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Interpolating asymptotic integration methods for second-order differential equations S. A. Stepin Lomonosov Moscow State University, Faculty of Mechanics and Mathematics
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2024, Volume 88, Issue 1, DOI: https://doi.org/10.4213/im9438 
Bibliographic databases:
   § 1. Introduction and statement of the resultsA wide class of second-order differential equations can be reduced (cf. [1]) to with $\lambda=1$ after passing to the new independent variable and substituting $u(t)=(Q(x(t)))^{1/4}y(x(t))$. Here, $-2\varphi(t)$ is the Schwarzian derivative $S[x(t)]$, so that
$$
\begin{equation*}
\varphi(t(x))=-\frac14\, \frac{Q''(x)}{Q(x)^2}+\frac5{16}\, \frac{Q'(x)^2}{Q(x)^3}.
\end{equation*}
\notag
$$
The above two substitutions are known as the Liouville transform. If $Q(x)$ behaves quite regularly at infinity and if the integral $\int^{\infty}\!\!\sqrt{Q(s)}\,ds$ diverges, the function $\varphi$ turns out to be small in a certain sense for large $t$. Sometimes several transformations of such a type are required to achieve the desired smallness of $\varphi$. From the perturbation theory point of view it is natural to expect that solutions to the transformed equation (2) and to the associated comparison equation corresponding to the case $\varphi(t)\equiv 0$ are asymptotically close at infinity, and, therefore, the original equation (1) has a fundamental system of solutions
$$
\begin{equation*}
y_{1,2}(x)\sim Q(x)^{-1/4} \exp\biggl(\mp\int_{x_0}^x\sqrt{Q(s)}\,ds\biggr),\qquad x\to\infty.
\end{equation*}
\notag
$$
The above asymptotic formulas are known as Liouville–Green approximations (or WKB-approximations; see [2]). Note that under the condition
$$
\begin{equation*}
\int^{\infty}\biggl(\frac{|Q''(x)|}{Q(x)^{3/2}} +\frac{Q'(x)^2}{Q(x)^{5/2}}\biggr)\,dx<\infty
\end{equation*}
\notag
$$
the integral $\int^{\infty}\!\!\sqrt{Q(s)}\,ds$ diverges (see [3]) and the function $\varphi(t)$ is integrable on the half-axis $\mathbb R_+$. Below, the function $\varphi(t)$ in (2) is assumed to be continuous. In order to study the asymptotic behaviour of solutions to equation (2) in the case $\varphi\in\mathrm{L}_1(\mathbb R_+)$, we reduce the corresponding first-order system
$$
\begin{equation*}
\begin{pmatrix}u\\u'\end{pmatrix}'=\begin{pmatrix}0&1 \\ \lambda^2+\varphi(t)&0\end{pmatrix} \begin{pmatrix}u\\u'\end{pmatrix}
\end{equation*}
\notag
$$
to the $L$-diagonal form
$$
\begin{equation}
Y'=\biggl\{\begin{pmatrix} \lambda &0 \\ 0 &-\lambda \end{pmatrix}+\frac{\varphi(t)}{2\lambda}\begin{pmatrix} 1 &1 \\ -1 &-1 \end{pmatrix}\biggr\}Y
\end{equation}
\tag{3}
$$
via the substitution $\left(\begin{smallmatrix}u\\u'\end{smallmatrix}\right) =\left(\begin{smallmatrix}1&1\\ \lambda&-\lambda\end{smallmatrix}\right)Y$, where $Y=\left(\begin{smallmatrix}\xi\\\eta\end{smallmatrix}\right)$. Now an application of the Levinson’s fundamental theorem (see [4]) produces the following result. Proposition 1. If $\varphi\in\mathrm{L}_1(\mathbb R_+)$, then there exists a pair of linearly independent solutions to equation (2) such that In the case $\varphi\in\mathrm{L}_2(\mathbb R_+)$, one can apply to equation (2) another method of asymptotic integration based on the substitution $u(t)=\exp\bigl(\mp\lambda t+\int_0^tw(s)\,ds\bigr)$ and transformation of (2) to the Riccati equation after which this equation is reduced to an integral equation with quadratic non-linear term (see [4]). The following result is obtained within this approach. Proposition 2. Let $\varphi\in\mathrm{L}_2(\mathbb R_+)$ and let $\varphi(t)\to 0$ as $t\to\infty$. Then equation (2) has a fundamental system of solutions The similarity between the asymptotic formulas for the solutions $u_{1,2}(t)$ with sharp estimates of the remainder terms, as given by Propositions 1 and 2, becomes even more vivid if the formula for $u_1(t)$ in Proposition 1 is written as
$$
\begin{equation*}
u_1(t)=\exp(-\lambda t)\biggl\{1+O\biggl(\int_t^{\infty}|\varphi(s)|\,ds +\int_t^{\infty}|e^{2\lambda(t-s)}\varphi(s)|\,ds\biggr)\biggr\}.
\end{equation*}
\notag
$$
The natural question here is whether the above formulas can be derived within a unified approach capable of interpolating the specified summability conditions for the function $\varphi$. It turns out that the desired interpolation in the classes $\mathrm{L}_p(\mathbb R_+)$, $p\in[1,2]$, can be achieved via an approach based on the known principle (see [5]) involving the topological notions of a retract and retraction (see § 2). Theorem 1. If $\varphi\in\mathrm{L}_p(\mathbb R_+)$, $p\in[1,2]$, then equation (2) has solutions The estimates for the remainder terms $\varepsilon_{1,2}(t)$ in formulas (4) coincide with those in Propositions 1 and 2 for $p=1$ and $p=2$, respectively, and can be actually considered as their deformations with respect to the parameter $p\in[1,2]$. The example $\varphi(t)=\sin t/\sqrt{t}$ shows that the condition of Theorem 1, which provides the asymptotic formulas
$$
\begin{equation*}
u_{1,2}(t)\sim\exp\biggl(\mp\lambda t \mp\frac1{2\lambda}\int_0^t\varphi(s)\,ds\biggr),\qquad t\to\infty,
\end{equation*}
\notag
$$
for the solutions to equation (2), is sharp in the scale of space of integrable functions $\mathrm{L}_p(\mathbb R_+)$. Indeed, the corresponding equation (2) for $\lambda=1$ turns out to have (see § 5) solutions behaving asymptotically as $u_{1,2}(t)\sim t^{\pm1/20}e^{\mp t}$. From the point of view of comparative analysis of various methods of asymptotic integration it is worth comparing the above assumptions on the function $\varphi(t)$ that provide the asymptotic equivalence as $t\to\infty$ of the solutions to equation (2) and to the corresponding model comparison equation with Hartman and Wintner type conditions (whose approach was developed in [6]–[9]). In [10], a number of results of this type (heterogeneous, at first glance) were unified by a method based on Leninson’s fundamental theorem. To be exact, we formulate here two such results (see [7], [8]), which we combine into a single proposition. For the sake of simplicity, we restrict to the case $\lambda>0$, which is non-elliptic in the language of [5]. Proposition 3. Let the integral $\int^{\infty}\varphi(t)\,dt$ converge (no matter absolutely or not) and let either of the conditions The complementary convergence conditions for $\Phi_0(t)$ and $\Phi_1(t)$ are actually obtained one from another by deformation and are included into a one-parameter family of conditions, which guarantee the equivalence of solutions to equation (2) and to the comparison equation as $t\to\infty$. The desired interpolation between $\Phi_0(t)$ and $\Phi_1(t)$ is secured by the method based on the asymptotic factorization of the fundamental matrix for the corresponding first-order system (see § 6). Theorem 2. Assuming that the integral $\int^{\infty}\varphi(t)\,dt$ converges, let the condition The present paper is organized as follows. Lyapunov type functions are defined in § 2, where we also give an appropriate version of the retraction principle which applies directly to system (3). In § 3, we derive some auxiliary integral estimates, which will be used below. Further, in § 4 and § 5, we obtain a priori estimates for the solutions to equation (2). These estimates are then used for asymptotic integration of this equation (see the proof of Theorem 1). Corresponding counterexamples are also given. The sketch of the reduction method based on a transformation of a first-order system to the $L$-diagonal form is outlined in § 6. Finally, this method is applied in § 7 to the asymptotic integration problem under consideration (see the proof of Theorem 2). As a result, we obtain new conditions for asymptotic equivalence for solutions to equation (2) and to the corresponding comparison equation. Examples show that the classes of functions $\varphi(t)$ satisfying the hypotheses of Theorems 1 and 2 are in general position. The corresponding results can be also given for equation (1) in the original representation (before application of the Liouville transform). Under proper conditions, these results provide effective estimates for the accuracy of Liouville–Green approximation for the corresponding solutions. Some of the results of the present paper were announced in [11]. § 2. Lyapunov type functions and their propertiesWe set
$$
\begin{equation*}
\sigma(t)=\int_0^te^{2\alpha(s-t)}|\varphi(s)|\,ds,\qquad \tau(t)= \int_t^{\infty}e^{2\alpha(t-s)}|\varphi(s)|\,ds,
\end{equation*}
\notag
$$
where $\alpha=\operatorname{Re}\lambda>0$, and, for $a\geqslant 2/|\lambda|$, define the Lyapunov type functions by
$$
\begin{equation*}
v^{(+)}(t,Y)=|\eta|^2-a^2\sigma(t)^2|\xi|^2,\qquad v^{(-)}(t,Y)=|\xi|^2-a^2\tau(t)^2|\eta|^2.
