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Existence of solutions for a class of nonlinear boundary value problems on half-line

Boundary Value Problems20122012:43

https://doi.org/10.1186/1687-2770-2012-43

• Accepted: 16 April 2012
• Published:

Abstract

Consider the infinite interval nonlinear boundary value problem

$\begin{array}{l}{\left(p\left(t\right){x}^{\prime }\right)}^{\prime }+q\left(t\right)x=f\left(t,x\right),\phantom{\rule{2.77695pt}{0ex}}t\ge {t}_{0}\ge 0,\phantom{\rule{2em}{0ex}}\\ x\left({t}_{0}\right)={x}_{0},\phantom{\rule{2em}{0ex}}\\ x\left(t\right)=a\phantom{\rule{0.3em}{0ex}}v\left(t\right)+b\phantom{\rule{0.3em}{0ex}}u\left(t\right)+o\left({r}_{i}\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\to \infty ,\phantom{\rule{2em}{0ex}}\end{array}$

where u and v are principal and nonprincipal solutions of (p(t)x')' + q(t)x = 0, r1(t) = o(u(t)(v(t)) μ ) and r2(t) = o(v(t)(u(t)) μ ) for some μ (0, 1), and a and b are arbitrary but fixed real numbers.

Sufficient conditions are given for the existence of a unique solution of the above problem for i = 1, 2. An example is given to illustrate one of the main results.

Mathematics Subject Classication 2011: 34D05.

Keywords

• Boundary value problem
• singular
• half-line
• principal
• nonprincipal

1. Introduction

Boundary value problems on half-line occur in various applications such as in the study of the unsteady flow of a gas through semi-infinite porous medium, in analyzing the heat transfer in radial flow between circular disks, in the study of plasma physics, in an analysis of the mass transfer on a rotating disk in a non-Newtonian fluid, etc. More examples and a collection of works on the existence of solutions of boundary value problems on half-line for differential, difference and integral equations may be found in the monographs [1, 2] For some works and various techniques dealing with such boundary value problems (we may refer to  and the references cited therein).

In this article by employing principal and nonprincipal solutions we introduce a new approach to study nonlinear boundary problems on half-line of the form
${\left(p\left(t\right){x}^{\prime }\right)}^{\prime }+q\left(t\right)x=f\left(t,x\right),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},$
(1.1)
$x\left({t}_{0}\right)={x}_{0},$
(1.2)
$x\left(t\right)=a\phantom{\rule{2.77695pt}{0ex}}v\left(t\right)+b\phantom{\rule{0.3em}{0ex}}u\left(t\right)+o\left(r\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\to \infty ,$
(1.3)
where a and b are any given real numbers, u and v are principal and nonprincipal solutions of
${\left(p\left(t\right){x}^{\prime }\right)}^{\prime }+q\left(t\right)x=0,\phantom{\rule{1em}{0ex}}t\ge 0$
(1.4)

and p C([0, ∞), (0, ∞)), q C([0, ∞), ) and f C([0, ∞) × , ).

We will show that the problem (1.1)-(1.3) has a unique solution in the case when
$r\left(t\right)=o\left(u\left(t\right){\left(v\left(t\right)\right)}^{\mu }\right)$
(1.5)
and
$r\left(t\right)=o\left(v\left(t\right){\left(u\left(t\right)\right)}^{\mu }\right),$
(1.6)

where μ (0, 1) is arbitrary but fixed real numbers.

