q}\left({\int}_{t}^{\mathrm{\infty}}f(s,{x}_{s},{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\right)\right|\\ \le & {\varphi}_{q}\left({\int}_{t}^{\mathrm{\infty}}\left|f(s,{x}_{s},{x}^{\prime}(s))\right|\phantom{\rule{0.2em}{0ex}}ds\right)\\ \le & {\varphi}_{q}\left({\int}_{0}^{\mathrm{\infty}}\left|f(s,{x}_{s},{x}^{\prime}(s))\right|\phantom{\rule{0.2em}{0ex}}ds\right)\\ \le & {\varphi}_{q}({\int}_{0}^{\mathrm{\infty}}F(s,{\eta}_{s},c)\phantom{\rule{0.2em}{0ex}}ds)\\ \le & {\varphi}_{q}({\varphi}_{p}(c))=c.\end{array}$

(3.9)

Inequality (3.9) means that ${(Mx)}^{\prime}$ is bounded on the interval $[0,\mathrm{\infty})$ and so *Mx* belongs to *E*. We have thus proved that $M\mathrm{\Omega}\subseteq E$.

Now, we shall prove that

*M* Ω is relatively compact. We observe that, for any function

$x\in \mathrm{\Omega}$, we have

$(Mx)(t)=\xi (t)$ for

$-r\le t\le 0$. By taking into account this fact as well as the definition of the norm

${\parallel \cdot \parallel}_{E}$, we can easily conclude that it suffices to show that the set

$U=\{{(Mx)}^{\prime}(t){|}_{t\in [0,\mathrm{\infty})}:x\in \mathrm{\Omega}\}$

is relatively compact in the Banach space

$BC([0,\mathrm{\infty}),\mathbb{R})$. Using (3.5), for any

$x\in \mathrm{\Omega}$ and any

${t}_{1}$,

${t}_{2}$ with

$0\le {t}_{1}\le {t}_{2}$, we obtain

$\begin{array}{rcl}|{\varphi}_{p}({(Mx)}^{\prime}({t}_{1}))-{\varphi}_{p}({(Mx)}^{\prime}({t}_{2}))|& =& |{\int}_{{t}_{1}}^{\mathrm{\infty}}f(s,{x}_{s},{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds-{\int}_{{t}_{2}}^{\mathrm{\infty}}f(s,{x}_{s},{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds|\\ =& \left|{\int}_{{t}_{1}}^{{t}_{2}}f(s,{x}_{s},{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\right|\\ \le & {\int}_{{t}_{1}}^{{t}_{2}}\left|f(s,{x}_{s},{x}^{\prime}(s))\right|\phantom{\rule{0.2em}{0ex}}ds\\ \le & {\int}_{{t}_{1}}^{{t}_{2}}F(s,{\eta}_{s},c)\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

In view of (3.2), this means that

${(Mx)}^{\prime}({t}_{1})\to {(Mx)}^{\prime}({t}_{2})$ as

${t}_{1}\to {t}_{2}$, and we can easily verify that

*U* is equicontinuous. Moreover, each function

$x\in \mathrm{\Omega}$ satisfies (3.9), where

*c* is independent of

*x*. This guarantees that

*U* is uniformly bounded. Furthermore, for any

$x\in \mathrm{\Omega}$, we have

$|{(Mx)}^{\prime}(t)|=\left|{\varphi}_{q}\left({\int}_{t}^{\mathrm{\infty}}f(s,{x}_{s},{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\right)\right|\le {\varphi}_{q}\left({\int}_{t}^{\mathrm{\infty}}\left|f(s,{x}_{s},{x}^{\prime}(s))\right|\phantom{\rule{0.2em}{0ex}}ds\right)\phantom{\rule{1em}{0ex}}\text{for}t\ge 0$

and hence, noting (3.5), it follows that

$|{(Mx)}^{\prime}(t)|\le {\varphi}_{q}({\int}_{t}^{\mathrm{\infty}}F(s,{\eta}_{s},c)\phantom{\rule{0.2em}{0ex}}ds)\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

(3.10)

Now (3.10) together with (3.2) implies that

$\underset{t\to \mathrm{\infty}}{lim}{(Mx)}^{\prime}(t)=0.$

By using (3.2) and (3.10) again, we immediately see that *U* is equiconvergent at ∞. It now follows from the given compactness criterion that the set *U* is relatively compact in $BC([0,\mathrm{\infty}),\mathbb{R})$.