\end{equation*}
\notag
$$
Statement 1. Let $Y(t)=\left(\begin{smallmatrix}\xi(t)\\\eta(t)\end{smallmatrix}\right)$ be a solution to system (3). Then on the null-level of the function $v^{(+)}(t,Y)$ for $t\geqslant T$ such that $\sigma(t)<1/a$. Proof. Let us evaluate the derivative
$$
\begin{equation*}
\frac12\,\frac{d}{dt}\, v^{(+)}(t,Y(t)) =\operatorname{Re}(\eta'\overline{\eta}) -a^2\sigma(t)\sigma'(t)|\xi|^2-a^2\sigma(t)^2\operatorname{Re}(\xi'\overline{\xi}),
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\xi'=\lambda\xi+\frac{\varphi(t)}{2\lambda}(\xi+\eta),\qquad \eta'=-\lambda\eta-\frac{\varphi(t)}{2\lambda}(\xi+\eta),
\end{equation*}
\notag
$$
and, in addition, $\sigma'=|\varphi(t)|-2\alpha\sigma$. As a result, on the null-level of the function $v^{(+)}(t,Y)$, that is, for $|\eta|=a\sigma(t)|\xi|$, we have
$$
\begin{equation*}
\begin{aligned} \, &\frac12\,\frac{d}{dt}\, v^{(+)}(t,Y(t)) =\operatorname{Re}\biggl(-\lambda|\eta|^2 -\frac{\varphi(t)}{2\lambda}(\xi+\eta)\overline{\eta}\biggr) \\ &\quad\qquad- a^2\sigma(t)\bigl(|\varphi(t)|-2\alpha\sigma(t)\bigr)|\xi|^2 -a^2\sigma(t)^2\operatorname{Re}\biggl(\lambda|\xi|^2 +\frac{\varphi(t)}{2\lambda}(\xi+\eta)\overline{\xi}\biggr) \\ &\quad= -a^2\sigma(t)|\varphi(t)||\xi|^2 -\operatorname{Re}\biggl(\frac{\varphi(t)}{2\lambda}(\xi+\eta)\overline{\eta}\biggr) -a^2\sigma(t)^2\operatorname{Re} \biggl(\frac{\varphi(t)}{2\lambda}(\xi+\eta)\overline{\xi}\biggr) \\ &\quad\leqslant a^2\sigma(t)|\xi|^2\biggl(\frac1{2a|\lambda|} \bigl(1+a\sigma(t)\bigr)^2-1\biggr)|\varphi(t)|\leqslant 0, \end{aligned}
\end{equation*}
\notag
$$
where $a\geqslant 2/|\lambda|$ and $\sigma(t)<1/a$ for $t\geqslant T$. This proves Statement 1. Statement 2. If $Y(t)=\left(\begin{smallmatrix}\xi(t)\\\eta(t)\end{smallmatrix}\right)$ is a solution to system (3), then on the null-level of the function $v^{(-)}(t,Y)$ one has $dv^{(-)}(t,Y(t))/dt\geqslant 0$ for $t\geqslant T$ such that $\tau(t)< 1/a$. Proof. Proceeding similarly and using the relation $\tau'=2\alpha\tau-|\varphi(t)|$, we estimate the derivative
$$
\begin{equation*}
\frac12\frac{d}{dt}\, v^{(-)}(t,Y(t))=\operatorname{Re} (\xi'\overline{\xi}) -a^2\tau(t)\tau'(t)|\eta|^2 -a^2\tau(t)^2\operatorname{Re}(\eta'\overline{\eta})
\end{equation*}
\notag
$$
on the null-level of the function $v^{(-)}(t,Y)$. As a result, under the condition $|\xi|=a\tau(t)|\eta|$, we have
$$
\begin{equation*}
\begin{aligned} \, &\frac12\,\frac{d}{dt}\, v^{(-)}(t,Y(t)) =\operatorname{Re} \biggl(\lambda|\xi|^2 +\frac{\varphi(t)}{2\lambda}(\xi+\eta)\overline{\xi}\biggr) \\ &\quad\qquad-a^2\tau(t)\bigl(2\alpha\tau(t)-|\varphi(t)|\bigr)|\eta|^2 + a^2\tau(t)^2\operatorname{Re} \biggl(\lambda|\eta|^2 +\frac{\varphi(t)}{2\lambda}(\xi+\eta)\overline{\eta}\biggr) \\ &\quad= a^2\tau(t)|\varphi(t)||\eta|^2 +\operatorname{Re} \biggl(\frac{\varphi(t)}{2\lambda}(\xi+\eta)\overline{\xi}\biggr) + a^2\tau(t)^2\operatorname{Re} \biggl(\frac{\varphi(t)}{2\lambda}(\xi+\eta)\overline{\eta}\biggr) \\ &\quad\geqslant a^2\tau(t)|\eta|^2 \biggl(1-\frac1{2a|\lambda|}\bigl(1+a\tau(t)\bigr)^2\biggr)|\varphi(t)|\geqslant 0, \end{aligned}
\end{equation*}
\notag
$$
where $a\geqslant 2/|\lambda|$ and $\tau(t)<1/a$ for $t\geqslant T$. This proves Statement 2. Below, in application to system (3), we give an appropriate version of a topological principle required for our purposes. This principle involves the notions of a retraction, that is, a continuous projection (an idempotent) in a topological space, and a retract — the range of such projection. A continuous vector field associated with system (3) defines the flow and the corresponding fibering $\Psi\colon X\mapsto\Psi(X)$ of the extended phase space, where $X=(t,Y)$, and the fibre $\Psi(X)$ is the integral trajectory passing through $X$. Consider the subsets of the extended phase space
$$
\begin{equation*}
\Omega=\{(t,Y)\colon v^{(-)}(t,Y)\leqslant 0,\, t\geqslant T\},\qquad \Gamma=\{(t,Y)\colon |\xi|=a\tau(t)|\eta|,\, t\geqslant T\},
\end{equation*}
\notag
$$
where $a=2/|\lambda|$ and $\tau(t)<1/a$ as $t\geqslant T$. By Proposition 2, for $t\geqslant T$ we have $dv^{(-)}(t,Y(t))/dt\geqslant 0$, and hence the field corresponding to (3) on $\Gamma$ is directed outside $\Omega$. Below, in § 4, we will define a set $\Sigma\subset\Omega|_{t=T}$ such that $\Sigma\cap\Gamma$ is a retract of $\Psi(\Sigma)\cap\Gamma$, and is not a retract of $\Sigma$ itself. Due to this fact, there exist initial data $\widehat{X}\in \Sigma$ such that $\Psi(\widehat{X})\subset\Omega$. While applying the construction outlined above (see § 4) we will follow the general scheme developed in [5], which in fact goes back to [12]. However, for our purposes, a certain modification of this scheme is required to be able to achieve the interpolation estimates for the remainder terms $\varepsilon_{1,2}(t)$ as in Theorem 1. § 3. Auxiliary estimatesWe set
$$
\begin{equation*}
\delta(t)=\sup_{s\geqslant t}\frac1{1+s-t}\int_t^s|\varphi(r)|\,dr.