The nonlinear boundary value problem (1.1)-(1.3) is also closely related to asymptotic integration of second order differential equations. Indeed, there are several important works in the literature, see , dealing with mostly the asymptotic integration of solutions of second order nonlinear equations of the form
${x}^{\prime \prime }=f\left(t,x\right).$
The authors are usually interested in finding conditions on the function f(t, x) which guarantee the existence of a solution asymptotic to linear function
$x\left(t\right)=a\phantom{\rule{0.3em}{0ex}}t+b,\phantom{\rule{1em}{0ex}}t\to \infty .$
(1.7)
We should point out that u(t) = 1 and v(t) = t are principal and nonprincipal solutions of the corresponding unperturbed equation
${x}^{\prime \prime }=0,$
and the function x(t) in (1.7) can be written as
$x=a\phantom{\rule{0.3em}{0ex}}v\left(t\right)+b\phantom{\rule{0.3em}{0ex}}u\left(t\right).$

Note that v(t) → ∞ as t → ∞ but u(t) is bounded in this special case. It turns out such information is crucial in investigating the general case. Our results will be applicable whether or not u(t) → ∞ (v(t) → ∞) as t → ∞.

2. Main results

It is well-known that [17, 18] if the second order linear Equation (1.4) has a positive solution or nonoscillatory at , then there exist two linearly independent solutions u(t) and v(t), called principal and nonprincipal solutions of the equation. The principal solution u is unique up to a constant multiple. Moreover, the following useful properties are satisfied:
$\underset{t\to \infty }{\text{lim}}\frac{u\left(t\right)}{v\left(t\right)}=0,\phantom{\rule{1em}{0ex}}\underset{{t}_{*}}{\overset{\infty }{\int }}\frac{1}{p\left(t\right){u}^{2}\left(t\right)}dt=\infty ,\phantom{\rule{1em}{0ex}}\underset{{t}_{*}}{\overset{\infty }{\int }}\frac{1}{p\left(t\right){v}^{2}\left(t\right)}dt<\infty ,$

where t* ≥ 0 is a sufficiently large real number.

Let v be a principal solution of (1.4). Without loss of generality we may assume that v (t) > 0 if tt1 for some t1 ≥ 0. It is easy to see that
$v\left(t\right)=u\left(t\right)\phantom{\rule{2.77695pt}{0ex}}\underset{{t}_{1}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}ds$
(2.1)

is a nonprincipal solution of (1.4), which is strictly positive for t > t1.

Theorem 2.1. Let t0 > t1. Assume that the function f satisfies
$\left|f\left(t,x\right)\right|\le {h}_{1}\left(t\right)g\left(\left|x\right|\right)+{h}_{2}\left(t\right),\phantom{\rule{1em}{0ex}}t\ge {t}_{0}$
(2.2)
and
$\left|f\left(t,{x}_{1}\right)-f\left(t,{x}_{2}\right)\right|\le \frac{k\left(t\right)}{v\left(t\right)}\left|{x}_{1}-{x}_{2}\right|,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\ge {t}_{0},$
(2.3)
where g C([0, ∞), [0, ∞)) is bounded; h1, h2, k C([t0, ∞), [0, ∞)). Suppose further that
$\underset{{t}_{0}}{\overset{\infty }{\int }}u\left(s\right)k\left(s\right)ds\le \mu$
(2.4)
and
$\frac{1}{p\left(t\right){u}^{2}\left(t\right)}\underset{t}{\overset{\infty }{\int }}u\left(s\right){h}_{i}\left(s\right)ds\le \beta \left(t\right),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},\phantom{\rule{1em}{0ex}}i=1,2$
(2.5)
for some β C([t0, ∞), [0, ∞)) such that
$\underset{{t}_{0}}{\overset{t}{\int }}\beta \left(s\right)ds=o\left({\left(v\left(t\right)\right)}^{\mu }\right),\phantom{\rule{1em}{0ex}}t\to \infty .$
(2.6)
If either
$v\left(t\right)\to \infty ,\phantom{\rule{1em}{0ex}}t\to \infty$
(2.7)
or else
$b=\frac{{x}_{0}}{u\left({t}_{0}\right)}-a\underset{{t}_{1}}{\overset{{t}_{0}}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}ds,$
(2.8)

then there is a unique solution x(t) of (1.1)-(1.3), where r is given by (1.5).

