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Existence of solutions of a nonlinear eigenvalue problem and their properties D. V. Valovik, S. V. Tikhov Penza State University, Penza, Russia
Russian version:
Matematicheskii Sbornik, 2024, Volume 215, Number 1, DOI: https://doi.org/10.4213/sm9892 
Bibliographic databases:
    § 1. Problem setupLet $\mathbb R=(-\infty,+\infty)$, $\mathbb R_+=(0,+\infty)$, $\mathbb R_+^0=\mathbb R_+ \cup \{0\}$, $\mathrm I=(0, 1)$, $\overline{\mathrm I}=[0, 1]$, $\mathrm I^\ast=(0, x^\ast)$ and $\overline{\mathrm I}^\ast=[0, x^\ast]$, where $x^\ast>0$ is some number, $\alpha>0$ is a parameter, $\lambda \in \mathbb R$ is the spectral parameter, and $a(s)$ and $f(s)$ are continuously differentiable functions of $s \in \mathbb R_+^0$ such that $a' \geqslant 0$, $f(0)=0$, $f' \geqslant 0$ and, moreover, where $\lim_{s \to+\infty} s^{-q} f_1(s)=0$ and $q>0$ is a constant.Consider the equation where $u \equiv u(x; \lambda)$ and $(x, \lambda) \in \overline{\mathrm I} \times \mathbb R$, with boundary conditions and the additional condition where $A>0$ is a (prescribed) constant; here we assume thatThroughout what follows we also use the notation
$$
\begin{equation}
a_\ast=\min_{x \in \overline{\mathrm I}^\ast} a(x)=a(0)\quad\text{and} \quad a^\ast=\max_{x \in \overline{\mathrm I}^\ast} a(x)=a(x^\ast);
\end{equation}
\tag{1.6}
$$
in particular, $a^\ast=a(1)$ for $x \in \overline{\mathrm I}$. We require the following definition. Definition 1. A number $\lambda=\widehat\lambda$ such that there exists a function $u \equiv u(x; \widehat\lambda)$ satisfying equation (1.2) and conditions (1.3)–(1.5) is called an eigenvalue, and the function $u(x; \widehat\lambda)$ corresponding to $\widehat\lambda$ is called an eigenfunction of problem (1.2)–(1.5). Given the specifics of the problem under consideration, it would be reasonable to call its eigenvalues $A$-eigenvalues. However, we will not mention this each time in what follows. We call problem (1.2)–(1.5) problem $\mathcal P$. We denote eigenvalues of problem $\mathcal P$ by $\widehat\lambda$ or $\widehat\lambda_i$, where the index of an eigenvalue is set to be equal to the number of zeros of the corresponding eigenfunction on $[0,1)$. We show below that there is an infinite number of pairs of (distinct) eigenvalues in problem $\mathcal P$ such that the (distinct) eigenfunctions corresponding to each pair have the same number of zeros. Starting from some index $i$ the eigenvalues in such pairs have different signs, and in our arguments below we consider such cases separately, so there is no need in introducing further notation to distinguish between positive and negative eigenvalues. Note that a special case of problem $\mathcal P$ arises in the propagation theory of electromagnetic waves in nonlinear waveguide structures. More precisely, positive eigenvalues $\widehat\lambda$ correspond to propagation constants in the problem of finding eigenwaves for a planar shielded dielectric waveguide filled by a medium with inhomogeneous nonlinear (dielectric) permittivity; for the statements of similar problems, see [1]–[3]. In electrodynamics authors usually consider the case $a>0$ (however, cases when $a<0$ also occur). Note that although the condition $a' \geqslant 0$ is possible in applications to electrodynamics, it is not at all necessary. On the other hand the condition $f' \geqslant 0$ covers many types of nonlinearity arising in nonlinear mathematical physics and nonlinear waveguide theory, in particular, cubic and polynomial nonlinearities and powerlike nonlinearity (see [1], [2] and [4]–[7]). Since conditions (1.3) and (1.4) involve the initial data of the Cauchy problem for equation (1.2), it is obvious that before investigating the solvability of problem $\mathcal P$, one must clear up the question of the existence of a unique solution of this Cauchy problem that is continuous for $x \in \overline{\mathrm I}$. Below we study the Cauchy problem for equation (1.2) with the initial data for $x \in \overline{\mathrm I}^\ast$, where $A>0$ is a constant, and for $u(x) \in C^2 (\overline{\mathrm I}^\ast)$.We purposely state problem (1.2), (1.7) on the interval $\overline{\mathrm I}^\ast$ (in place of $\overline{\mathrm I}$) to reduce some reasoning related to problem $\mathcal P$. Just for this reason the function $a(x)$ introduced at the beginning of this section is defined on $\mathbb R_+^0$. Considering an auxiliary Cauchy problem on $\overline{\mathrm I}$ would mean no loss of generality of the results obtained below, but would require additional calculations in the verification of these results, Note that we collect the proofs of theorems and propositions in § 5. In the conclusion of this section we present briefly the main ideas and results of our investigations. To problem $\mathcal P$ we apply the method of integral characteristic equations. Using this method we obtain an integral characteristic equation for problem $\mathcal P$ (see (3.14)), which is equivalent to the original problem (see Theorem 1), that is, any solution of it is an eigenvalue of problem $\mathcal P$ and vice versa. By analysing (3.14) we show that problem $\mathcal P$ is solvable and reveal some properties of its eigenvalues: the existence of infinite numbers of negative and positive eigenvalues, and the asymptotic behaviour of the eigenvalues and maxima of eigenfunctions (see Theorems 2 and 3). In addition, we prove comparison theorems, which establish connections between solutions of two problems (with different parameters) of type $\mathcal P$ (see Theorems 4 and 5), and some results establishing connections between solutions of the original problem and the corresponding linearized problem (see Propositions 8 and 9). In this paper we do not address questions of completeness and the basis property of the set of eigenfunctions (for fixed $A$ and the set of all $A$) in natural functional spaces (these questions were treated, for instance, in [8]). § 2. Linear problemWe call the linear problem (for $\alpha=0$) problem $\mathcal P_0$ and denote eigenvalues of problem $\mathcal P_0$ by $\widetilde\lambda$ or $\widetilde\lambda_i$, where the index of an eigenvalue is equal to the number of zeros of the corresponding eigenfunction on $[0,1)$. Note that, by contrast with problem $\mathcal P$, in problem $\mathcal P_0$ there exists at most one eigenfunction with a prescribed number of zeros. Problem $\mathcal P_0$ consists in finding $\lambda=\widetilde{\lambda}$ such that the equation has a nontrivial solution $v \equiv v (x; \lambda) \in C^2 (\overline{\mathrm I})$ satisfying the boundary conditions Note that condition (1.4) is necessary for the definition of discrete eigenvalues of problem $\mathcal P$, but we do not need it in the linear problem, and therefore we drop it.Problem $\mathcal P_0$ is classical in the Sturm–Liouville theory, and it is well understood (see [9]–[12]). Taking some known results into account we state the following proposition. Proposition 1. Problem $\mathcal P_0$ has an infinite number of negative eigenvalues $\widetilde\lambda$ and finitely many (maybe zero) positive eigenvalues $\widetilde\lambda$; in addition, all solutions of problem $\mathcal P_0$ have multiplicity one. The asymptotic behaviour of the eigenvalues $\widetilde\lambda_n$ is described by the formula $\widetilde\lambda_n= O^\ast(n^2)$, where the coefficient of the leading term is $-\pi^2$. The result of Proposition 1 also holds in the case when the property $a'(x) \geqslant 0$ fails. For $a'(x) \geqslant 0$ Proposition 1 can be refined. In fact, in addition to Proposition 1 we have the following result. Proposition 2. If $a'(x) \geqslant 0$, then each eigenvalue $\widetilde\lambda$ of problem $\mathcal P_0$ satisfies $\widetilde\lambda<a^\ast$. § 3. Integral characteristic equationWe begin this section by stating a result on the solvability of the Cauchy problem for equation (1.2) with initial data (1.7). Multiplying (1.2) by $u'$ and integrating we obtain
$$
\begin{equation}
u'^2+(a-\lambda) u^2+\alpha F(u^2)-\int_{0}^x a' (s) u^2 (s)\,ds \equiv C,
\end{equation}
\tag{3.1}
$$
where $\displaystyle F(u^2)=\int_0^{u^2}\!\!\! f(t)\,dt$ and $C$ is a constant. Using (1.7) we find that ${C=A^2>0}$. Taking (1.1) into account we obtain where $\displaystyle F_1(u^2)=\int_{0}^{u^2} f_1(s)\,ds$ and $\lim_{u^2 \to+\infty} (F_1(u^2)/u^{2(q+1)})=0$. Using the expression (3.1) we can prove the following. Proposition 3. The Cauchy problem (1.2), (1.7) has a unique solution $u \equiv u(x; \lambda)$ which is defined and continuous for $(x, \lambda, \alpha) \in \overline{\mathrm I}^\ast \times \mathbb R \times \mathbb R_+$. We turn to the derivation of the integral characteristic equation. We introduce new variables by the formulae For brevity we often drop the dependence of the quantities under consideration under parameters. Using (3.3) we write equation (1.2) as
$$
\begin{equation}
\tau' =2\tau\eta, \qquad \eta'=-(a-\lambda+\alpha f(\tau)+\eta^2).