\end{equation*}
\notag
$$
If $\varphi\in\mathrm{L}_p(\mathbb R_+)$, $p\geqslant 1$, then by Hölder’s inequality,
$$
\begin{equation*}
\frac1{1+s-t}\int_t^s|\varphi(r)|\,dr \leqslant \frac{(p-1)^{1/q}}{p}\biggl( \int_t^{\infty}|\varphi(r)|^p\, dr\biggr)^{1/p},
\end{equation*}
\notag
$$
where $q=p/(p-1)$, and therefore, $\delta(t)\to 0$ as $t\to\infty$. Indeed, on the one hand, integrating by parts, we get
$$
\begin{equation*}
\begin{aligned} \, \tau(t) &= 2\alpha\int_t^{\infty} \biggl(\int_t^s|\varphi(r)|\, dr\biggr)e^{2\alpha(t-s)}\,ds \\ &\leqslant 2\alpha\delta(t)\int_t^{\infty}(1+s-t)e^{2\alpha(t-s)}\,ds =\biggl(1+\frac1{2\alpha}\biggr)\delta(t) \end{aligned}
\end{equation*}
\notag
$$
and, on the other hand, since $\tau'(r)=2\alpha\tau(r)-|\varphi(r)|$, we have, for arbitrary $s\geqslant t$,
$$
\begin{equation*}
\frac1{1+s-t}\int_t^s|\varphi(r)|\, dr =\frac{2\alpha}{1+s-t}\int_t^s\tau(r)\, dr + \frac{\tau(t)-\tau(s)}{1+s-t}\leqslant 2\alpha\sup_{r\geqslant t}\tau(r)+\tau(t).
\end{equation*}
\notag
$$
Lemma 2. $\delta(t)\leqslant (1+2\alpha)\sup_{r\geqslant t}\sigma(r)$. In addition, if $t\geqslant A$, then Indeed, since $\sigma'(r)=|\varphi(r)|-2\alpha\sigma(r)$, it follows that
$$
\begin{equation*}
\frac1{1+s-t}\int_t^s|\varphi(r)|\, dr=\frac{2\alpha}{1+s-t}\int_t^s\sigma(r)\, dr+ \frac{\sigma(s)-\sigma(t)}{1+s-t}\leqslant 2\alpha\sup_{r\geqslant t}\sigma(r)+\sigma(s)
\end{equation*}
\notag
$$
for arbitrary $s\geqslant t$. On the other hand, integrating by parts for $t\geqslant A$, we obtain
$$
\begin{equation*}
\begin{aligned} \, \sigma(t) &= \int_0^Ae^{2\alpha(s-t)}|\varphi(s)|\, ds +e^{2\alpha(A-t)}\int_A^t|\varphi(r)|\, dr \\ &\qquad+ 2\alpha\int_A^t\biggl(\int_s^t|\varphi(r)|\, dr\biggr)e^{2\alpha(s-t)}\, ds \\ &\leqslant e^{2\alpha(A-t)}\int_0^t|\varphi(r)|\, dr + 2\alpha\sup_{s\geqslant A} \delta(s)\int_A^t(1+t-s)e^{2\alpha(s-t)}\, ds \\ &\leqslant e^{2\alpha(A-t)}\int_0^t|\varphi(s)|\, ds + \biggl(1+\frac1{2\alpha}\biggr)\sup_{s\geqslant A} \delta(s). \end{aligned}
\end{equation*}
\notag
$$
Lemma 3. Assume that $\varphi\in\mathrm{L}_p(\mathbb R_+),p\in[1,2]$. Then $\tau\in\mathrm{L}_p(\mathbb R_+)$, $\tau\varphi\in\mathrm{L}_1(\mathbb R_+)$, $\tau(t)\to 0$ as $t\to\infty$, and for sufficiently large $t$, Proof. Multiplying the equality $2\alpha\tau(s)=|\varphi(s)|+\tau'(s)$ by $\tau(s)^{p-1}$ and integrating over $[t,A]$, we have
$$
\begin{equation*}
2\alpha\int_t^A\tau(s)^p\, ds =\int_t^A|\varphi(s)|\tau(s)^{p-1}\, ds+ \frac{\tau(A)^p-\tau(t)^p}{p}.
\end{equation*}
\notag
$$
Since $\varphi\in\mathrm{L}_p(\mathbb R_+)$, it follows that $\tau(t)\to 0$ as $t\to\infty$ by Lemma 1. As a result,
$$
\begin{equation*}
2\alpha\int_t^A\tau(s)^p\, ds \leqslant \int_t^A|\varphi(s)|\tau(s)^{p-1}\, ds \leqslant \biggl(\int_t^A|\varphi(s)|^pds\biggr)^{1/p} \biggl(\int_t^A\tau(s)^p\, ds\biggr)^{1/q}
\end{equation*}
\notag
$$
for sufficiently large $A$, where $q=p/(p-1)$. Consequently, $\tau\in\mathrm{L}_p(\mathbb R_+)$, and
$$
\begin{equation*}
\int_t^{\infty}\tau(s)^p\,ds \leqslant (2\alpha)^{-p}\int_t^{\infty}|\varphi(s)|^p\, ds.
\end{equation*}
\notag
$$
Note that in this setting $q\geqslant p$, and, in addition, $\tau(t)\to 0$ as $t\to\infty$. Hence $\tau\in\mathrm{L}_q(\mathbb R_+)$. In view of the condition $\varphi\in\mathrm{L}_p(\mathbb R_+)$, this implies integrability of the function $\tau\varphi$ on the half-axis $\mathbb R_+$. Therefore, by Hölder’s inequality, we have, for sufficiently large $t$,
$$
\begin{equation*}
\begin{aligned} \, &\int_t^{\infty}|\varphi(s)|\tau(s)\, ds \leqslant \biggl(\int_t^{\infty}|\varphi(s)|^p\, ds\biggr)^{1/p} \biggl(\int_t^{\infty}\tau(s)^q\, ds\biggr)^{1/q} \\ &\qquad\leqslant \biggl(\int_t^{\infty}|\varphi(s)|^p\, ds\biggr)^{1/p} \biggl(\int_t^{\infty}\tau(s)^p\, ds\biggr)^{1/q} \leqslant (2\alpha)^{-p/q}\int_t^{\infty}|\varphi(s)|^p\, ds. \end{aligned}
\end{equation*}
\notag
$$
This proves Lemma 3. Lemma 4. Let $\varphi\in\mathrm{L}_p(\mathbb R_+)$, $p\in[1,2]$. Then $\sigma\in\mathrm{L}_p(\mathbb R_+)$, $\sigma\varphi\in\mathrm{L}_1(\mathbb R_+)$, $\sigma(t)\to 0$ as $t\to\infty$, and for sufficiently large $t$. Proof. Multiplying the equality $2\alpha\sigma(s)=|\varphi(s)|-\sigma'(s)$ by $\sigma(s)^{p-1}$ and integrating over $[0,t]$, we have
$$
\begin{equation*}
\begin{aligned} \, 2\alpha\int_0^t\sigma(s)^p\, ds &=\int_0^t|\varphi(s)|\sigma(s)^{p-1}\, ds- \frac{\sigma(t)^p}{p} \\ &\leqslant \biggl(\int_0^t|\varphi(s)|^p\, ds\biggr)^{1/p} \biggl(\int_0^t\sigma(s)^p\, ds\biggr)^{1/q}, \end{aligned}
\end{equation*}
\notag
$$
where $q=p/(p-1)\geqslant 2$. Consequently,
$$
\begin{equation*}
\int_0^t\sigma(s)^p\, ds\leqslant (2\alpha)^{-p}\int_0^t|\varphi(s)|^p\, ds.