Proof. Denote by M the supremum of the function g over [0, ∞). Let X be a space of functions defined by
$X=\left\{x\phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}C\left(\left[{t}_{0},\infty \right),ℝ\right)|\phantom{\rule{1em}{0ex}}\left|x\left(t\right)\right|\le {l}_{1}v\left(t\right)+{l}_{2}u\left(t\right),\phantom{\rule{1em}{0ex}}\forall t\ge {t}_{0}\right\},$
where
${l}_{1}=\left(M+1\right)p\left({t}_{0}\right){u}^{2}\left({t}_{0}\right)\beta \left({t}_{0}\right)+\left|a\right|$
and
${l}_{2}=\frac{\left|{x}_{0}\right|}{u\left({t}_{0}\right)}+\left|a\right|\underset{{t}_{1}}{\overset{{t}_{0}}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}ds.$
Note that X is a complete metric space with the metric d defined by
$d\left({x}_{1},{x}_{2}\right)=\underset{t\ge {t}_{0}}{\text{sup}}\frac{1}{v\left(t\right)}\left|{x}_{1}\left(t\right)-{x}_{2}\left(t\right)\right|,\phantom{\rule{1em}{0ex}}{x}_{1},{x}_{2}\phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}X.$
Define an operator F on X by
$\begin{array}{ll}\hfill \left(Fx\right)\left(t\right)& =-u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}\underset{s}{\overset{\infty }{\int }}u\left(\tau \right)f\left(\tau ,x\left(\tau \right)\right)d\tau ds+av\left(t\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left[\frac{{x}_{0}}{u\left({t}_{0}\right)}-a\underset{{t}_{1}}{\overset{{t}_{0}}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}ds\right]u\left(t\right).\phantom{\rule{2em}{0ex}}\end{array}$
In view of conditions (2.2) and (2.5) we see that F is well defined. Next we show that F X X. Indeed, let x X, then
$\begin{array}{ll}\hfill \left|\left(Fx\right)\left(t\right)\right|& \le u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}\underset{s}{\overset{\infty }{\int }}u\left(\tau \right)\left|f\left(\tau ,x\left(\tau \right)\right)\right|d\tau ds+\left|a\right|v\left(t\right)+{l}_{2}u\left(t\right)\phantom{\rule{2em}{0ex}}\\ \le u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}\underset{{t}_{0}}{\overset{\infty }{\int }}u\left(\tau \right)\left|f\left(\tau ,x\left(\tau \right)\right)\right|d\tau ds+\left|a\right|\upsilon \left(t\right)+{l}_{2}u\left(t\right)\phantom{\rule{2em}{0ex}}\\ \le u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}\underset{{t}_{0}}{\overset{\infty }{\int }}u\left(\tau \right)\left({h}_{1}\left(\tau \right)g\left(\left|x\left(\tau \right)\right|\right)+{h}_{2}\left(\tau \right)\right)d\tau ds+\left|a\right|v\left(t\right)+{l}_{2}u\left(t\right)\phantom{\rule{2em}{0ex}}\\ \le u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}\underset{{t}_{0}}{\overset{\infty }{\int }}u\left(\tau \right)\left(M{h}_{1}\left(\tau \right)+{h}_{2}\left(\tau \right)\right)d\tau ds+\left|a\right|\upsilon \left(t\right)+{l}_{2}u\left(t\right)\phantom{\rule{2em}{0ex}}\\ \le \left(M+1\right)p\left({t}_{0}\right){u}^{2}\left({t}_{0}\right)\beta \left({t}_{0}\right)u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}ds+\left|a\right|v\left(t\right)+{l}_{2}u\left(t\right)\phantom{\rule{2em}{0ex}}\\ \le {l}_{1}v\left(t\right)+{l}_{2}u\left(t\right),\phantom{\rule{2em}{0ex}}\end{array}$

which means that F x X.