\end{equation}
\tag{3.4}
$$
In the variables (3.4), under assumptions (1.7) relation (3.1) takes the following form:
$$
\begin{equation}
(\eta^2+a-\lambda)\tau+\alpha F(\tau)-\int_{0}^x a' (s) \tau (s) \,ds=A^2.
\end{equation}
\tag{3.5}
$$
Since (3.5) defines implicitly the function $\tau \equiv \tau(x, \eta; \lambda)$, we can write the second equation in (3.4) as $\eta'=-w(x, \eta; \lambda)$, where
$$
\begin{equation}
w(x, \eta; \lambda)=\eta^2+a(x)-\lambda+\alpha f(\tau(x, \eta; \lambda)).
\end{equation}
\tag{3.6}
$$
It is easy to verify that $w>0$ for all $\lambda \in \mathbb R$. In fact, let $w$ vanish at some point $x=x'$. Then from $w=0$ we find that $\eta^2+a- \lambda=- \alpha f(\tau)$. Substituting this into (3.5) we obtain
$$
\begin{equation*}
A^2=\alpha F(\tau)-\alpha f(\tau)\tau-\int_{0}^{x'} a'(s) \tau(s)\,ds= -\alpha\int_{0}^{\tau} f'(s)s\,ds-\int_{0}^{x'} a'(s) \tau(s)\,ds.
\end{equation*}
\notag
$$
Since $f' \geqslant 0$ and $a ' \geqslant 0$, it follows that $A^2 \leqslant 0$, which clearly contradicts the condition $A>0$. This contradiction shows that $w$ does not vanish and thus has constant sign for $\lambda \in \mathbb R$. Calculating $w$ for $\lambda \leqslant a_\ast$ we see that $w>0$. Hence $\eta'=-w(x, \eta; \lambda)< 0$, so that $\eta(x)$ is monotonically decreasing for $x \in \overline{\mathrm I}^\ast$. It follows from the second formula (3.3) that $\eta$ is continuous if and only of $u$ does not vanish. Assume that $u(x)$ has $n \geqslant 0$ zeros $x_1, \dots, x_{n} \in \mathrm I^\ast$ arranged in ascending order, and let it vanish for $x=0$; then $\eta(x)$ has discontinuities at some points $x_0, \dots, x_{n}$, where $x_0=0$. We assume that $u(x)$ does not vanish on the interval $(x_n, x^\ast)$, while $x=x^\ast$ can be a zero of $u(x)$. In the latter case we set $x^\ast=x_{n+1}$. It is obvious that if $u \not\equiv 0$, then $u'(x_i) \ne 0$ for all $i=0,\dots, n+1$. In fact, if the solution $u$ of (1.2) vanishes at some point together with its derivative $u'$, then it follows from the classical results on the (local) existence and uniqueness of the solution of the Cauchy problem for an ordinary differential equation that $u \equiv 0$ (see [13]). Thus, we have essential discontinuities at all points of discontinuity. Since $\eta(x)$ is monotonically decreasing, it follows from (1.3) and (1.4) that
$$
\begin{equation}
\eta(x_0+0)=+\infty\quad\text{and} \quad \eta(x_{n+1}-0)=- \infty.
\end{equation}
\tag{3.7}
$$
The first relation in (3.7) follows from (1.3), while the second holds in the case when $x^\ast=x_{n+1}$. Consider the intervals
$$
\begin{equation}
\mathrm I_j=(x_j, x_{j+1})\quad\text{and} \quad \mathrm I_n=(x_n, x^\ast),
\end{equation}
\tag{3.8}
$$
where $j=0,\dots, n-1$ and $x_0=0$. At the endpoints of $\mathrm I_j$ we have The function $\eta$ is monotonically decreasing from $+\infty$ to $-\infty$ for $x \in \mathrm I_{j}$, $j= 0,\dots, n-1$, while $\eta$ is monotonically increasing from $+\infty$ to $\eta (x^\ast)$ for $x \in \mathrm I_{n}$, but it is not necessary that $\eta (x^\ast)=-\infty$ (clearly, $\eta(x^\ast)$ depends on $\lambda$). This means in its turn that there exist continuous natural bijections
$$
\begin{equation}
g_{j}\colon \mathbb R \to \mathrm I_{j}\colon \eta \mapsto x
\end{equation}
\tag{3.10}
$$
of the line $\mathbb R$ onto $\mathrm I_{j}$, $j=0,\dots, n-1$, and a continuous bijection
$$
\begin{equation}
g_{n}\colon (\eta(x^\ast),+\infty) \to \mathrm I_{n}\colon \eta \mapsto x.
\end{equation}
\tag{3.11}
$$
Using these bijections we can, in particular, represent $a$ on each interval $\mathrm I_j$ as a function of $\eta$, namely, $a(x)\equiv a(g_j(\eta))$. Now for the function $w$ defined by (3.6) and considered on the interval $\mathrm I_j$ we use the notation here $\tau \equiv \tau (s; g_j (s); \lambda)$ is determined from (3.5) for $\eta=s$ and $x=g_j (s)$, and $g_j (s)$ is one of the above bijections.We have the following result. Proposition 4. Assume that the solution $u \equiv u(x; \lambda)$ of Cauchy problem (1.2), (1.7) has $n \geqslant 0$ zeros for $x \in \overline{\mathrm I}^\ast$. Then where the quantity $\eta(x^\ast)$ depends on $\lambda$. If $n=0$, then this means that $u$ does not vanish for $x \in \overline{\mathrm I}^\ast$, and the left-hand side of (3.13) contains only the last term. The following result holds. Proposition 5. The function defined by the left-hand side of (3.13) exists and is positive and continuous for $\lambda \in \mathbb R$. Note that we do not fix $n$ and $\eta(x^\ast)$ in Proposition 5. Provided that $x^\ast=x_{n+1}=1$, under the second condition in (3.7) we obtain from (3.13) the integral characteristic equation of problem $\mathcal P$. It has the form
$$
\begin{equation}
\Phi_{\alpha}(\lambda; n) \equiv \sum_{j=0}^{n} T_j (\lambda)-1=0,
\end{equation}
\tag{3.14}
$$
where $n=0, 1,\dots$ and
$$
\begin{equation*}
T_j(\lambda)=\int_{-\infty}^{+\infty} \frac{ds}{w_j(s; \lambda)}.