\end{equation*}
\notag
$$
Therefore, $\sigma\in\mathrm{L}_p(\mathbb R_+)$. By Lemma 2, in our setting $\sigma(t)\to 0$ as $t\to\infty$, and hence, $\sigma\in\mathrm{L}_q(\mathbb R_+)$. Since $\varphi\in\mathrm{L}_p(\mathbb R_+)$, the function $\sigma\varphi$ is integrable on the half-axis $\mathbb R_+$. Next, multiplying the equation $2\alpha\sigma(s)=|\varphi(s)|-\sigma'(s)$ by $\sigma(s)^{p-1}$ and integrating over $[t,\infty)$, we find that
$$
\begin{equation*}
\begin{aligned} \, 2\alpha\int_t^{\infty}\sigma(s)^p\, ds &=\int_t^{\infty}|\varphi(s)|\sigma(s)^{p-1}\, ds+ \frac{\sigma(t)^p}{p} \\ &\leqslant \biggl(\int_t^{\infty}|\varphi(s)|^p\, ds\biggr)^{1/p} \biggl(\int_t^{\infty}\sigma(s)^p\, ds\biggr)^{1/q}+ \sigma(t)^p. \end{aligned}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\biggl(\int_t^{\infty}\sigma(s)^p\, ds\biggr)^{1/q} \leqslant (2\alpha)^{-p/q} \biggl(\int_t^{\infty}|\varphi(s)|^p\, ds\biggr)^{1/q} +(2\alpha)^{-1/q}\sigma(t)^{p/q}.
\end{equation*}
\notag
$$
Here, we used the fact that if $a,b,c\geqslant 0$ and $a^q\leqslant ab+c$, where $q\geqslant 2$, then $a\leqslant b^{1/(q-1)}+c^{1/q}$. Finally, applying Hölder’s and Young’s inequalities, we have, for sufficiently large $t$,
$$
\begin{equation*}
\begin{aligned} \, \int_t^{\infty}|\varphi(s)|\sigma(s)\, ds &\leqslant \biggl(\int_t^{\infty}|\varphi(s)|^p\, ds\biggr)^{1/p} \biggl(\int_t^{\infty}\sigma(s)^q\, ds\biggr)^{1/q} \\ &\leqslant (2\alpha)^{-p/q}\int_t^{\infty}|\varphi(s)|^p\, ds + (2\alpha)^{-1/q}\sigma(t)^{p/q} \biggl(\int_t^{\infty}|\varphi(s)|^p\, ds\biggr)^{1/p} \\ &\leqslant \biggl((2\alpha)^{-p/q}+\frac1{p}\biggr) \int_t^{\infty}|\varphi(s)|^p\, ds+ \frac{\sigma(t)^{p}}{2\alpha q}. \end{aligned}
\end{equation*}
\notag
$$
This proves Lemma 4. § 4. Asymptotic integrationStatement 3. Let $\sigma(t)<|\lambda|/2$ if $t\geqslant T$ and let $Y(t)=\left(\begin{smallmatrix}\xi(t)\\\eta(t)\end{smallmatrix}\right)$ be any solution to system (3) such that $|\eta(T)|\leqslant (2/|\lambda|)\sigma(T)|\xi(T)|$. Then, for all $t\geqslant T$, In the definition of the Lyapunov type function $v^{(+)}(t,Y)$, we set $a=2/|\lambda|$, and note that, for an arbitrary trajectory of system (3) with initial data satisfying the condition $v^{(+)}(T,Y(T))\leqslant 0$, from Statement 1 we have, for all $t\geqslant T$, Statement 4. Let $\tau(t)<|\lambda|/2$ if $t\geqslant T$. Then system (3) has a solution $\widehat{Y}(t)=\left(\begin{smallmatrix}\widehat{\xi}(t) \\ \widehat{\eta}(t)\end{smallmatrix} \right)$ such that $\widehat{\eta}(t)\ne 0$ and, for all $t\geqslant T$, Proof. Proceeding as in § 2, we define the set
$$
\begin{equation*}
\Sigma=\biggl\{\biggl(T,\xi,\frac{R}{a\tau(T)}\biggr),|\xi|\leqslant R\biggr\} \subset\Omega|_{t=T},
\end{equation*}
\notag
$$
where $ R > 0$. The set $\Psi(\Sigma)\cap\,\Gamma$ consists of segments of the integral curves $\Psi(X)\cap \,\Gamma$, where $X\in\Sigma$ so that the corresponding mapping $\gamma\colon\Sigma\to\Psi(\Sigma)\cap\Gamma$ is continuous, and the transform is single-valued on $\Psi(X)\cap\Gamma$. Indeed, if a segment of the integral curve $(t,Y(t))\subset\Psi(X)$ lies in $\Gamma$, then, on this segment, so that $\varphi(t)\equiv 0$. Hence, $\arg (\xi/\overline{\eta})$ is constant on this segment. Therefore, we have $\pi(\Psi(X)\cap\Gamma)=\mathrm{const}$, as far as $|\xi|=a\tau(t)|\eta|$ on $\Gamma$.
Now let us show that there exist initial data $\widehat{X}\in \Sigma$ such that $\Psi(\widehat{X})\subset\Omega$, and so, $v^{(-)}(t,\widehat{Y}(t))\leqslant 0$ for $t\geqslant T$. Assume that $\Psi(X)\cap\Gamma\ne\varnothing$ for arbitrary $X\in\Sigma$. Then the mapping is continuous on $\Sigma$ and is identical on $\partial\Sigma$. Hence the composition $\pi\circ\gamma$ is a retraction. However, $\Sigma\cap\Gamma=\partial\Sigma$ is a sphere, which by no means can be a retract of the ball $\Sigma$ according to the Brouwer’s fixed-point theorem (see, for example, [13]). Next, $\Psi(\Sigma)$ does not contain the stationary point of system (3), and hence $\widehat{\eta}(t)\ne 0$ for $t\geqslant T$. This proves Statement 4. Proof of Theorem 1. Let $\tau(t)\to 0$ as $t\to\infty$ and let $\tau\varphi\in\mathrm{L}_1(\mathbb R_+)$. Integrating the equality
$$
\begin{equation*}
\widehat{\eta}'(t)=-\lambda\widehat{\eta}(t) -\frac{\varphi(t)}{2\lambda} \bigl(\widehat{\xi}(t)+\widehat{\eta}(t)\bigr),
\end{equation*}
\notag
$$
and normalizing, we have
$$
\begin{equation*}
\begin{aligned} \, \widehat{\eta}(t) &=\exp\biggl(-\int_0^t\biggl(\lambda +\frac{\varphi(s)}{2\lambda}\biggr)\, ds\biggr) \exp\biggl(\frac1{2\lambda}\int_t^{\infty}\varphi(s) \frac{\widehat{\xi}(s)}{\widehat{\eta}(s)}\, ds\biggr) \\ &=\exp\biggl(-\lambda t-\frac1{2\lambda} \int_0^t\varphi(s)\, ds\biggr)\biggl\{1 + O\biggl(\int_t^{\infty}|\varphi(s)|\tau(s)\, ds\biggr)\biggr\} \end{aligned}
\end{equation*}
\notag
$$
for the solution $\widehat{Y}(t)=\left(\begin{smallmatrix}\widehat{\xi}(t) \\ \widehat{\eta}(t) \end{smallmatrix} \right)$ of system (3), which was constructed in Statement 4. Here, we used the estimate $\widehat{\xi}(t)=O(\tau(t)\widehat{\eta}(t))$. The solution $u_1(t)=\widehat{\xi}(t)+\widehat{\eta}(t)$ to the original equation (2) obtained in this way behaves asymptotically as