Finally, we shall prove that the mapping

$M:\mathrm{\Omega}\to E$ is continuous. Let

$x,{\{{x}^{n}\}}_{n\ge 1}\in \mathrm{\Omega}$ with

${\parallel {x}^{n}-x\parallel}_{E}\to 0$ as

$n\to \mathrm{\infty}$. It is not difficult to verify that

${lim}_{n\to \mathrm{\infty}}{x}^{n}(t)=x(t)$ uniformly for

$t\in [-r,\mathrm{\infty})$ and

${lim}_{n\to \mathrm{\infty}}{({x}^{n})}^{\prime}(t)={x}^{\prime}(t)$ uniformly for

$t\in [0,\mathrm{\infty})$. On the other hand, using (3.5) we have

$\left|f(t,{x}_{t}^{n},{\left({x}^{n}\right)}^{\prime}(t))\right|\le F(t,{\eta}_{t},c)\phantom{\rule{1em}{0ex}}\text{for}t\ge 0\text{and}n\ge 1.$

Thus, by taking into account the fact that

${\int}_{0}^{t}{\varphi}_{q}({\int}_{\theta}^{\mathrm{\infty}}F(s,{\eta}_{s},c)\phantom{\rule{0.2em}{0ex}}ds)\phantom{\rule{0.2em}{0ex}}d\theta <\mathrm{\infty},$

we can apply the Lebesgue dominated convergence theorem to obtain, for every

$t\ge 0$,

$\underset{n\to \mathrm{\infty}}{lim}{\int}_{0}^{t}{\varphi}_{q}\left({\int}_{\theta}^{\mathrm{\infty}}f(s,{x}_{s}^{n},{\left({x}^{n}\right)}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\right)\phantom{\rule{0.2em}{0ex}}d\theta ={\int}_{0}^{t}{\varphi}_{q}\left({\int}_{\theta}^{\mathrm{\infty}}f(s,{x}_{s},{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\right)\phantom{\rule{0.2em}{0ex}}d\theta .$

This, together with the fact that

$\left(M{x}^{n}\right)(t)=(Mx)(t)=\xi (t)\phantom{\rule{1em}{0ex}}\text{for}-r\le t\le 0\text{and}n\ge 1,$

guarantees the pointwise convergence

$\underset{n\to \mathrm{\infty}}{lim}\left(M{x}^{n}\right)(t)=(Mx)(t)\phantom{\rule{1em}{0ex}}\text{for}t\ge -r.$

It remains to show that this convergence is also convergence in the sense of

${\parallel \cdot \parallel}_{E}$,

*i.e.*,

$\underset{n\to \mathrm{\infty}}{lim}{\parallel M{x}^{n}-Mx\parallel}_{E}=0.$

(3.11)

For this purpose, we consider an arbitrary subsequence $\{M{x}^{k}\}$ of $\{M{x}^{n}\}$. Since *M* Ω is relatively compact, there exists a subsequence $\{M{x}^{j}\}$ of the sequence $\{M{x}^{k}\}$ and a function *u* in *E* so that ${lim}_{j\to \mathrm{\infty}}{\parallel M{x}^{j}-u\parallel}_{E}=0$. As the convergence in the sense of ${\parallel \cdot \parallel}_{E}$ implies the pointwise convergence to the same limit function, we must have $u=Mx$, therefore (3.11) holds. Consequently, *M* is continuous. The proof is complete. □

*Proof of Theorem 2.1* We shall apply the Schauder fixed point theorem. Let Ω be the subset of the Banach space

*E* defined as in Lemma 3.1. Clearly, Ω is a nonempty convex and closed subset of