Using (2.1), (2.3) and (2.4) we also see that
$\begin{array}{ll}\hfill \left|\left(F{x}_{1}\right)\left(t\right)-\left(F{x}_{2}\right)\left(t\right)\right|& \le u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}\underset{s}{\overset{\infty }{\int }}u\left(\tau \right)\left|f\left(\tau ,{x}_{1}\left(\tau \right)\right)-f\left(\tau ,{x}_{2}\left(\tau \right)\right)\right|d\tau ds\phantom{\rule{2em}{0ex}}\\ \le u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}\underset{s}{\overset{\infty }{\int }}u\left(\tau \right)\frac{k\left(\tau \right)}{v\left(\tau \right)}\left|{x}_{1}\left(\tau \right)-{x}_{2}\left(\tau \right)\right|d\tau ds\phantom{\rule{2em}{0ex}}\\ \le d\left({x}_{1},{x}_{2}\right)u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}\underset{s}{\overset{\infty }{\int }}u\left(\tau \right)k\left(\tau \right)d\tau ds\phantom{\rule{2em}{0ex}}\\ \le d\left({x}_{1},{x}_{2}\right)u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}\underset{{t}_{0}}{\overset{\infty }{\int }}u\left(\tau \right)k\left(\tau \right)d\tau ds\phantom{\rule{2em}{0ex}}\\ \le d\left({x}_{1},{x}_{2}\right)v\left(t\right)\underset{{t}_{0}}{\overset{\infty }{\int }}u\left(\tau \right)k\left(\tau \right)d\tau \phantom{\rule{2em}{0ex}}\\ \le \mu d\left({x}_{1},{x}_{2}\right)v\left(t\right),\phantom{\rule{2em}{0ex}}\end{array}$

where x1, x2 X arbitrary. This implies that F is a contracting mapping.

Thus according to Banach contraction principle F has a unique fixed point x. It is not difficult to see that the fixed point solves (1.1) and (1.2). It remains to show that x(t) satisfies (1.3) as well. It is not difficult to show that
$\begin{array}{ll}\hfill \left|x\left(t\right)-av\left(t\right)-bu\left(t\right)\right|& \le u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}\underset{s}{\overset{\infty }{\int }}u\left(\tau \right)\left|f\left(\tau ,x\left(\tau \right)\right)\right|d\tau ds+\left|c\right|u\left(t\right)\phantom{\rule{2em}{0ex}}\\ \le u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}\underset{s}{\overset{\infty }{\int }}u\left(\tau \right)\left(M{h}_{1}\left(\tau \right)+{h}_{2}\left(\tau \right)\right)d\tau ds+\left|c\right|u\left(t\right)\phantom{\rule{2em}{0ex}}\\ \le \left(M+1\right)u\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\beta \left(s\right)ds+\left|c\right|u\left(t\right),\phantom{\rule{2em}{0ex}}\end{array}$
where
$c=\frac{{x}_{0}}{u\left({t}_{0}\right)}-a\underset{{t}_{1}}{\overset{{t}_{0}}{\int }}\frac{1}{p\left(s\right){u}^{2}\left(s\right)}ds-b.$

If (2.7) is satisfied, then in view (2.6) and the above inequality we easily obtain (1.3). In case (2.8) holds, then c = 0 and hence we still have (1.3).

From Theorem 2.1 we deduce the following Corollary.