\end{equation*}
\notag
$$
Equation (3.14) and the functions $T_j$ introduced above are the central objects under investigation here. Our study of (3.14), and therefore of the original eigenvalue problem, is based on an analysis of the functions $T_j$. In particular, using the properties of the $T_j$ as $\lambda \to \pm \infty$ we will prove that problem $\mathcal P$ is solvable. Relation (3.14) is a family (but not a system) of equations corresponding to different $n$. Note that the numbering of eigenvalues fixed at the beginning of the paper has an obvious link with the number of terms in (3.14). In fact, the index of each eigenvalue $\widehat\lambda_{n+1}$ is equal to the number of zeros of the corresponding eigenfunction $u \equiv u(x; \widehat\lambda_{n+1})$ on the half-open interval $[0, 1)$. It is clear that this is equal to the number of terms in (3.14). The following result allows us to go over from problem $\mathcal P$ to the examination of equation (3.14). Theorem 1 (on equivalence). Equation (3.14) is equivalent to problem $\mathcal P$ in the following sense: $\widehat\lambda \in \mathbb R$ is an eigenvalue of problem $\mathcal P$ if and only if there exists an integer $\widehat n \geqslant 0$ such that $\lambda= \widehat\lambda$ solves equation (3.14) for $n=\widehat n$; in this case the eigenfunction $u \equiv u(x; \widehat\lambda)$ has $\widehat n$ (simple) zeros $x_m \in \mathrm I$, where $x_m=\sum_{j=0}^{m} T_j(\widehat\lambda)$ and $m=1,\dots,\widehat n$. In connection with Theorem 1 we can introduce the following definition. Definition 2. An eigenvalue $\widehat\lambda$ of problem $\mathcal P$ has multiplicity $p$ if $\lambda=\widehat\lambda$ is a root of equation (3.14) of multiplicity $p$. In Definition 2 we deal with the situation when we can speak of the multiplicity of a root of equation (3.14). § 4. Main resultsIn what follows we require some properties of the $T_j$ as functions of $\lambda$; see (3.14). We begin this section by discussing them. We represent $T_{j}$ in a form that will occasionally be more convenient for us: in $T_{j}$ we change the variable of integration from $s=\eta$ to $\tau$. Then we must understand how to calculate $x$ in terms of $\tau$. Since we consider $T_{j}$ for $x \in \mathrm I_j$, and the function $\eta \equiv \eta (x)$ is monotone and continuous on this interval, we can calculate $x$ in terms of $\eta$ using the map $g_j$ (see (3.10)), which gives meaning to $T_j$. Now we establish a bijective correspondence between $x$ and $\tau$. It is clear that $\tau$, in contrast to $\eta$, is not a monotone function of ${x \in \mathrm I_j}$. However, we can subdivide $\mathrm I_j$ into two subintervals on which $\tau$ is monotone. In fact, $\eta$ is continuous and monotone for $x \in \mathrm I_{j}$, where it decreases from $+\infty$ to $-\infty$ (or to $\eta(x^\ast)$). Let $x_j^{\ast} \in \mathrm I_j$ be the point at which $\eta (x_j^{\ast})=0$; then $\eta>0$ for $x \in (x_j, x_j^{\ast})$ and $\eta<0$ for $x \in (x_j^{\ast}, x_{j+1})$. By definition $\eta=u' / u$; hence, for $\eta>0$ the function $\tau=u^2$ increases monotonically, and for $\eta<0$ it decreases monotonically. Setting $\tau_j^{\ast}=\tau (x_j^{\ast})$ and bearing in mind that $x_j$ and $x_{j+1}$ are zeros of $u$, we conclude that $\tau$ increases monotonically from $0$ to $\tau_j^{\ast}$ for $x \in (x_j, x_j^{\ast})$ and decreases monotonically from $\tau_j^{\ast}$ to $0$ for $x \in (x_j^{\ast}, x_{j+1})$. Thus we conclude that there exists a continuous bijection which takes the interval $(0, \tau_j^{\ast})$ to $\mathrm I_{j}^{(1)}=(x_j, x_j^{\ast})$; in a similar way there exists a continuous bijection that maps $(0, \tau_j^{\ast})$ onto $\mathrm I_{j}^{(2)}=(x_j^{\ast}, x_{j+1})$. For $x \in (x_n, x^\ast)$ the relevant maps are similarly defined.We have the following result. Proposition 6. The following formula holds: We are interested in the behaviour of $T_j(\lambda)$ as $\lambda \to \pm\infty$. To find the asymptotic behaviour of $T_j(\lambda)$ as $\lambda \to -\infty$, we use the expression provided in (3.14) for this integral, and as $\lambda \to+\infty$, we use the (equivalent) expression (4.3). The following result characterizes the behaviour of the functions $T_j (\lambda)$. Proposition 7. For each $j$ the function $T_j (\lambda)$ is defined, positive and continuous for all sufficiently large $|\lambda|$. Furthermore, if $\lambda$ is negative and sufficiently large in absolute value, then where $t_{1,2}=|\lambda|^{-q-r_{1,2}}$, $r_{1,2}>0$ depend on the properties of $f_1$, and for sufficiently large $\lambda>0$ By Theorem 1 formulae (4.5) and (4.6) yield estimates for the distance between consecutive zeros of eigenfunctions in the case when $|\lambda|$ is sufficiently large (or, equivalently, when there are sufficiently many of these zeros on $\overline{\mathrm I}$). For problem $\mathcal P_0$ an estimate for the distance between consecutive zeros of the corresponding eigenfunction is given by formula (4.14). An estimate for the distance between consecutive zeros of solutions is a classical problem in Sturm–Liouville theory [11], [14]). The following two theorems ensure that problem $\mathcal P$ is solvable. Theorem 2. Problem $\mathcal P$ has an infinite number of negative eigenvalues $\widehat\lambda_n$ with accumulation point at infinity. In addition, for sufficiently large indices $n$, and the following estimate holds for the maximum of the eigenfunction: where $\delta>0$ depends on the properties of $f_1$. Note that $O^\ast$ in (4.7) means that we know the coefficient of $n^{-2q}$ (see the proof); the function $f_1$ was defined in (1.1). Theorem 3. Problem $\mathcal P$ has an infinite number of positive eigenvalues $\widehat\lambda_n$ with accumulation point at infinity. In addition, for all sufficiently large $n$ where $\Delta>0$ is an arbitrary constant, $\lambda_n= g^{-1}\bigl(\frac{q}{n(q+1)}\bigr)$, $g^{-1}$ is the inverse function of $g(t)=t^{-1/2}{\log t}$, and the following estimate holds for the maximum of the eigenfunction: where $\delta>0$ depends on the properties of $f_1$. We can partition the set $\Sigma$ of eigenvalues $\widehat\lambda$ of problem $\mathcal P$ into two subsets: $\Sigma=\Sigma_1 \cup \Sigma_2$, where $\Sigma_1 \cap \Sigma_2= \varnothing$. Eigenvalues $\widehat\lambda \in \Sigma_1$ have the following property: each $\widehat\lambda \in \Sigma_1$ turns to an eigenvalue $\widetilde\lambda$ of problem $\mathcal P_0$ as $\alpha \to+0$. Eigenvalues $\widehat\lambda \in \Sigma_2$ do not have this property (cannot be linearized). We will see below that the set $\Sigma_1$ contains an infinite number of elements $\widehat\lambda_n$ and $\lim_{n \to \infty} \widehat\lambda_n=-\infty$. In the particular case when $a \equiv \mathrm{const}$ and $f(s) \equiv s$, using elementary estimates we can show that for each $\lambda=\widehat\lambda>a+\delta$, where $\delta>0$ is an arbitrary constant (for estimates required for this, see [15]). Hence $\lim_{\alpha \to+0} \widehat\lambda(\alpha)=+\infty$, where $\widehat\lambda=\widehat\lambda(\alpha)$ is an eigenvalue for which we have (4.11). Hence each eigenvalue $\widehat\lambda$ such that (4.11) holds is nonlinearizable (there are infinitely many such values).Using inequalities (4.5) and (4.7) for negative and (4.6) and (4.9) for positive eigenvalues of problem $\mathcal P$ we can obtain comparison results for eigenvalues of two nonlinear problems with distinct parameters, as well as for ones of a nonlinear problem and the corresponding linearized one. Let $\mathcal P_1$ and $\mathcal P_2$ denote two problems of type $\mathcal P$; we assume that all parameters and functions specifying these problems also have these indices. Theorem 2 claims that each of $\mathcal P_1$ and $\mathcal P_2$ has an infinite number of negative eigenvalues $\lambda=\widehat\lambda^{(1)}_{i}$ and $\lambda=\widehat\lambda^{(2)}_{i}$, respectively, where $\lim_{i\to\infty} \widehat\lambda^{(j)}_{i}=-\infty$ ($j=1,2$). Then the following result holds. Theorem 4. If the functions $a_j$ in problems $\mathcal P_1$ and $\mathcal P_2$ satisfy then $\widehat\lambda^{(1)}_{i}< \widehat\lambda^{(2)}_{i}$ for sufficiently large $i$. Note that if $\max_{x\in \overline{\mathrm I}} a_1(x)=\min_{x\in \overline{\mathrm I}} a_2(x)$, then we also need that $\alpha_1 A_1^{q_1}<\alpha_2 A_2^{q_2}$ for the assertion of the theorem to hold. Theorem 3 states that each of problems $\mathcal P_1$ and $\mathcal P_2$ has an infinite number of positive eigenvalues $\lambda=\widehat\lambda^{(1)}_{i}$ and $\lambda=\widehat\lambda^{(2)}_{i}$, respectively, where $\lim_{i\to\infty} \widehat\lambda^{(j)}_{i}=+\infty$ ($j=1,2$). Then the following result holds. Theorem 5. If $q_1<q_2$, then $\widehat\lambda^{(1)}_{i}>\widehat\lambda^{(2)}_{i}$ for sufficiently large $i$. We go over to stating some results connecting eigenvalues of problems $\mathcal P$ and $\mathcal P_0$. Although problem $\mathcal P_0$ was thoroughly investigated long ago using various approaches (see [9]–[11] and [16]), we apply to it also the approach developed above. It is yet another method for the derivation of an asymptotic estimate for eigenvalues of problem $\mathcal P_0$. An equation of type (3.14) can rigorously be deduced for problem $\mathcal P_0$; however, we suppress a large part of the calculations involved because they simply repeat the derivation of (3.14). Setting $\alpha=0$ in equation (3.14) we obtain the integral characteristic equation of problem $\mathcal P_0$. It has the following form:
$$
\begin{equation}
\Phi_0(\lambda; n) \equiv \sum_{j=0}^{n} \widetilde T_j (\lambda)-1=0,
\end{equation}
\tag{4.13}
$$
where
$$
\begin{equation*}
\widetilde T_j(\lambda)=\int_{-\infty}^{+\infty} \frac{ds}{\widetilde w_j(s; \lambda)},\qquad \widetilde w_j(s; \lambda)=s^2+a-\lambda,\quad \lambda \in (-\infty, a^\ast),\quad n=0, 1,\dots\,.