*E*. By Lemma 3.1, the mapping

$M:\mathrm{\Omega}\to E$ is continuous and

*M* Ω is relatively compact. We shall show that

*M* maps Ω into itself,

*i.e.*,

$M\mathrm{\Omega}\subseteq \mathrm{\Omega}$. Let us consider an arbitrary function

$x\in \mathrm{\Omega}$. Following the argument in the proof of Lemma 3.1, we see that

*x* satisfies (3.10), which together with (2.8) provides

$|{(Mx)}^{\prime}(t)|\le {\varphi}_{q}({\int}_{t}^{\mathrm{\infty}}F(s,{\eta}_{s},c)\phantom{\rule{0.2em}{0ex}}ds)\le {\varphi}_{q}({\varphi}_{p}(c))=c\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

(3.12)

Now, (3.12) and the fact that $(Mx)(t)=\xi (t)$ for $-r\le t\le 0$ imply that $Mx\in \mathrm{\Omega}$. We have thus proved that $M\mathrm{\Omega}\subseteq \mathrm{\Omega}$.

By the Schauder fixed point theorem, there exists an $x\in \mathrm{\Omega}$ such that $x=Mx$. Hence, *x* has the expression (3.1), which coincides with (2.1). It follows from Lemma 2.1 that *x* is a solution on $[0,\mathrm{\infty})$ of the boundary value problem (1.1)-(1.3). Also, since $x\in \mathrm{\Omega}$, clearly *x* satisfies (2.11). Moreover, since $x(0)=\xi (0)=0$, it follows from (2.11) that *x* also fulfills (2.10). This completes the proof of Theorem 2.1. □

*Proof of Theorem 2.2* We shall employ the Banach contraction principle. Let Ω be the subset of the Banach space *E* defined in Lemma 3.1. Clearly, Ω is a nonempty closed subset of *E*. Following the argument in the proof of Theorem 2.1, we have $M:\mathrm{\Omega}\to \mathrm{\Omega}$.

Now, we shall prove that the mapping

*M* is a contraction. For this purpose, let us consider two arbitrary functions

*x* and

$\tilde{x}$ in Ω. From (3.1), we have

$(Mx)(t)=(M\tilde{x})(t)=\xi (t)$ for

$-r\le t\le 0$, and consequently,

$\underset{-r\le t\le 0}{max}|(Mx)(t)-(M\tilde{x})(t)|=0.$

(3.13)

Furthermore, by using (2.12), from (3.1) we obtain for

$t\ge 0$ $\begin{array}{rl}|{(Mx)}^{\prime}(t)-{(M\tilde{x})}^{\prime}(t)|& =|{\varphi}_{q}\left({\int}_{t}^{\mathrm{\infty}}f(s,{x}_{s},{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\right)-{\varphi}_{q}\left({\int}_{t}^{\mathrm{\infty}}f(s,{\tilde{x}}_{s},{\tilde{x}}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\right)|\\ \le K(t)max\{\parallel {x}_{s}-{\tilde{x}}_{s}\parallel ,|{x}^{\prime}(s)-{\tilde{x}}^{\prime}(s)|\}.\end{array}$

This gives

$\underset{t\ge 0}{sup}|{(Mx)}^{\prime}(t)-{(M\tilde{x})}^{\prime}(t)|\le \underset{t\ge 0}{sup}K(t)max\{\parallel {x}_{t}-{\tilde{x}}_{t}\parallel ,|{x}^{\prime}(t)-{\tilde{x}}^{\prime}(t)|\}.$

By the definition of the norm

${\parallel \cdot \parallel}_{E}$, the last inequality and (3.13) imply

${\parallel Mx-M\tilde{x}\parallel}_{E}\le \underset{t\ge 0}{sup}K(t)max\{\parallel {x}_{t}-{\tilde{x}}_{t}\parallel ,|{x}^{\prime}(t)-{\tilde{x}}^{\prime}(t)|\}.$

(3.14)

Next, from the definition of Ω, we have

$x(t)=\tilde{x}(t)=\xi (t)$ for

$-r\le t\le 0$, and so

$|x(t)-\tilde{x}(t)|=0\phantom{\rule{1em}{0ex}}\text{for}-r\le t\le 0.$

(3.15)