Now let $\sigma(t)\to 0$ as $t\to\infty$ and $\sigma\varphi\in\mathrm{L}_1(\mathbb R_+)$. Consider the solution $\widetilde{Y}(t)=\left(\begin{smallmatrix}\widetilde{\xi}(t) \\ \widetilde{\eta}(t) \end{smallmatrix}\right)$ to system (3) specified in Proposition 3 such that $\widetilde{\xi}(T)\ne 0$. Hence $\widetilde{\xi}(t)\ne 0$ for all $t\geqslant T$. Proceeding as above and integrating the equality
$$
\begin{equation*}
\widetilde{\xi}'(t)=\lambda\widetilde{\xi}(t) + \frac{\varphi(t)}{2\lambda}\bigl(\widetilde{\xi}(t)+\widetilde{\eta}(t)\bigr),
\end{equation*}
\notag
$$
we have, after a suitable normalization,
$$
\begin{equation*}
\begin{aligned} \, \widetilde{\xi}(t) &=\exp\biggl(\int_0^t\biggl(\lambda +\frac{\varphi(s)}{2\lambda}\biggr)\, ds\biggr) \exp\biggl(-\frac1{2\lambda} \int_t^{\infty}\varphi(s) \frac{\widetilde{\eta}(s)}{\widetilde{\xi}(s)}\, ds\biggr) \\ &=\exp\biggl(\lambda t+\frac1{2\lambda}\int_0^t\varphi(s)\, ds\biggr) \biggl\{1+ O\biggl(\int_t^{\infty}|\varphi(s)|\sigma(s)\, ds\biggr)\biggr\}, \end{aligned}
\end{equation*}
\notag
$$
where $\widetilde{\eta}(t)=O(\sigma(t)\widetilde{\xi}(t))$. Thus we have constructed the second desired solution $u_2(t)=\widetilde{\xi}(t)+\widetilde{\eta}(t)$ to equation (2) with the asymptotics
$$
\begin{equation*}
\begin{aligned} \, u_2(t) &=\widetilde{\xi}(t)\bigl(1+O(\sigma(t))\bigr) \\ &=\exp\biggl(\lambda t+\frac1{2\lambda}\int_0^t\varphi(s)\, ds\biggr) \biggl\{1+ O\biggl(\sigma(t)+\int_t^{\infty}|\varphi(s)|\sigma(s)\, ds\biggr)\biggr\}. \end{aligned}
\end{equation*}
\notag
$$
If $\varphi\in\mathrm{L}_p(\mathbb R_+)$, then $\tau(t)\to0$ and $\sigma(t)\to0$ as $t\to\infty$ by Lemmas 1 and 2. In turn, from Lemmas 3 and 4 we have the conditions $\tau\varphi\in\mathrm{L}_1(\mathbb R_+)$ and $\sigma\varphi\in\mathrm{L}_1(\mathbb R_+)$ with sharp estimates for the corresponding integrals. Consequently, under the assumption $\varphi\in\mathrm{L}_p(\mathbb R_+),p\in[1,2]$, equation (2) has the fundamental system of solutions
$$
\begin{equation*}
u_{1,2}(t)=\exp\biggl(\mp\lambda t\mp\frac1{2\lambda} \int_0^t\varphi(s)\, ds\biggr) \bigl(1+\varepsilon_{1,2}(t)\bigr),
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\varepsilon_1(t)= O\biggl(\tau(t) +\int_t^{\infty}|\varphi(s)|^p\, ds\biggr),\qquad \varepsilon_2(t)= O\biggl(\sigma(t)+\int_t^{\infty}|\varphi(s)|^p\, ds\biggr).
\end{equation*}
\notag
$$
This proves Theorem 1. The set of functions $\varphi(t)$ satisfying the conditions
$$
\begin{equation*}
\begin{alignedat}{5} \tau\varphi &\in \mathrm{L}_1(\mathbb R_+), &\qquad \tau(t)&\to 0, &\quad t&\to\infty, \\ \sigma\varphi &\in \mathrm{L}_1(\mathbb R_+), &\qquad \sigma(t) &\to 0, &\quad t&\to\infty, \end{alignedat}
\end{equation*}
\notag
$$
under which equation (2) has solutions with the asymptotics
$$
\begin{equation}
u_{1,2}(t)\sim\exp\biggl(\mp\lambda t\mp\frac1{2\lambda}\int_0^t\varphi(s)\, ds\biggr),
\end{equation}
\tag{6}
$$
is somewhat wider than the class $\mathrm{L}_p(\mathbb R_+)$, $p\in[1,2]$. Let us construct a function $\varphi\notin\mathrm{L}_p(\mathbb R_+)$ such that $\tau(t)\to 0$ as $t\to\infty$ and $\tau\varphi\in\mathrm{L}_1(\mathbb R_+)$. To this end, we set $\lambda=1/2$, $a_n=n^{3/2}$, $h_n=(\ln n)^{-1}, \varepsilon_n=n^{-4/5}$, where $n=2,3,\dots$, and define
$$
\begin{equation*}
\varphi(t)= \begin{cases} h_n, &t\in(a_n-\varepsilon_n,a_n), \\ 0, &t\notin(a_n-\varepsilon_n,a_n). \end{cases}
\end{equation*}
\notag
$$
Then $\lim_{t\to\infty}\tau(t)=0$ since $\varphi(t)\to 0$ as $t\to\infty$, and, at the same time,
$$
\begin{equation*}
\int_0^{\infty}\varphi(t)^p\, dt=\sum_{n=2}^{\infty}\frac1{n^{4/5}(\ln n)^p}=\infty
\end{equation*}
\notag
$$
for arbitrary $p\geqslant 1$. Now let us show that the integral
$$
\begin{equation*}
\begin{aligned} \, \int_1^{\infty}\tau(t)^2\, dt &= \sum_{n=2}^{\infty}\int_{a_{n-1}}^{a_n}e^{2t} \biggl(\int_t^{\infty}e^{-s}\varphi(s)\, ds\biggr)^2\, dt \\ &\leqslant \sum_{n=2}^{\infty}e^{2a_n} \biggl(\int_{a_n-\varepsilon_n}^{\infty}e^{-s}\varphi(s)\, ds\biggr)^2 (a_n-a_{n-1}) \end{aligned}
\end{equation*}
\notag
$$
converges, where $(a_n-a_{n-1})\sim3\sqrt{n}/2$. Indeed, since $\sum_{k=n}^{\infty}e^{-a_k}\sim e^{-a_n}$, we have
$$
\begin{equation*}
\int_{a_n-\varepsilon_n}^{\infty}e^{-s}\varphi(s)\, ds \leqslant \sum_{k=n}^{\infty}e^{-a_k+\varepsilon_k}h_k\varepsilon_k \leqslant h_n\varepsilon_ne^{\varepsilon_n}\sum_{k=n}^{\infty}e^{-a_k} =O\bigl(h_n\varepsilon_ne^{-a_n}\bigr).
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
e^{2a_n}\biggl(\int_{a_n-\varepsilon_n}^{\infty} e^{-s}\varphi(s)\, ds\biggr)^2 (a_n-a_{n-1}) =O\bigl(n^{-11/10}(\ln n)^{-2}\bigr)
\end{equation*}
\notag
$$
and, therefore, $\int^{\infty}\tau(t)^2\, dt<\infty$. Now an integration by parts
$$
\begin{equation*}
\int_0^{\infty}\varphi(t)\tau(t)\,dt =\frac12\biggl(\int_0^{\infty}e^{-t}\varphi(t)\, dt\biggr)^2 + \int_0^{\infty}\tau(t)^2\, dt,
\end{equation*}
\notag
$$
where $\tau(\infty)=0$, shows that $\tau\varphi\in\mathrm{L}_1(\mathbb R_+)$ in our setting.
§ 5. Bellman asymptotic solutionsUnder the condition $\varphi\in\mathrm{L}_3(\mathbb R_+)$, the method based on the transformation of (2) to the Riccati equation can be applied (see [4]) to construct solutions with the asymptotics
$$
\begin{equation*}
u_{1,2}(t)\sim\exp\biggl(\mp\lambda t\mp\frac1{2\lambda} \int_0^t\varphi(s)\, ds\pm \frac1{4\lambda^2} \int_0^t\varphi(s)\int_0^se^{2\lambda(r-s)}\varphi(r)\, dr\, ds\biggr).