\end{equation*}
\notag
$$
Here $a \equiv a(x)$ and $x=\widetilde g_j(s)$, where the maps $\widetilde g_j$ are defined similarly to the definition of the maps $g_j$ in problem $\mathcal P$. We do not consider separately the question of the solvability of the Cauchy problem for (2.1) because this question is also well understood (see [13] and [17]). It is important to note that (4.13) is equivalent to problem $\mathcal P_0$ in the sense of Theorem 1. We do not prove this fact here, but its proof can be obtained similarly to the proof of Theorem 1; for the more complicated (though linear with respect to the unknown function) Sturm–Liouville problem the proofs can be found in [18] and [19]. For all $\lambda<a_\ast$ the integral $\widetilde T_j(\lambda)$ has an elementary estimate
$$
\begin{equation}
\frac{\pi}{\sqrt{-\lambda+a^\ast}} \leqslant \widetilde T_j(\lambda) \leqslant \frac{\pi}{\sqrt{-\lambda+a_\ast}}.
\end{equation}
\tag{4.14}
$$
Taking (4.14) into account, from (4.13) we obtain
$$
\begin{equation*}
\frac{\pi(n+1)}{\sqrt{-\lambda+a^\ast}} \leqslant \sum_{j=0}^{n} \widetilde T_j (\lambda) \leqslant \frac{\pi(n+1)}{\sqrt{-\lambda+a_\ast}}.
\end{equation*}
\notag
$$
This inequality, in view of (4.13), yields
$$
\begin{equation}
-\pi^2 n^2+a_\ast \leqslant \widetilde \lambda_n \leqslant -\pi^2 n^2+a^\ast,
\end{equation}
\tag{4.15}
$$
which provides the estimate $\widetilde \lambda_n=O^\ast(n^2)$, where the coefficients of the leading term in the asymptotic expression is $-\pi^2$. We can conclude from (4.7) and (4.15) that for problems $\mathcal P$ and $\mathcal P_0$ asymptotic estimates for negative eigenvalues with large index have the same leading term, so that the nonlinearity $f(u^2)$, which can have an arbitrary powerlike growth, influences the behaviour of the eigenvalues only in the second and higher-order terms of the asymptotic expansion. Let $v \equiv v(x; \lambda)$ be a nontrivial solution of the Cauchy problem for equation (2.1) which satisfies $v|_{x=0}=0$, and let $u \equiv u(x; \lambda, \alpha)$ be the solution of the Cauchy problem (1.2), (1.7). Then $v(1; \lambda)$ is called the characteristic function of problem $\mathcal P_0$, and the characteristic equation $v(1; \lambda)=0$ defines the eigenvalues of this problem (see [10]). By analogy we call $u(1; \lambda, \alpha)$ the characteristic function of problem $\mathcal P$. Then we can show that the characteristic equation defines the eigenvalues of problem $\mathcal P$.Consider the equation obtained by subtracting $v(1; \lambda)$ from both sides of (4.16).Using relation (4.17) we obtain. Proposition 8. Each neighbourhood of any eigenvalue $\widetilde\lambda$ of problem $\mathcal P_0$ contains at least one eigenvalue $\widehat\lambda= \widehat\lambda(\alpha)$ of problem $\mathcal P$, provided that $\alpha>0$ is sufficiently small. In addition, these eigenvalues satisfy The proof of the existence of eigenvalues in close problems, which is based on an equation of the same type as (4.17), was carried out in [3]. Consider equation (3.14). Subtracting $\Phi_0(\lambda; n)$ from both sides of it we obtain
$$
\begin{equation}
\Phi_0(\lambda; n)-\Phi_{\alpha}(\lambda; n)=\Phi_0(\lambda; n).
\end{equation}
\tag{4.19}
$$
Using relation (4.19) we arrive at the following result. Proposition 9. For each $\alpha>0$ there exists a positive constant $\lambda_0$ (which can be quite large) for which the following property holds: in each neighbourhood of any eigenvalue $\widetilde\lambda_i \in (-\infty, -\lambda_0)$ of problem $\mathcal P_0$ there is at least one eigenvalue $\widehat\lambda_i$ of problem $\mathcal P$; in addition, these eigenvalues satisfy Note that, given $\alpha$, the result of Proposition 9 holds for an infinite number of eigenvalues $\widetilde\lambda_i \in (-\infty, -\lambda_0)$ of problem $\mathcal P_0$. We also stress that the neighbourhoods mentioned in Proposition 9 can have the same radius for each eigenvalue in this infinite family (it follows from the proof of Proposition 9 that these estimates decrease with $n$). The proofs of Propositions 8 and 9 are realizations of some versions of the perturbation method. A result which is similar to Proposition 8 in a certain sense can be obtained using a fundamentally different approach, based on the inversion of the linear part of the differential operator defined by (1.2) by use of the Green’s function: see, for instance, [20]. § 5. Proofs Proof of Proposition 2. Let $a'(x) \geqslant 0$, and let $\lambda \geqslant a^\ast$. Multiplying (2.1) by $v'$, integrating and using that $v (0)=0$ we obtain
$$
\begin{equation}
v'^2(x)=v'^2 (0)+(\lambda-a(x))v^2(x)+\int_0^x a' (s) v^2 (s)\, ds.
\end{equation}
\tag{5.1}
$$
Since $v'(0) \ne 0$ when the solution is nontrivial, the right-hand side of (5.1) is strictly positive. Hence $v'$ does not vanish, and since $v(x)$ is negative for $x \in \overline{\mathrm I}$, $v'$ has constant sign. We see that in this case $v(x)$ is monotone for $x \in \overline{\mathrm I}$. However, then the function $v$ cannot vanish at $x=1$.
The proof is complete. Proof of Proposition 3. Integrating by parts in (3.1) we obtain
Assume that the function $u$, as a solution of the Cauchy problem (1.2), (1.7), is defined for $x \in [0, x_{\ast})$, where $x_{\ast}>0$ is some constant. Assume that $u$ is unbounded, that is, $|u| \to+\infty$ and $|u'| \to+\infty$ as $x \to x_{\ast}$. Then it is clear that $u (x) u' (x) \to {+}\infty$ as $x \to x_{\ast}$. Let $ x'<x_{\ast}$ be a point such that $u(x) u'(x)>0$ for all $x \in [x', x_{\ast})$. Let $\lambda<0$. Then for $x \in [x', x_{\ast})$ all terms on the right-hand side of (5.2), except the first term, are negative. Assuming that $u$ is unbounded, we see that, starting from some $x'' \in [x', x_{\ast})$ the right-hand side of (5.2) is negative, while the left-hand side is always nonnegative. This is a contradiction. Now let $\lambda>0$. In (5.2) we set $x=x_{\ast}-\delta$, where $0<\delta< x_{\ast}-x'$. We write (5.2) as
$$
\begin{equation}
u'^2=A^2\,{-}\,2 \int_0^{x'} a(s) u(s) u'(s) \,ds\,{+}\,\lambda u^2\,{-}\,\alpha F(u^2)\,{-}\,2 \int_{x'}^{x_{\ast}-\delta} a(s) u(s) u'(s)\,ds.
\end{equation}
\tag{5.3}
$$
Applying the mean-value theorem to the integral $\displaystyle\int_{x'}^{x_{\ast}-\delta} a(s) u(s) u'(s)\, ds$ we obtain
$$
\begin{equation*}
2 \int_{x'}^{x_{\ast}-\delta} a(s) u(s) u'(s) \,ds=a(\xi) \int_{x'}^{x_{\ast}- \delta} 2 u(s) u'(s)\, ds=a(\xi) \bigl( u^2 (x_{\ast}-\delta)-u^2 (x') \bigr),
\end{equation*}
\notag
$$
where $\xi \in [x', x_{\ast}-\delta]$. Substituting this into (5.3) we find that
$$
\begin{equation*}
u'^2=A^2-2 \int_0^{x'} a(s) u(s) u'(s)\, ds+a(\xi) u^2 (x')+\lambda u^2-\alpha F(u^2)-a(\xi) u^2(x_{\ast}-\delta).