Corollary 2.2. Assume that the function f satisfies (2.2) and
$\left|f\left(t,{x}_{1}\right)-f\left(t,{x}_{2}\right)\right|\le \frac{k\left(t\right)}{t}\left|{x}_{1}-{x}_{2}\right|,\phantom{\rule{1em}{0ex}}t\ge {t}_{0},$
where k C([t0, ∞), [0, ∞)). Suppose further that
$\underset{{t}_{0}}{\overset{\infty }{\int }}k\left(s\right)ds\le \mu ;\phantom{\rule{1em}{0ex}}\underset{t}{\overset{\infty }{\int }}{h}_{i}\left(s\right)ds\le \beta \left(t\right),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},\phantom{\rule{1em}{0ex}}i=1,2$
for some μ (0, 1) and β C([t0, ∞), [0, ∞)), where
$\underset{{t}_{0}}{\overset{t}{\int }}\beta \left(s\right)ds=o\left({t}^{\mu }\right),\phantom{\rule{1em}{0ex}}t\to \infty .$
Then for each a, b the boundary value problem
$\begin{array}{l}{x}^{\prime \prime }=f\left(t,x\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\ge {t}_{0},\phantom{\rule{2em}{0ex}}\\ x\left({t}_{0}\right)={x}_{0},\phantom{\rule{2em}{0ex}}\\ x\left(t\right)=at+b+o\left({t}^{\mu }\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\to \infty \phantom{\rule{2em}{0ex}}\end{array}$

has a unique solution.

Let υ be a nonprincipal solution of (1.4). Without loss of generality we may assume that v(t) > 0, if tt2 for some t2 ≥ 0. It is easy to see that [17, 18]
$u\left(t\right)=v\left(t\right)\underset{t}{\overset{\infty }{\int }}\frac{1}{p\left(s\right){v}^{2}\left(s\right)}ds$
(2.9)
is a principal solution of (1.4) which is strictly positive. Take t2 large enough so that
$\underset{t}{\overset{\infty }{\int }}\frac{1}{p\left(s\right){v}^{2}\left(s\right)}ds\le 1.$

Then from (2.9), we have v (t) ≥ u(t) for tt2, which is needed in the proof of the next theorem.

Theorem 2.3. Let t0t2. Assume that the function f satisfies (2.2) and (2.3). Suppose further that
$\underset{{t}_{0}}{\overset{\infty }{\int }}v\left(s\right)k\left(s\right)ds\le \mu$
(2.10)
and
$\frac{1}{p\left(t\right){v}^{2}\left(t\right)}\underset{t}{\overset{\infty }{\int }}v\left(s\right){h}_{i}\left(s\right)ds\le \beta \left(t\right),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},\phantom{\rule{1em}{0ex}}i=1,2$
(2.11)
for some β C([t0, ∞), [0, ∞)) such that
$\underset{{t}_{0}}{\overset{t}{\int }}\beta \left(s\right)ds=o\left({\left(u\left(t\right)\right)}^{\mu }\right),\phantom{\rule{1em}{0ex}}t\to \infty .$
(2.12)
If either
$u\left(t\right)\to \infty ,\phantom{\rule{1em}{0ex}}t\to \infty$
(2.13)
or else
$a=\frac{{x}_{0}}{v\left({t}_{0}\right)}-b\underset{{t}_{0}}{\overset{\infty }{\int }}\frac{1}{p\left(s\right){v}^{2}\left(s\right)}ds,$
(2.14)

then there is a unique solution x(t) of (1.1) - (1.3), where r is given by (1.6).

Proof. Let X be a space of functions defined by
$X=\left\{x\phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}C\left(\left[{t}_{0},\infty \right),ℝ\right)|\phantom{\rule{1em}{0ex}}\left|x\left(t\right)\right|\le {l}_{1}v\left(t\right)+{l}_{2}u\left(t\right),\phantom{\rule{1em}{0ex}}\forall t\ge {t}_{0}\right\},$
where
${l}_{1}=\left(M+1\right)p\left({t}_{0}\right)u\left({t}_{0}\right)v\left({t}_{0}\right)\beta \left({t}_{0}\right)+\frac{\left|{x}_{0}\right|}{v\left({t}_{0}\right)}+\left|b\right|\underset{{t}_{0}}{\overset{\infty }{\int }}\frac{1}{p\left(s\right){v}^{2}\left(s\right)}ds\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{and}}\phantom{\rule{2.77695pt}{0ex}}{l}_{2}=\left|b\right|.$

Again, X is a complete metric space with the metric d defined in the proof of the previous theorem.