\end{equation*}
\notag
$$
Lemma 5. If $\varphi(t)=t^{-\kappa}\sin t$, where $\kappa\in(0,1)$, then the limit exists and is finite. Proof. Integrating by parts two times, we have, after an appropriate regularization,
$$
\begin{equation*}
\begin{aligned} \, I(s) &:=\int_0^se^{2r}\varphi(r)\, dr=\frac12\,\varphi(s)e^{2s} + \frac{\kappa}2\int_0^s\frac{\sin r}{r^{\kappa+1}}\,e^{2r}\, dr \\ &\,\qquad-\frac14\lim_{\varepsilon\to 0} \biggl\{e^{2r}\frac{\cos r}{r^{\kappa}} \bigg|_{\varepsilon}^s + \int_{\varepsilon}^s e^{2r}\biggl(\frac{\sin r}{r^{\kappa}} +\kappa\frac{\cos r}{r^{\kappa+1}}\biggr)\, dr\biggr\} \\ &\,=\frac12\,\varphi(s)e^{2s}{+}\, \frac{\kappa}2\int_0^s\frac{\sin r}{r^{\kappa+1}}\, e^{2r}\, dr -\frac{I(s)}4+\frac{1\,{-}\,e^{2s}\cos s}{4s^{\kappa}} +\frac{\kappa}4\int_0^s \! \frac{1\,{-}\,e^{2r}\cos r}{r^{\kappa+1}}\, dr. \end{aligned}
\end{equation*}
\notag
$$
Hence where
$$
\begin{equation*}
A(s)=\frac{1-e^{2s}\cos s}{s^{\kappa}},\quad B(s) =2\kappa\int_0^s\frac{\sin r}{r^{\kappa+1}}\, e^{2r}\, dr,\quad C(s)=\kappa\int_0^s\frac{1-e^{2r}\cos r}{r^{\kappa+1}}\, dr.
\end{equation*}
\notag
$$
By the Dirichlet’s test, the integral $\int^{\infty}\varphi(s)e^{-2s}A(s)\, ds$ converges conditionally, and $\varphi(s)e^{-2s}B(s)$ and $\varphi(s)e^{-2s}C(s)$ possess absolutely integrable majorants, since
$$
\begin{equation*}
\int_1^s\frac{e^{2r}}{r^{\kappa+1}}\, dr \sim \frac{e^{2s}}{2s^{\kappa+1}},\qquad s\to\infty.
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\int_0^t\varphi(s)e^{-2s}I(s)\, ds =\frac25\int_0^t\varphi(s)^2\, ds+\mathrm{const} +o(1),\qquad t\to\infty.
\end{equation*}
\notag
$$
This proves Lemma 5. Corollary 1. If $\varphi(t)=t^{-\kappa}\sin t$, where $\kappa>1/3$, then equation (2) for $\lambda=1$ has a fundamental system of solutions which behave asymptotically as On the other hand, the following result holds (see [14]). Statement 5. Let $\varphi(t)=\varphi_1(t)+\varphi_2(t)$, where $\varphi_1'(t)\in\mathrm{L}_1(\mathbb R_+)$ and $\varphi_2'(t)=O(t^{-\alpha})$, $\alpha>1/2$. Then the limit exists and is finite, provided that $\varphi(t)\to 0$ as $t\to\infty$. § 6. The asymptotic reduction methodIn parallel with the equation where the function $f(t)$ is (in general)complex-valued and $r(t)>0$, we consider the comparison equation with a real-valued function $g(t)$. Equation (8) is regarded as an (unperturbed) model with respect to (7). It is assumed that (8), being non-oscillatory, possesses a fundamental system of solutions $\{y_1(t),y_2(t)\}$ which are positive for sufficiently large $t$ and such that
$$
\begin{equation*}
y_1(t)y_2'(t)-y_1'(t)y_2(t)=\frac1{r(t)},\qquad \rho(t):=\frac{y_2(t)}{y_1(t)}\nearrow\infty, \quad t\to\infty.
\end{equation*}
\notag
$$
The solution $y_1(t)$ is called principal (or subdominant); this solution is determined uniquely up to a constant factor (see [5]). Let us write equation (7) as the system
$$
\begin{equation}
X'=\begin{pmatrix} 0 &\dfrac1{r} \\ -f &0 \end{pmatrix}X,
\end{equation}
\tag{9}
$$
where $X=\left(\begin{smallmatrix} x\\ rx'\end{smallmatrix}\right)$, which can be reduced to the form
$$
\begin{equation*}
Y'=(g-f)\begin{pmatrix} -y_1y_2&-y_2^2\\ y_1^2&y_1y_2\end{pmatrix}Y
\end{equation*}
\notag
$$
by the substitution $X=\left(\begin{smallmatrix}y_1&y_2\\ ry_1'&ry_2'\end{smallmatrix}\right)Y$. The transformation $Y=\left(\begin{smallmatrix}1&0\\ 0&1/\rho\end{smallmatrix}\right)\widetilde{Y}$ enables one to balance the off-diagonal matrix entries of the system
$$
\begin{equation*}
\widetilde{Y}'=\biggl\{\frac{\rho'}{\rho}\begin{pmatrix} 0&0\\ 0&1\end{pmatrix} +(g-f)y_1y_2\begin{pmatrix} -1&-1\\ 1&1\end{pmatrix}\biggr\} \widetilde{Y} =\{\Lambda(t)+\widetilde{V}(t)\}\widetilde{Y}.
\end{equation*}
\notag
$$
The goal of the current procedure (cf. [6]) is to reduce system (9) to the $L$-diagonal form. If $(g-f)y_1y_2\notin\mathrm{L}_1(\mathbb R_+)$, then one can try to achieve integrability of the perturbation $\widetilde{V}(t)$ by making an appropriate invertible transformation $\widetilde{Y}=(I+ \widetilde{Q}(t))\widehat{Y}$ so that
$$
\begin{equation*}
\widehat{Y}'=\bigl\{\Lambda+(I+\widetilde{Q})^{-1}\bigl([\Lambda,\widetilde{Q}] +\widetilde{V}(I+\widetilde{Q})-\widetilde{Q}'\bigr)\bigr\}\widehat{Y},
\end{equation*}
\notag
$$
where $[\Lambda,\widetilde{Q}]=\Lambda \widetilde{Q}-\widetilde{Q}\Lambda$. If $\widetilde{Q}(t)=q(t)\left(\begin{smallmatrix} -1&-1\\ 1&1\end{smallmatrix}\right)$, then $(I+\widetilde{Q})^{-1}=I-\widetilde{Q}$, and so the reduced system takes the form
$$
\begin{equation*}
\widehat{Y}'=\biggl\{\frac{\rho'}{\rho}\begin{pmatrix} 0&0\\ 0&1\end{pmatrix} +q\frac{\rho'}{\rho}\begin{pmatrix} 0&1\\ 1&0\end{pmatrix} +\biggl[(g-f)y_1y_2-q'-q^2\frac{\rho'}{\rho}\biggr] \begin{pmatrix} -1&-1\\ 1&1\end{pmatrix} \biggr\}\widehat{Y}.
\end{equation*}
\notag
$$
As the functional parameter of this transformation we take $q(t)$ from the interpolation (one-parameter) family
$$
\begin{equation*}
q(t)=\rho(t)^{\kappa}\int_t^{\infty}(f-g)y_1y_2\rho^{-\kappa}\, ds,\qquad 0\leqslant \kappa\leqslant 1,
\end{equation*}
\notag
$$
so that $q'=(g-f)y_1y_2+\kappa q\rho'/\rho$. As a result, we obtain the system
$$
\begin{equation*}
\begin{aligned} \, \widehat{Y}' &=\biggl\{\frac{\rho'}{\rho}\begin{pmatrix} 0&0\\ 0&1\end{pmatrix} +q\frac{\rho'}{\rho}\begin{pmatrix} \kappa&1+\kappa\\ 1-\kappa &-\kappa\end{pmatrix} +q^2\frac{\rho'}{\rho}\begin{pmatrix} 1&1\\ -1&-1\end{pmatrix}\biggr\}\widehat{Y} \\ &=\{\Lambda(t)+\widehat{V}(t)+R(t)\}\widehat{Y}. \end{aligned}
\end{equation*}
\notag
$$
Finally, in order to provide the desired decay of $\widehat{V}(t)$ at infinity, we carry out the additional transformation
$$
\begin{equation*}
\widehat{Y}=(I+\widehat{Q}(t))Z=\begin{pmatrix} 1 +\widehat{q}_{11}&\widehat{q}_{12}\\ \widehat{q}_{21}&1+\widehat{q}_{22} \end{pmatrix}Z.