\end{equation*}
\notag
$$
It is clear that $\displaystyle\int_0^{x'} a(s) u(s) u'(s)\,ds$ and $a(\xi) u^2 (x')$ are bounded quantities. They play no role in the analysis that follows, so it is convenient to write the last relation in the form where $\displaystyle C_1=A^2-2 \int_0^{x'} a(s) u(s) u'(s) \,ds+a(\xi) u^2 (x')$.
Substituting $F(u^2)$ from (3.2) into (5.4) we obtain
$$
\begin{equation*}
u'^2=C_1+\lambda u^2-\frac{\alpha}{q+1} u^{2q+2}-F_1(u^2)-a(\xi) u^2.
\end{equation*}
\notag
$$
As $\delta$ decreases, the quantity $|u|=|u(x_{\ast}-\delta)|$ increases without limit. In the las equation the term $\frac{\alpha}{q+1} u^{2q+2}$ shows the strongest growth, so starting from some $x=x_{\ast}-\delta'$ the right-hand side becomes negative, while the left-hand side is positive. This contradiction shows that our assumption that $U$ is unbounded also fails for $\lambda>0$.
It follows from the above arguments that the right-hand side of the expression (5.2) remains bounded, so that $u'$ also remains bounded. Thus the function $u$ is defined, continuous and bounded for $x \in [0, x_{\ast})$ and any fixed $\lambda \in \mathbb R$. In the above argument $x=x_\ast$ is an arbitrary point, so we can set $x_\ast=x^\ast$. Since the right-hand side of (1.2) is continuous in $x \in \overline{\mathrm I}^\ast$, $u \in \mathbb R$, $\alpha \in \mathbb R_+$ and $\lambda \in \mathbb R$, the solution $u$ of the Cauchy problem (1.2), (1.7) us unique and depends smoothly on $x$ and the parameters $\lambda$ and $\alpha$ for $(x, \lambda, \alpha) \in \overline{\mathrm I}^\ast \times \mathbb R \times \mathbb R_+$. Proposition 3 is proved. Proof of Proposition 4. Assume that $u$ has $n \geqslant 0$ zeros $x_1, \dots,x_n \in (0, x^\ast)$; then $\eta(x)$ has $n+1$ discontinuity points $x_0, x_1, \dots, x_n$, where $x_0=0$, and all these discontinuities are of the second kind.
Integrating the equation $\eta'=-w_j (\eta; \lambda)$ on each interval $\mathrm I_j$ and taking (3.7) and (3.9) into account we obtain
$$
\begin{equation}
\begin{gathered} \, 0<x_{i+1}-x_i=\int_{-\infty}^{+\infty} \frac{ds}{w_i (s; \lambda)}, \qquad i= 0,\dots, n-1, \\ 0<x^\ast-x_n=\int_{\eta(x^\ast)}^{+\infty} \frac{ds}{w_n (s; \lambda)}. \end{gathered}
\end{equation}
\tag{5.5}
$$
Formulae (5.5) mean that all improper integrals under consideration are convergent. Moreover, these formulae provide explicit expressions for the distances between zeros of $u$. Now, adding all relations (5.5) we obtain
$$
\begin{equation*}
\begin{aligned} \, &x_1+x_2-x_1+x_3-x_2+\dots+x_{n-1}-x_{n-2}+x_n-x_{n-1}+x^\ast-x_n \\ &\qquad =\sum_{i=0}^{n-1} \int_{-\infty}^{+\infty} \frac{ds}{w_i (s; \lambda)}+ \int_{\eta(x^\ast)}^{+\infty} \frac{ds}{w_n (s; \lambda)}. \end{aligned}
\end{equation*}
\notag
$$
This implies (3.13) trivially.
The proof is complete. Proof of Proposition 5. The function $\tau \equiv \tau(\eta; \lambda) \geqslant 0$ exists for all $\lambda \in \mathbb R$ and depends continuously on $(\eta, \lambda) \in \mathbb R \times \mathbb R$. Since the functions $w_j (\eta; \lambda)$ are strictly positive and depend continuously on $\lambda \in \mathbb R$, the proposition follows. We note separately that the values of $n$ and $\eta(x^\ast)$ are not fixed in these arguments.
The proof is complete. Proof of Theorem 1. Since (3.14) is a consequence of problem $\mathcal P$, each eigenvalue of the problem is also a root of this equation.
Now we show the converse. Let $\lambda=\widehat\lambda$ be a solution of equation (3.14) for $n= n'$. It corresponds to a solution $u \equiv u(x; \widehat\lambda)$ of the Cauchy problem for equation (1.2) with initial data (1.7). The fact that there exists a unique continuous solution $u(x; \widehat\lambda)$, which is defined for $x \in \overline{\mathrm I}$, follows from Proposition 3. Using this solution $u$ of the Cauchy problem we construct the functions $\tau=u^2$ and $\eta=u'/u$. It is clear that $ \tau(0; \widehat\lambda)=0$ and $\eta(0; \widehat\lambda)=+\infty$. Assume that $\eta(1; \widehat\lambda) \ne -\infty$. For definiteness, let $\eta(1; \widehat\lambda)=-\delta>-\infty$. Using $\tau$ and $\eta$ we can write the equality
$$
\begin{equation*}
\int_{-\delta}^{+\infty} \frac{ds}{w_{n'} (s; \widehat\lambda)}+ \sum_{j=0}^{n'-1} \int_{-\infty}^{+\infty} \frac{ds}{w_j (s; \widehat\lambda)}=1,
\end{equation*}
\notag
$$
which is analogous to (3.14). This is equivalent to
$$
\begin{equation*}
\int_{-\infty}^{+\infty} \frac{ds}{w_{n'} (s; \widehat\lambda)}- \int_{-\infty}^{-\delta} \frac{ds}{w_{n'} (s; \widehat\lambda)}+ \sum_{j=0}^{n'-1} \int_{-\infty}^{+\infty} \frac{ds}{w_j (s; \widehat\lambda)}=1.
\end{equation*}
\notag
$$
Since $\lambda=\widehat\lambda$ is a solution of equation (3.14) for $n=n'$, it follows that
$$
\begin{equation*}
\int_{-\infty}^{+\infty} \frac{ds}{w_{n'} (s; \widehat\lambda)}+\sum_{j=0}^{n'-1} \int_{-\infty}^{+\infty} \frac{ds}{w_j (s; \widehat\lambda)}=1.
\end{equation*}
\notag
$$
Taking the difference of the last two expressions we obtain
$$
\begin{equation*}
\int_{-\infty}^{-\delta} \frac{ds}{w_{n'} (s; \widehat\lambda)}=0.
\end{equation*}
\notag
$$
Since $w_{n'} (s; \lambda)>0$, the assumption that ${-\delta\!>\!-\infty}$ obviously fails. Hence ${-\delta\!=\!-\infty}$.
By construction the function $u \equiv u(x; \widehat\lambda)$ satisfies the first condition in (1.3). As $\eta(1)=-\infty$, this means that $u$ also satisfies the second condition in (1.3). But then $u$ is an eigenfunction and $\lambda= \widehat\lambda$ is an eigenvalue of problem $\mathcal P$. Thus we have shown that problem $\mathcal P$ is spectrally equivalent to equation (3.14). Formulae (5.5) provide explicit expressions for the distances between zeros of $u$; in particular, we can deduce a formula for the $i$th zero $x_i$ of $u$. Theorem 1 is proved. Proof of Proposition 6. We represent $T_{j}$ as a sum:
$$
\begin{equation*}
T_{j}=\int_{-\infty}^{0} \frac{ds}{w_j (s; \lambda)}+\int_{0}^{+\infty} \frac{ds}{w_j (s; \lambda)}.
\end{equation*}
\notag
$$
Expressing $\eta=s$ from (3.5) we obtain and
The limits of integration $s=\pm \infty$ correspond to $\tau=0$ because $s \to+\infty$ and $s \to -\infty$ as $x \to x_{j'}$ and $x \to x_{j'+1}$, respectively, and $\tau$ vanishes at these points. The limit of integration corresponding to $s=0$ can be determined as the solution of equation (4.4), where $x=x_{j}^{\ast}$ is a point on $\mathrm I_{j'}$ at which $\eta (x_{j}^{\ast})=0$ (we obtain equation (4.4) from the above expression for $s$ by setting $s=0$). Combining the above results we arrive at (4.3). The proof is complete. Proof of Proposition 7. The fact that the continuous positive functions $T_j (\lambda)$ exist, including for all sufficiently large $|\lambda|$, is ensured by Proposition 5.