We define an operator F on X by
$\begin{array}{ll}\hfill \left(Fx\right)\left(t\right)& =-v\left(t\right)\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{p\left(s\right){v}^{2}\left(s\right)}\underset{s}{\overset{\infty }{\int }}v\left(\tau \right)f\left(\tau ,x\left(\tau \right)\right)d\tau ds\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left[\frac{{x}_{0}}{v\left({t}_{0}\right)}-b\underset{{t}_{0}}{\overset{\infty }{\int }}\frac{1}{p\left(s\right){v}^{2}\left(s\right)}ds\right]v\left(t\right)+bu\left(t\right).\phantom{\rule{2em}{0ex}}\end{array}$

The remainder of the proof proceeds similarly as in that of Theorem 2.1 by using (2.2), (2.3), (2.9)-(2.14).

Corollary 2.4. Assume that the function f satisfies (2.2) and (2.3). Suppose further that
$\underset{{t}_{0}}{\overset{\infty }{\int }}sk\left(s\right)ds\le \mu ;\phantom{\rule{1em}{0ex}}\frac{1}{{t}^{2}}\underset{t}{\overset{\infty }{\int }}s{h}_{i}\left(s\right)ds\le \beta \left(t\right),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},\phantom{\rule{1em}{0ex}}i=1,2$
for some μ (0, 1) and β C([t0, ∞), [0, ∞)), where
$\underset{1}{\overset{t}{\int }}\beta \left(s\right)ds=o\left(1\right),\phantom{\rule{1em}{0ex}}t\to \infty .$
If for any given a, b the condition (2.14) holds then the boundary value problem
$\begin{array}{l}{x}^{\prime \prime }=f\left(t,x\right),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},\phantom{\rule{2em}{0ex}}\\ x\left({t}_{0}\right)={x}_{0},\phantom{\rule{2em}{0ex}}\\ x\left(t\right)=at+b+o\left(t\right),\phantom{\rule{1em}{0ex}}t\to \infty \phantom{\rule{2em}{0ex}}\end{array}$

has a unique solution.

3. An example

Consider the boundary value problem
${\left(t{x}^{\prime }\right)}^{\prime }=\frac{1}{{t}^{2}}\text{arctan}\phantom{\rule{2.77695pt}{0ex}}x+{t}^{\nu },\phantom{\rule{1em}{0ex}}t\ge {t}_{0},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\nu <-2,$
(3.1)
$x\left({t}_{0}\right)={x}_{0},$
(3.2)
$x\left(t\right)=a\phantom{\rule{2.77695pt}{0ex}}\text{ln}t+b+o\left({\left(\text{ln}t\right)}^{\mu }\right),\phantom{\rule{1em}{0ex}}t\to \infty .$
(3.3)
where t0 > t1 = 1 and μ (0, 1) are chosen to satisfy
$\frac{1+\text{ln}{t}_{0}}{{t}_{0}}\le \mu .$
(3.4)
Note that since
$\underset{{t}_{0}\to \infty }{\text{lim}}\frac{1+\text{ln}{t}_{0}}{{t}_{0}}=0$

for any given μ (0, 1) there is a t0 such that (3.4) holds.