\end{equation*}
\notag
$$
After a corresponding substitution, the system takes the form
$$
\begin{equation*}
Z'=\bigl\{\Lambda+(I+\widehat{Q})^{-1}\bigl([\Lambda,\widehat{Q}] + \widehat{V}(I+\widehat{Q})-\widehat{Q}'+R(I+\widehat{Q})\bigr)\bigr\}Z,
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\widehat{V}(I+\widehat{Q})=q\frac{\rho'}{\rho}\begin{pmatrix} \kappa(1+\widehat{q}_{11}) +(1+\kappa)\widehat{q}_{21} &\kappa\widehat{q}_{12}+(1+\kappa)(1+\widehat{q}_{22}) \\ (1-\kappa)(1+\widehat{q}_{11})-\kappa\widehat{q}_{21} &(1-\kappa)\widehat{q}_{12}-\kappa(1+\widehat{q}_{22}) \end{pmatrix}
\end{equation*}
\notag
$$
and $[\Lambda,\widehat{Q}]\,{=}\,(\rho'/\rho) \left(\begin{smallmatrix} 0&-\widehat{q}_{12}\\ \widehat{q}_{21}&0 \end{smallmatrix}\right)$. We next set $L(t) = \int_t^{\infty}q(s)\rho'(s)/\rho(s)\, ds$ and define $\widehat{q}_{11}=e^{-\kappa L}-1$, so that $(\widehat{q}_{11})'=\kappa(1+\widehat{q}_{11})q\rho'/\rho$. As the off-diagonal entries
$$
\begin{equation*}
\widehat{q}_{12}=(\kappa+1)\frac1{\rho(t)}\, e^{-\kappa L(t)}\int_0^tq\rho'e^{\kappa L}\, ds, \qquad \widehat{q}_{21}=(\kappa-1)\rho(t)e^{\kappa L(t)} \int_t^{\infty}q\frac{\rho'}{\rho^2}\, e^{-\kappa L}\,ds
\end{equation*}
\notag
$$
we take the solutions to the equations
$$
\begin{equation*}
(\widehat{q}_{12})'=(\kappa q-1)\widehat{q}_{12}\frac{\rho'}{\rho} +(1+\kappa)q\frac{\rho'}{\rho},\qquad (\widehat{q}_{21})' =(1-\kappa q)\widehat{q}_{21}\frac{\rho'}{\rho} +(1-\kappa)q\frac{\rho'}{\rho},
\end{equation*}
\notag
$$
and $\widehat{q}_{22}=e^{\kappa L}+\widehat{q}_{12}\widehat{q}_{21}-1$ is chosen according to the relation
$$
\begin{equation*}
(\widehat{q}_{12}\widehat{q}_{21})' =\bigl((1-\kappa)\widehat{q}_{12}+ (1+\kappa)\widehat{q}_{21}\bigr)q\frac{\rho'}{\rho}.
\end{equation*}
\notag
$$
Proceeding in this manner, we have
$$
\begin{equation*}
\begin{aligned} \, Z' &=\biggl\{\Lambda+q\frac{\rho'}{\rho}(I+\widehat{Q})^{-1} \begin{pmatrix} (1+\kappa)\widehat{q}_{21} &(1+\kappa)\widehat{q}_{22} \\ (1-\kappa)\widehat{q}_{11} &-(1+\kappa(1+\widehat{q}_{12}))\widehat{q}_{21} \end{pmatrix} \\ &\qquad+(I+\widehat{Q})^{-1}R(I+\widehat{Q})\biggr\}Z= \{\Lambda(t)+V(t)\}Z. \end{aligned}
\end{equation*}
\notag
$$
Lemma 6. If $q(t)\to 0$ as $t\to\infty$ and the integral Note that $\widehat{q}_{11}(t)=O(\kappa|L(t)|)$ as $t\to\infty$ and, besides, for sufficiently large $t$, we have
$$
\begin{equation*}
|\widehat{q}_{21}(t)|\leqslant 2\rho(t)M(t)\int_t^{\infty}\frac{\rho'(s)}{\rho(s)^2}\, ds =2M(t).
\end{equation*}
\notag
$$
We next derive the estimate
$$
\begin{equation*}
|\widehat{q}_{12}(t)| \leqslant \frac2{\rho(t)}\max_{s\geqslant 0}e^{\kappa L(s)} \int_0^t|q(s)|\rho'(s)\, ds,
\end{equation*}
\notag
$$
where the right-hand side tends to zero as $t\to\infty$, since $\rho(t)\nearrow\infty$ and $q(t)\to 0$. Finally, by the above, $\widehat{q}_{22}(t)=O(\kappa|L(t)|+M(t))$ as $t\to\infty$. Corollary 2. Under the hypothesis of Lemma 6, The above reduction scheme was employed earlier (see [10]) in the cases $\kappa=0$ and $\kappa=1$ for the derivation of certain well-known (cf. Proposition 3) conditions for asymptotic equivalence for the solutions to equations (7) and (8). § 7. Hartman and Wintner type conditionsIf $V\in\mathrm{L}_1(\mathbb R_+)$, then by Levinson’s theorem the reduced $L$-diagonal system has the fundamental matrix
$$
\begin{equation*}
\begin{pmatrix} 1+\varepsilon_{11}(t)&\varepsilon_{12}(t) \\ \varepsilon_{21}(t)&1+\varepsilon_{22}(t) \end{pmatrix}\begin{pmatrix} 1&0 \\ 0&\rho(t) \end{pmatrix},
\end{equation*}
\notag
$$
where
$$
\begin{equation}
\varepsilon_{ii}(t)= O\biggl(\int_t^{\infty}\|V(s)\|\, ds\biggr),\qquad i=1,2,
\end{equation}
\tag{12}
$$
and
$$
\begin{equation}
\varepsilon_{12}(t)=O\biggl(\frac1{\rho(t)}\int_0^t\rho(s)\|V(s)\|\, ds\biggr),\qquad \varepsilon_{21}(t)=O\biggl(\rho(t)\int_t^{\infty}\frac1{\rho(s)}\|V(s)\|\, ds\biggr).