Consider the case of negative $\lambda$ with sufficiently large modulus. We can estimate the integral $T_j(\lambda)$ from below and above by replacing $\tau$ in $T_j(\lambda)$ by the larger quantity $\tau^\ast$ and the smaller quantity $\tau_\ast$, respectively. We can determine $\tau^\ast$ and $\tau_\ast$ from the equations obtained by ‘coarsening’ several terms in (3.5). In particular, $\tau=\tau^\ast$ is defined as the solution of the equation
$$
\begin{equation}
(\eta^2+a-\lambda)\tau+\alpha F(\tau)=A^2+\tau_{\max}\int_{0}^x a' (s) \,ds,
\end{equation}
\tag{5.6}
$$
where $\tau_{\max}=\max u^2$ is the maximum value of $\tau$. This maximum value is determined by formula (4.8).
The quantity $\tau=\tau_\ast$ is determined as the solution of the equation It turns out that the asymptotic representations for $\tau^\ast$ and $\tau_\ast$ coincide in the leading term, but are different starting with the second terms. Thus, we obtain
$$
\begin{equation}
\tau^\ast=A^2|\lambda|^{-1}+O(|\lambda|^{-1-r_1})\quad\text{and} \quad \tau_\ast= A^2|\lambda|^{-1}+O(|\lambda|^{-1-r_2}),
\end{equation}
\tag{5.8}
$$
where $r_{1,2}>0$ depend on the properties of $f_1$. This accuracy is sufficient for our aims.
Now plugging $a^\ast$ for $a$ and $\tau^\ast$ for $\tau$ into $T_j(\lambda)$ and using (1.1) we obtain a lower estimate for $T_j(\lambda)$. In a similar way, plugging $a_\ast$ for $a$ and $\tau_\ast$ for $\tau$ into $T_j(\lambda)$ and using (1.1) we obtain an upper estimate for $T_j(\lambda)$. Note that the integrals in these estimates can easily be calculated and produce formula (4.5). Now we turn to the case of sufficiently large positive $\lambda$. We can see that the function $\tau \equiv \tau(\eta; \lambda)$ defined by (3.5) is unbounded for $\lambda \to+\infty$. We ‘normalize’ (3.5) so as to avoid large values of $\tau(\eta; \lambda)$. Using the ‘normalized’ variables $\tau= \lambda^{1/q}\overline\tau$ and $\eta=\lambda^{1/2}\overline\eta$ we rewrite (3.5) as
$$
\begin{equation}
\overline C+(1-\overline a-\overline\eta^{\,2})\overline\tau-\alpha \lambda^{-1-1/q} F(\lambda^{2/q}\overline\tau)+\int_{0}^x \overline a' \overline \tau \,dt=0,
\end{equation}
\tag{5.9}
$$
where $\overline a=a\lambda^{-1}$ and $\overline C=A^2 \lambda^{-1-1/q}$. The implicit function $\overline\tau \equiv \overline\tau(\overline\eta; \lambda)$ defined by (5.9) is positive and bounded for all possible values of $\lambda$. Note that $\lim_{\overline\eta \to \pm\infty} \overline\tau(\overline\eta; \lambda)=0$ independently of $\lambda$.
Using Proposition 6 we obtain $T_j=T_{j}^{(1)}+T_{j}^{(2)}$, where
$$
\begin{equation}
T_j^{(1)}=\frac{1}{2 \sqrt{\lambda}} \int_{0}^{\overline\tau_j^{\,\ast}} \frac{d \overline \tau}{\overline v_{j}^{\,(1)} (\overline \tau; \lambda)}
\end{equation}
\tag{5.10}
$$
and
$$
\begin{equation}
T_j^{(2)}=\frac{1}{2 \sqrt{\lambda}} \int_{0}^{\overline\tau_j^{\,\ast}} \frac{d \overline \tau}{\overline v_{j}^{\,(2)} (\overline \tau; \lambda)};
\end{equation}
\tag{5.11}
$$
here
$$
\begin{equation*}
\overline v_j^{(r)} (\tau; \lambda)=\sqrt{\overline \tau}\sqrt{\overline C-(\overline a-1) \overline \tau+\int_{0}^x \overline a' \overline \tau \,dt-\alpha \lambda^{-1-1/q} F(\lambda^{2/q}\overline\tau)},
\end{equation*}
\notag
$$
$\overline a=\overline a(x)$, $x=g_{j}^{(r)} (\overline \tau)$, $ g_{j}^{(r)}$ is one of the bijections (4.1) and (4.2), $\tau=\overline\tau_j^{\,\ast}$ is the (unique) positive root of equation (5.9) for $\overline\eta=0$, and $x=x_j^{\ast}$, where $x_j^{\ast}$ is the point at which $\overline \eta (x_j^{\ast})=0$. Note that $\overline\tau_j^{\,\ast}= \lambda^{-1/q}\tau_j^{\ast}$.
Consider, for instance, $T_j^{(1)}$ separately. The expression under the ‘large’ root sign in (5.10) is just the left-hand side of (5.9) for $\overline\eta=0$ and $x= x_j^{\ast}$. Setting $\overline\eta=0$ (and $x=x_j^{\ast}$), using (3.2) and taking the limit as $\lambda \to+\infty$, from (5.9) we obtain the equation
$$
\begin{equation*}
\overline\tau\biggl(1-\frac{\alpha}{q+1} \overline\tau^{\,q}\biggr)=0,
\end{equation*}
\notag
$$
which has at least two real roots, $\tau_-^0=0$ and $\tau_+^0= ((q+1)/\alpha)^{1/q}$. Hence for $\overline\eta=0$ and sufficiently large $\lambda$ equation (5.9) has at least two real roots, which we denote by $\overline\tau_j^{\,-}$ and $\overline\tau_j^{\,+}=\overline\tau_j^{\,\ast}$, where $\overline\tau_j^{\,+}>0$.
We can show that $\lim_{\gamma \to \infty} \overline\tau_j^{\,-}=\tau_-^0$ and $\lim_{\gamma \to \infty} \overline\tau_j^{\,+}=\tau_+^0$. However, $\overline \tau_j^{\,-}$ can also be negative; then we let $\overline \tau_j^{\,-}$ be the largest negative root. Since $\tau=u^2$, negative values of $\tau$ correspond to purely imaginary values of $u$. To use negative $\tau$ below we extend the expression (5.9) to the domain $\overline\tau<0$. We find asymptotic expressions for $\overline\tau_j^{\,-}$ and $\overline\tau_j^{\,+}$ for large $\lambda$. The quantities $\tau_-^0$ and $\tau_+^0$ are the first approximations to $\overline\tau_j^{\,-}$ and $\overline\tau_j^{\,+}$, respectively. Using (3.2) we can rewrite (5.9) as
$$
\begin{equation*}
(\overline\eta^{\,2}+\overline{a}-1)\overline\tau=\overline C-\frac{\alpha}{q+1}|\overline\tau|^{q}\overline\tau-\lambda^{-1- 1/q} F_1(\lambda^{1/q}\overline\tau)-\int_0^{x} \overline a' \overline \tau \,dt.
\end{equation*}
\notag
$$
Here we use the modulus sign to extend the right-hand side into the domain of negative values of $\overline\tau$. Since the variable of integration in the integral $T_j^{(1)}$ is nonnegative, this sign does not affect calculations for $\overline\tau \geqslant 0$, but it allows one to use properly the value of $\overline\tau_j^{\,-}$. Taking (3.1) into account, it is clear that our extension is natural.
Using (5.9) for $\overline\eta=0$ and the approximations obtained we find that
$$
\begin{equation*}
\overline\tau_j^{\,-}=-A^2 \lambda^{-1-1/q}+O(\lambda^{-r_-})\quad\text{and} \quad \overline\tau_j^{\,\ast}=\biggl(\frac{q+1}{\alpha}\biggr)^{1/q}+O(\lambda^{-r_+}),
\end{equation*}
\notag
$$
where $r_->1+q^{-1}$ and $r_+>0$ depend on the properties of $F_1$. In the general case, when $a$ is not a constant, the $\overline\tau_j^{\,\pm}$ are distinct for distinct $j$. However, we will see below that the form of $a$ does not influence the leading term of the asymptotic estimate; in what follows we drop the index $j$ of the roots obtained for convenience and simply write $\overline \tau_-$ and $\overline \tau_+$.