Comparing with the boundary value problem (1.1)-(1.3) we see that p(t) = t, q(t) = 0, and f(t, x) = (1/t2) arctan x + t υ . The corresponding linear equation becomes
${\left(t{x}^{\prime }\right)}^{\prime }=0,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\ge {t}_{0}.$
Clearly, we may take
$u\left(t\right)=1\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{and}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}v\left(t\right)=\text{ln}\phantom{\rule{2.77695pt}{0ex}}t.$
Let
${h}_{1}\left(t\right)=\frac{1}{{t}^{2}},\phantom{\rule{1em}{0ex}}{h}_{2}\left(t\right)={t}^{\nu },\phantom{\rule{1em}{0ex}}g\left(x\right)=\text{arctan}\phantom{\rule{2.77695pt}{0ex}}x,\phantom{\rule{1em}{0ex}}k\left(t\right)=\frac{\text{ln}\phantom{\rule{2.77695pt}{0ex}}t}{{t}^{2}},\phantom{\rule{1em}{0ex}}\beta \left(t\right)=\frac{1}{{t}^{2}},$
then it is easy to see that
$\begin{array}{l}\left|f\left(t,x\right)\right|\le \frac{1}{{t}^{2}}\text{arctan}\left|x\right|+{t}^{\nu }={h}_{1}\left(t\right)g\left(\left|x\right|\right)+{h}_{2}\left(t\right),\phantom{\rule{2em}{0ex}}\\ \left|f\left(t,{x}_{1}\right)-f\left(t,{x}_{2}\right)\right|\le \frac{1}{{t}^{2}}\left|{x}_{1}-{x}_{2}\right|=\frac{k\left(t\right)}{v\left(t\right)}\left|{x}_{1}-{x}_{2}\right|,\phantom{\rule{2em}{0ex}}\\ \underset{{t}_{0}}{\overset{\infty }{\int }}k\left(s\right)ds=\underset{{t}_{0}}{\overset{\infty }{\int }}\frac{\text{ln}\phantom{\rule{2.77695pt}{0ex}}s}{{s}^{2}}ds=\frac{1+\text{ln}\phantom{\rule{2.77695pt}{0ex}}{t}_{0}}{{t}_{0}}\le \mu \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{by}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{(3}}\mathsf{\text{.4)}}\phantom{\rule{2em}{0ex}}\\ \frac{1}{t}\underset{t}{\overset{\infty }{\int }}{h}_{1}\left(s\right)ds\le \frac{1}{t}\underset{t}{\overset{\infty }{\int }}\frac{1}{{s}^{2}}ds=\frac{1}{{t}^{2}}=\beta \left(t\right),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},\phantom{\rule{2em}{0ex}}\\ \frac{1}{t}\underset{t}{\overset{\infty }{\int }}{h}_{2}\left(s\right)ds=-\frac{{t}^{\nu }}{\nu +1}\le \beta \left(t\right),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},\phantom{\rule{2em}{0ex}}\\ \underset{{t}_{0}}{\overset{t}{\int }}\beta \left(s\right)ds=\underset{{t}_{0}}{\overset{t}{\int }}\frac{1}{{s}^{2}}ds=\frac{1}{{t}_{0}}-\frac{1}{t}=o\left({\left(\text{ln}t\right)}^{\mu }\right),\phantom{\rule{1em}{0ex}}t\to \infty ,\phantom{\rule{1em}{0ex}}\mu \in \left(0,1\right),\phantom{\rule{2em}{0ex}}\end{array}$
and
$v\left(t\right)=\text{ln}t\to \infty ,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\to \infty ,$

i.e., all the conditions of Theorem 2.1 are satisfied. Therefore we may conclude that if (3.4) holds, then the boundary value problem (3.1)-(3.3) has a unique solution.

Furthermore, we may also deduce that there exist solutions x1(t) and x2(t) such that
${x}_{1}\left(t\right)=1+o\left({\left(\text{ln}\phantom{\rule{2.77695pt}{0ex}}t\right)}^{\mu }\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\to \infty$
and
${x}_{2}\left(t\right)=\text{ln}\phantom{\rule{2.77695pt}{0ex}}t+o\left({\left(\text{ln}\phantom{\rule{2.77695pt}{0ex}}t\right)}^{\mu }\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}t\to \infty .$

by taking (a, b) = (0, 1) and (a, b) = (1, 0), respectively.

Declarations

Acknowledgements

This work was supported by the Scientific and Technological Research Council of Turkey (TUBITAK) under project number 108T688.

Authors’ Affiliations

(1)
Department of Mathematics, Middle East Technical University, 06800 Ankara, Turkey

References 