\end{equation}
\tag{13}
$$
After inverse transformations, we obtain an asymptotic factorization of the fundamental matrix for the initial system (9), and, as a result, we arrive at the conclusion that the solutions to equations (7) and (8) are asymptotically equivalent, with sharp estimates for the accuracy of the corresponding approximations (cf. [14] and [15]). Proposition 4. Let $q(t)\to 0$ as $t\to\infty$, let the integral For a proof, it suffices to evaluate the product
$$
\begin{equation*}
\begin{pmatrix} x_1 &x_2 \\ rx_1' &rx_2' \end{pmatrix}=\begin{pmatrix} y_1 &y_2 \\ ry_1' &ry_2' \end{pmatrix}\begin{pmatrix} 1 &0 \\ 0 &\dfrac1{\rho} \end{pmatrix}\begin{pmatrix} 1+q_{11} &q_{12} \\ q_{21} &1+q_{22} \end{pmatrix}\begin{pmatrix} 1+\varepsilon_{11} &\varepsilon_{12} \\ \varepsilon_{21} &1+\varepsilon_{22} \end{pmatrix}\begin{pmatrix} 1 &0 \\ 0 &\rho \end{pmatrix},
\end{equation*}
\notag
$$
where the entries $q_{ij}$ of the matrix $Q=(I+\widetilde{Q})(I+\widehat{Q})-I$ are given by
$$
\begin{equation*}
\begin{alignedat}{3} q_{11} &=\widehat{q}_{11}-q(1+\widehat{q}_{11}+\widehat{q}_{21}), &\qquad q_{12} &=\widehat{q}_{12}-q(1+\widehat{q}_{12}+\widehat{q}_{22}), \\ q_{21} &=\widehat{q}_{21}+q(1+\widehat{q}_{11}+\widehat{q}_{21}), &\qquad q_{22} &=\widehat{q}_{22}+q(1+\widehat{q}_{12}+\widehat{q}_{22}), \end{alignedat}
\end{equation*}
\notag
$$
so that $q_{11}+q_{21}=\widehat{q}_{11}+\widehat{q}_{21}$ and $q_{21}+q_{22}=\widehat{q}_{21}+\widehat{q}_{22}$. Multiplying the above matrices, we find that
$$
\begin{equation*}
\begin{aligned} \, x_1(t) &=y_1(t)\bigl(1+q_{11}(t)+q_{21}(t)\bigr)\bigl(1+\varepsilon_{11}(t)\bigr)+ y_1(t)\bigl(1+q_{12}(t)+q_{22}(t)\bigr)\varepsilon_{21}(t) \\ &=y_1(t)\{1+\widehat{q}_{11}(t)+\widehat{q}_{21}(t) +O(\varepsilon_{11}(t)) +O(\varepsilon_{21}(t))\}, \\ x_2(t) &=y_2(t)\bigl(1+q_{11}(t)+q_{21}(t)\bigr)\varepsilon_{12}(t)+ y_2(t)\bigl(1+q_{12}(t)+q_{22}(t)\bigr)\bigl(1+\varepsilon_{22}(t)\bigr) \\ &=y_2(t)\{1+\widehat{q}_{12}(t)+\widehat{q}_{22}(t) +O(\varepsilon_{12}(t)) +O(\varepsilon_{22}(t))\}. \end{aligned}
\end{equation*}
\notag
$$
This gives us the desired fundamental system of solutions $\{x_1(t),x_2(t)\}$ to equation (7) with asymptotics (14) and with estimates (10)–(13) for the remainder terms. Proof of Theorem 2. Equation (2) with $\varphi(t)\equiv 0$ will be chosen in our context as an unperturbed comparison equation within the framework of the above scheme, so that $y_{1,2}(t)=e^{\mp\lambda t}$ and $\rho(t)=e^{2\lambda t}$.
We claim first that if $\int^{\infty}\varphi(t)\, dt<\infty$, then both integrals
$$
\begin{equation*}
q(t)=\int_t^{\infty}\varphi(s)e^{2\kappa\lambda(t-s)}\, ds,\qquad \kappa L(t)=2\kappa\lambda\int_t^{\infty}q(s)\, ds
\end{equation*}
\notag
$$
also converge for $\kappa>0$ and behave as $O(\mu(t))$, where $\mu(t)=\sup_{s\geqslant t} \bigl|\int_s^{\infty}\varphi(r)\, dr\bigr|$. Indeed, an integration by parts
$$
\begin{equation*}
\int_t^{\infty}\varphi(s)e^{-2\kappa\lambda s}\, ds =e^{-2\kappa\lambda t} \int_t^{\infty}\varphi(s)\, ds -2\kappa\lambda\int_t^{\infty}e^{-2\kappa\lambda s} \biggl(\int_s^{\infty}\varphi(r)\, dr\biggr)\, ds
\end{equation*}
\notag
$$
shows that
$$
\begin{equation*}
\biggl|\int_t^{\infty}\varphi(s)e^{-2\kappa\lambda s}\, ds\biggr| \leqslant 2\mu(t)e^{-2\kappa\lambda t},
\end{equation*}
\notag
$$
and, hence, $|q(t)|\leqslant 2\mu(t)$. Similarly we apply integration by parts to deal with the expression
$$
\begin{equation*}
\int_t^{\infty}q(s)\, ds=\int_t^{\infty}e^{2\kappa\lambda s} \biggl(\int_s^{\infty}\varphi(r)e^{-2\kappa\lambda r}\, dr\biggr)\, ds =\frac1{2\kappa\lambda}\biggl(\int_t^{\infty}\varphi(s)\, ds-q(t)\biggr),
\end{equation*}
\notag
$$
so that $\kappa|L(t)|\leqslant 3\mu(t)$.
We next note that condition (5) guarantees that the function
$$
\begin{equation*}
\bigl(\kappa|L(t)|+M(t)\bigr)|q(t)|\frac{\rho'(t)}{\rho(t)}
\end{equation*}
\notag
$$
is integrable on the half-axis $\mathbb R_+$. Indeed, in our setting, $\rho'(t)/\rho(t)=2\lambda$, and
$$
\begin{equation*}
\kappa|L(t)q(t)|\leqslant 3|q(t)|\mu(t),\qquad |M(t)q(t)|\leqslant 2|q(t)|\mu(t).
\end{equation*}
\notag
$$
Now an application of Proposition 4 shows that equation (2) has required solutions with asymptotics
$$
\begin{equation*}
\begin{aligned} \, u_1(t)&= \exp(-\lambda t)\{1+\widehat{q}_{11}(t)+\widehat{q}_{21}(t) + O(\varepsilon_{11}(t))+O(\varepsilon_{21}(t))\}, \\ u_2(t)&= \exp(\lambda t)\{1+\widehat{q}_{12}(t)+\widehat{q}_{22}(t) + O(\varepsilon_{12}(t))+O(\varepsilon_{22}(t))\}, \end{aligned}
\end{equation*}
\notag
$$
where, by Lemma 6,
$$
\begin{equation*}
\begin{aligned} \, \widehat{q}_{11}(t) &= O\bigl(\kappa|L(t)|\bigr)=O(\mu(t)),\qquad \widehat{q}_{21}(t)=O(M(t))=O(\mu(t)), \\ \widehat{q}_{12}(t) &= O\biggl(\frac1{\rho(t)}\int_0^t|q(s)|\rho'(s)\, ds\biggr) = O\biggl(\int_0^te^{2\lambda(s-t)}\mu(s)\, ds\biggr), \\ \widehat{q}_{22}(t) &= O\bigl(\kappa|L(t)|+M(t)\bigr) = O\biggl(\int_0^te^{2\lambda(s-t)}\mu(s)\, ds\biggr). \end{aligned}
\end{equation*}
\notag
$$
In addition, we have
$$
\begin{equation*}
\begin{gathered} \, \begin{aligned} \, |\varepsilon_{11}(t)|+|\varepsilon_{21}(t)|+|\varepsilon_{22}(t)| &= O\biggl(\int_t^{\infty}\|V(s)\|\, ds\biggr) \\ &=O\biggl(\int_t^{\infty}\bigl(\kappa|L(s)|+M(s)\bigr)|q(s)|\, ds\biggr) = O\bigl(\Phi_{\kappa}(t)\bigr), \end{aligned} \\ \varepsilon_{12}(t) =O\biggl(\frac1{\rho(t)}\int_0^t\rho(s)\|V(s)\|\, ds\biggr) =O\biggl(\int_0^te^{2\lambda(s-t)}\mu(s)\, ds\biggr). \end{gathered}
\end{equation*}
\notag
$$
This proves Theorem 2. To conclude, we give an illustrative example of equation (2) to which Theorem 2 applies, but Theorem 1 does not. Setting $\varphi(t)=\sin t^2$, we have
$$
\begin{equation*}
\tau(t)=\int_t^{\infty}e^{2\lambda(t-s)}|\varphi(s)|\, ds \geqslant \int_t^{\infty}e^{2\lambda(t-s)}\varphi(s)^2\, ds =\frac1{4\lambda}+O(t^{-1}).
\end{equation*}
\notag
$$
Hence $\tau\varphi\notin\mathrm{L}_1(\mathbb R_+)$, and so the hypotheses of Theorem 1 are not met. At the same time,
$$
\begin{equation*}
\frac12|q(t)|\leqslant \mu(t)=\sup_{s\geqslant t} \biggl|\int_s^{\infty}\sin r^2\, dr\biggr|\leqslant \frac1{t},
\end{equation*}
\notag
$$
and, therefore, $\Phi_{\kappa}(t)\leqslant 2/t$. We also have $\int_0^te^{2\lambda(s-t)}\mu(s)\, ds=O(t^{-1})$, and so by Theorem 2 the corresponding equation (2) has linearly independent solutions of the form
Citation in format AMSBIB
\Bibitem{Ste24} |
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