Moreover, in view of the fact that

$x(0)=\tilde{x}(0)=\xi (0)=0$, we get, for

$t\ge 0$,

$|x(t)-\tilde{x}(t)|=\left|{\int}_{0}^{t}({x}^{\prime}(s)-{\tilde{x}}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}ds\right|\le {\int}_{0}^{t}|{x}^{\prime}(s)-{\tilde{x}}^{\prime}(s)|\phantom{\rule{0.2em}{0ex}}ds.$

(3.16)

But, by the definition of the norm

${\parallel \cdot \parallel}_{E}$, we have

$|{x}^{\prime}(t)-{\tilde{x}}^{\prime}(t)|\le {\parallel x-\tilde{x}\parallel}_{E}\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

(3.17)

Thus, using (3.17) in (3.16) yields

$|x(t)-\tilde{x}(t)|\le t{\parallel x-\tilde{x}\parallel}_{E}\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

(3.18)

Combining (3.15) and (3.18), we get

$|x(t)-\tilde{x}(t)|\le \mu (t){\parallel x-\tilde{x}\parallel}_{E}\phantom{\rule{1em}{0ex}}\text{for}t\ge -r,$

(3.19)

where the function

*μ* is defined by

$\mu (t)=\{\begin{array}{cc}0\hfill & \text{for}-r\le t\le 0,\hfill \\ t\hfill & \text{for}t\ge 0.\hfill \end{array}$

We can rewrite (3.19) as

$|x(t+\tau )-\tilde{x}(t+\tau )|\le \mu (t+\tau ){\parallel x-\tilde{x}\parallel}_{E}\phantom{\rule{1em}{0ex}}\text{for}t\ge 0\text{and}-r\le \tau \le 0,$

*i.e.*,

$|{x}_{t}(\tau )-{\tilde{x}}_{t}(\tau )|\le \mu (t+\tau ){\parallel x-\tilde{x}\parallel}_{E}\phantom{\rule{1em}{0ex}}\text{for}t\ge 0\text{and}-r\le \tau \le 0.$

It follows that

$\underset{-r\le \tau \le 0}{max}|{x}_{t}(\tau )-{\tilde{x}}_{t}(\tau )|\le [\underset{-r\le \tau \le 0}{max}\mu (t+\tau )]{\parallel x-\tilde{x}\parallel}_{E}\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

(3.20)

But, since

*μ* is nondecreasing on

$[-r,\mathrm{\infty})$, we have

$\underset{-r\le \tau \le 0}{max}\mu (t+\tau )=\mu (t)=t\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

So, from (3.20) we have

$\parallel {x}_{t}-{\tilde{x}}_{t}\parallel \le t{\parallel x-\tilde{x}\parallel}_{E}\phantom{\rule{1em}{0ex}}\text{for}t\ge 0.$

(3.21)

Now, using (3.17) and (3.21) in (3.14), we get

$\begin{array}{rl}{\parallel Mx-M\tilde{x}\parallel}_{E}& \le \underset{t\ge 0}{sup}K(t)max\{t{\parallel x-\tilde{x}\parallel}_{E},{\parallel x-\tilde{x}\parallel}_{E}\}\\ =\underset{t\ge 0}{sup}K(t)max\{t,1\}{\parallel x-\tilde{x}\parallel}_{E}\\ <{\parallel x-\tilde{x}\parallel}_{E},\end{array}$

where the last inequality is due to (2.13). Hence, we have shown that the mapping $M:\mathrm{\Omega}\to \mathrm{\Omega}$ is a contraction.

Finally, by the Banach contraction principle, the mapping $M:\mathrm{\Omega}\to \mathrm{\Omega}$ has a unique fixed point $x\in \mathrm{\Omega}$ having the expression (3.1), which coincides with (2.1). It follows from Lemma 2.1 that *x* is the unique solution on $[0,\mathrm{\infty})$ of the boundary value problem (1.1)-(1.3). Furthermore, as in the proof of Theorem 2.1, we conclude that this unique solution *x* of the boundary value problem (1.1)-(1.3) satisfies (2.10) and (2.11). The proof of Theorem 2.2 is now complete. □