Clearly, the ‘large’ root sign in (5.10) vanishes for $\overline\tau=\overline\tau_-$ and ${\overline\tau=\overline\tau_+}$; furthermore, $\overline\tau_- \to 0$ as $\lambda \to+\infty$. In other words, as $\lambda \to+\infty$, we obtain in (5.10) a logarithmic singularity in a neighbourhood of $\overline\tau=0$. The expression under the ‘large’ root sign in (5.10) can be written as
$$
\begin{equation}
\overline C+(1-\overline a)\overline\tau-\alpha\lambda^{-1-1/q} F(\lambda^{1/q}\overline\tau)+ \int_0^{x} \overline a' \overline \tau\, dt=(\overline\tau-\overline\tau_-) f_0(\overline\tau),
\end{equation}
\tag{5.12}
$$
where $f_0(\overline\tau_+)=0$ and $\lim_{\lambda \to+\infty} f_0(\overline\tau)=1-(\alpha/(q+1)) |\overline\tau|^q$. Let $g(\overline\tau)={1}/{\sqrt{f_0(\overline\tau)}}$; then we can write (5.10) as
$$
\begin{equation*}
T_j^{(1)}=\frac{1}{2 \sqrt{\lambda}} \int_{0}^{\overline\tau_j^{\,\ast}} \frac{g(\overline\tau)\,d\overline\tau}{\sqrt{\overline\tau(\overline\tau-\overline\tau_-)}}.
\end{equation*}
\notag
$$
Let $g_0(\overline\tau)={1}/{\sqrt{\lim_{\lambda \to+\infty}f_0(\overline\tau)}}$. Then
$$
\begin{equation}
\begin{aligned} \, \notag T_j^{(1)} &=\frac{1}{2 \sqrt{\lambda}} \int_{0}^{\overline\tau_j^{\,\ast}} \frac{g(\overline\tau)\,d\overline\tau}{\sqrt{\overline\tau(\overline\tau-\overline\tau_-)}} \\ &=\frac{1}{2 \sqrt{\lambda}} \int_{0}^{\overline\tau_j^{\,\ast}} \frac{g(\overline\tau)- g_0(\overline\tau_-)}{\sqrt{\overline\tau(\overline\tau-\overline\tau_-)}}\,d\overline\tau+\frac{g_0(\overline\tau_-)}{2 \sqrt{\lambda}} \int_{0}^{\overline\tau_j^{\,\ast}} \frac{d\overline\tau}{\sqrt{\overline\tau(\overline\tau- \overline\tau_-)}}. \end{aligned}
\end{equation}
\tag{5.13}
$$
We estimate the first term on the right in (5.13) as follows:
$$
\begin{equation*}
\int_{0}^{\overline\tau_+} \frac{g(\overline\tau)-g_0(\overline\tau_-)}{\sqrt{\overline\tau(\overline\tau- \overline\tau_-)}}\,d\overline\tau=\lim_{\lambda \to+\infty} \int_{0}^{\overline\tau_+} \frac{g(\overline\tau)-g_0(\overline\tau_-)} {\sqrt{\overline\tau(\overline\tau-\overline\tau_-)}}\,d\overline\tau+ O(\lambda^{-r_1}),
\end{equation*}
\notag
$$
where $r_1>0$. Next, we have
$$
\begin{equation*}
\begin{aligned} \, \lim_{\lambda \to+\infty} \int_{0}^{\overline\tau_+} \frac{g(\overline\tau)- g_0(\overline\tau_-)}{\sqrt{\overline\tau(\overline\tau-\overline\tau_-)}}\,d\overline\tau &= \int_{0}^{((q+1)/\alpha)^{1/q}} \biggl(\frac{1}{\sqrt{1-\frac{\alpha}{q+1}\, \overline\tau^{\,q}}}-1\biggr)\,\frac{d\overline\tau}{\overline\tau} \\ &=\int_{0}^{((q+1)/\alpha)^{1/q}} \frac{\frac{\alpha}{q+1}\, \overline\tau^{\,q-1}}{\sqrt{1-\frac{\alpha}{q+1}\, \overline\tau^{\,q}} \biggl(1+\sqrt{1-\frac{\alpha}{q+1}\, \overline\tau^{\,q}} \biggr)} \,d\overline\tau \\ &=\frac 2q \int_{0}^{1} \frac{dv}{1+v}=\frac{2\log2}{q}. \end{aligned}
\end{equation*}
\notag
$$
The second term on the right in (5.13) can be calculated explicitly, namely,
$$
\begin{equation*}
\int_{0}^{\overline\tau_+} \frac{d\overline\tau}{\sqrt{\overline\tau(\overline\tau-\overline\tau_-)}}=2 \log\bigl(\sqrt{\overline\tau_+}+\sqrt{\overline\tau_+-\overline\tau_-}\bigr)-2\log\sqrt{-\overline\tau_-}.
\end{equation*}
\notag
$$
Now, using the formulae for $\overline\tau_-$ and $\overline\tau_+$, we find that
$$
\begin{equation*}
\begin{aligned} \, \log\bigl(\sqrt{\overline\tau_+}+\sqrt{\overline\tau_+-\overline\tau_-}\bigr) &=\log2+ \frac{1}{2q}\log\frac{q+1}{\alpha} +O(\lambda^{-\delta}),\\ \log\sqrt{-\overline\tau_-} &=-\frac{1}{2}\biggl(1+\frac{1}{q}\biggr) \log\lambda+\log A+ O(\lambda^{-r_-+1+1/q}), \end{aligned}
\end{equation*}
\notag
$$
where $\delta=\min\{r_+, 1+q^{-1}\}$. Using (5.12) we obtain
$$
\begin{equation*}
g_0(\overline\tau_-)=\frac{1}{\sqrt{1-\alpha |\overline\tau_-|^q}}=1+O(\lambda^{-1-q}).
\end{equation*}
\notag
$$
Combining our results, we obtain
$$
\begin{equation*}
T_j^{(1)}=\frac{1}{2}\biggl( 1+\frac{1}{q} \biggr) \frac{\log \lambda}{\sqrt{\lambda}}+O(\lambda^{-1/2}).
\end{equation*}
\notag
$$
The same estimate holds for $T_j^{(2)}$. This yields (4.6).
Proposition 7 is proved. Proof of Theorem 2. It is clear from (4.5) that $T_j(\lambda) \to 0$ as $\lambda \to -\infty$; this holds for all $1 \leqslant j \leqslant n$. Hence there exists a nonnegative integer $m_0$ such that equation (3.14) has at least one solution $\widehat\lambda_n$ for each $n=m_0, m_0+1,\dots$; it is clear here that $\lim_{i\to \infty} \widehat\lambda_i=-\infty$. Thus, problem $\mathcal P$ has an infinite set of negative eigenvalues $\widehat\lambda_i$.
Inequalities (4.7) follow from (3.14) and estimates (4.5). We can find an estimate for $\max_{x\in[0,1]} u^2(x; \widehat\lambda)$ as follows. If $u(x; \widehat\lambda)$ has more than one zero on $(0, 1)$, then there exists a point $z \in (0, 1)$ such that $u'(z)=0$. The required maximum is $u^2(z)$. Setting $u'=0$ in (3.1) we obtain
$$
\begin{equation}
(a-\lambda) u^2+\alpha F(u^2)-\int_{0}^z a' (s) u^2 (s) \,ds=A^2.
\end{equation}
\tag{5.14}
$$
We can show that for negative $\lambda$ with large modulus we must seek the required maximum $u^2(z)$ in the form $u^2=|\lambda|^p c+O(|\lambda|^{p_1})$, where the constants $p$, $p_1$ and $c$ are to be determined and $p_1<p$. Substituting this representation for $u^2$ into (5.14), using (3.2) and equating the powers of the leading terms we obtain $p=-1$ and $c=A^2$. Calculating the power of the term next to the leading one we obtain $p_1=-1-\delta$, where $\delta>0$ depends on the properties of $f_1$. This yields (4.8). Theorem 2 is proved. Proof of Theorem 3. It is clear from (4.6) that $T_j(\lambda) \to 0$ as $\lambda \to+\infty$; this holds for all $1 \leqslant j \leqslant n$. Hence there exists a nonnegative integer $n_0 \geqslant 0$ such that equation (3.14) has at least one solution $\widehat\lambda_n$ for each $n=n_0, n_0+1,\dots$; moreover, it is clear that $\lim_{i\to \infty} |\widehat\lambda_i|=+\infty$. Thus, problem $\mathcal P$ has an infinite number of positive eigenvalues $\widehat\lambda_i$.
We see from (4.6) that the leading term of the asymptotic expansion is independent of $\alpha$, so that for each $\alpha>0$ problem $\mathcal P$ has an infinite set of positive eigenvalues. Furthermore, problem $\mathcal P_0$ can only have a finite number of positive eigenvalues. Hence infinitely many positive eigenvalues of problem $\mathcal P$ do not tend to eigenvalues of problem $\mathcal P_0$ even for $\alpha \to+0$. Inequalities (4.9) follow from (3.14) and (4.6). We seek an estimate for $\max_{x\in[0,1]}u^2(x; \widehat\lambda)$ using the same method as in the proof of Theorem 2. Assuming that $z \in (0,1)$ is a point where ${u'(z)= 0}$, we see that the required maximum is $u^2(z)$, where $u^2(z)$ satisfies (5.14). We can show that for sufficiently large $\lambda$ one must seek the required maximum of $u^2(z)$ in the form $u^2=|\lambda|^p c+O(|\lambda|^{p_1})$, where $p$, $p_1$ and $c$ are some constants to be determined and $p_1<p$. Substituting this representation for $u^2$ into (5.14), using (3.2) and equating the powers of the leading terms we obtain
$$
\begin{equation*}
p=\frac{1}{q}\quad\text{and} \quad c=\biggl(\frac{q+1}{\alpha}\biggr)^{1/q}.
\end{equation*}
\notag
$$
Calculating the power of the term following the leading one we obtain $p_1=1/q-\delta$, where $\delta>0$ depends on the properties of $f_1$. This yields (4.10).
Theorem 3 is proved. Proof of Theorem 4. Using formula (4.7) for problems $\mathcal P_1$ and $\mathcal P_2$ and also using (4.12) we obtain This inequality yields the main assertion.
The proof is complete. Proof of Theorem 5. Using formula (4.6) for problems $\mathcal P_1$ and $\mathcal P_2$ we obtain where $j=1,2$ and $\lambda>0$ is assumed to be sufficiently large. Hence it is clear that if $q_1< q_2$, then $T_1>T_2$ for sufficiently large $\lambda$, and therefore $\widehat\lambda^{(1)}_{i}>\widehat\lambda^{(2)}_{i}$.
The proof is complete. Proof of Proposition 8. Consider equation (4.17). Its left-hand side depends of $\alpha$, and it can be shown that it tends to zero as $\alpha \to+0$, provided that $\lambda \in \Lambda$, where $\Lambda$ is a prescribed bounded set. This is because the solutions of Cauchy problems that we use have the property
$$
\begin{equation*}
\lim_{\alpha \to+0}u(x; \lambda, \alpha)=v(x; \lambda)
\end{equation*}
\notag
$$
for $\lambda \in \Lambda$. The right-hand side of (4.17) is independent of $\alpha$ and vanishes for $\lambda=\widetilde\lambda$ (that is, at eigenvalues of problem $\mathcal P_0$).
Since all eigenvalues $\widetilde\lambda$ of problem $\mathcal P_0$ are simple (see Proposition 1), taking account of the equivalence between equation $v(1; \lambda)=0$ and problem $\mathcal P_0$, we obtain that each eigenvalue $\widetilde\lambda$ of problem $\mathcal P_0$ can be surrounded by a neighbourhood ${U_{\delta}= [\widetilde\lambda-\delta, \widetilde\lambda+\delta]}$ that contains no other eigenvalues of this problem and such that $\Phi_0(\lambda; n)$ takes values of different sign at the two endpoints of the interval $U_{\delta}$. Next, bearing in mind that the left- and right-hand sides of (4.17) are continuous in $\lambda$, it follows from a classical result in analysis that a neighbourhood of any zero of the right-hand side must contain a root of equation (4.17), provided that $|\alpha|$ is sufficiently small. Clearly, the diameter of this neighbourhood can be taken to be the smaller, the closer $\alpha$ is to zero. This shows that the limit relation in the proposition holds indeed. The proof is complete. Proof of Proposition 9. We transform as follows the difference on the left-hand side of (4.19):
$$
\begin{equation}
\begin{aligned} \, \notag \Phi_0(\lambda; n)-\Phi_{\alpha}(\lambda; n)&=\sum_{j=0}^n (\widetilde T_j(\lambda)- T_j(\lambda)) \\ \notag &=\sum_{j=0}^n \biggl(\int_{-\infty}^{+\infty}\frac{ds}{s^2+a-\lambda}- \int_{-\infty}^{+\infty}\frac{ds}{s^2+a-\lambda+\alpha f(\tau)}\biggr) \\ \notag &=\sum_{j=0}^n \int_{-\infty}^{+\infty}\frac{a(\widetilde g_j(s))- a(g_j(s))+\alpha f(\tau)}{(s^2+a-\lambda)\bigl(s^2+a-\lambda+\alpha f(\tau)\bigr)}\,ds \\ \notag &\leqslant \sum_{j=0}^n \int_{-\infty}^{+\infty}\frac{a(\widetilde g_j(s))- a(g_j(s))+\alpha f(\tau)}{(s^2+a-\lambda)^2}\,ds \\ &\leqslant (a^\ast-a_\ast+\alpha f(t)|_{t=\max \tau})\sum_{j=0}^n \int_{-\infty}^{+\infty}\frac{ds}{(s^2+a-\lambda)^2}. \end{aligned}
\end{equation}
\tag{5.15}
$$
For each fixed $\lambda$ the quantity $f(t)|_{t=\max \tau}$ is bounded, and if $a-\lambda>0$, then the integrals converge. We examine the right-hand side of (5.15). For negative $\lambda$ with sufficiently large modulus we have $\max \tau=A^2|\lambda|^{-1}+ O(|\lambda|^{-1-\delta})$, where $\delta>0$: see (4.8). Using estimate $a \geqslant a_\ast$, we can easily calculate the improper integral on the right-hand side of (5.15). Thus, formula (5.15) assumes the following form:
$$
\begin{equation}
\begin{aligned} \, \notag \Phi_0(\lambda; n)-\Phi_{\alpha}(\lambda; n) &\leqslant \frac{a^\ast-a_\ast+\alpha \bigl(A^2 |\lambda|^{-1}+O(|\lambda|^{-1-\delta})\bigr)}{2(|\lambda|+a_\ast)^{3/2}}\pi(n+1) \\ &\leqslant \frac{(a^\ast-a_\ast)\pi(n+1)}{2(|\lambda|+a_\ast)^{3/2}}, \end{aligned}
\end{equation}
\tag{5.16}
$$
where $\delta>0$.
It follows from (5.16) that the difference $\Phi_0(\lambda; n)- \Phi_{\alpha}(\lambda; n)$ decreases with $\lambda$ and increases with $n$. It is important here that $n$ and $\lambda$ are not independent variables: $\lambda$ grows with $n$. Since we are looking for a solution of (4.19) in a neighbourhood of a solution $\widetilde\lambda$ of problem $\mathcal P_0$, we can substitute the estimate $\widetilde\lambda_n=-\pi^2n^2+O(1)$, which follows from (4.15), into the right-hand side of (5.16). Plugging $\widetilde\lambda_n=-\pi^2n^2+O(1)$ for $\lambda$ into the right-hand side of (5.16) we obtain
$$
\begin{equation}
\Phi_0(\lambda; n)-\Phi_{\alpha}(\lambda; n)=O^\ast((n+1)^{-2}),
\end{equation}
\tag{5.17}
$$
where the coefficient of the leading term of the expansion is $(a^\ast-a_\ast)/(2\pi^2)$.
It follows from the above formula that the difference under consideration becomes less than any prescribed quantity (independently of $\alpha$), starting from some $\lambda= \lambda_0$ (which can be quite large in modulus). However, then using the same arguments as in the proof of Proposition 8 we see that equation (4.19) has at least one solution $\widehat\lambda$ in a neighbourhood of each solution $\widetilde\lambda<-\lambda_0$. In other words, in a neighbourhood of each zero $\lambda=\widetilde\lambda \in (-\infty, -\lambda_0)$ of the right-hand side of (4.19) (of which there are infinitely many), there exists at least one solution $\widehat\lambda$ of problem $\mathcal P$. In addition, $\alpha>0$ can be taken to be the same for all (that is, infinitely many) solutions. It is clear that each of these neighbourhoods decreases with $\lambda$. Hence the limit relation required in the proposition holds. Proposition 9 is proved. AcknowledgementThe authors are grateful to anonymous referees for their useful comments. 
Citation in format AMSBIB
\Bibitem{ValTik24} |
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