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On summable solutions of a class of nonlinear integral equations on the whole line Kh. A. Khachatryanab, H. S. Petrosyanbc a Yerevan State University b Lomonosov Moscow State University c National Agrarian University of Armenia
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2022, Volume 86, Issue 5, DOI: https://doi.org/10.4213/im9211 
Bibliographic databases:
    § 1. Introduction and problem statementConsider the following class of nonlinear integral equations on the whole line with a monotone Hammerstein–Nemytskii operator:
$$
\begin{equation}
f(x)\,{=}\,G_0(x,f(x))\,{+}\,\mu(x)\int_{-\infty}^\infty K(x-t)G_1(t,f(t))\,dt,\qquad x\,{\in}\,\mathbb{R}\,{:=}\,(-\infty,+\infty),
\end{equation}
\tag{1}
$$
with respect to unknown measurable nonnegative function $f(x)$. Equation (1) arises in various areas of natural science. In particular, such equations arise in mathematical biology (in the theory of spatio-temporal spread of an epidemic), in econometrics (in theory of national income distribution), in astrophysics (in the theory of radiative transfer in spectral lines), (see [1]–[7] and references therein). A distinctive feature of equation (1) is the non-compactness of the corresponding nonlinear Hammerstein–Nemytskii operator in the space of essentially bounded on $\mathbb{R}$ functions. The latter circumstance, the criticality condition for equation (1) (i.e., the presence of trivial (constant) solutions), and unboundedness of the integration domain in (1) lead to certain rather subtle difficulties in constructing nontrivial solutions to equations (1). It should be noted that the general classical principles of fixed points do not work for equation (1) due to the properties of the Hammerstein–Nemytskii operator listed above. In the case when the kernel $K$ is an even conservative function and $0\leqslant\mu(x)\leqslant1$, $x\in\mathbb{R}$, $(1-\mu(x))x\in L_1(\mathbb{R})$, under sufficiently strong restrictions on $G_0$ and $G_1$, equation (1) was studied in [8]. In the case, when there exists $\nu(K):=\int_{-\infty}^\infty x K(x)\,dx\neq0$, and mapping $G_1(t, u)$ is contractive in $u$ uniformly with respect to the variable $t$, under certain conditions on $G_0(x, u)$, equation (1) and the corresponding vector analog (a system of nonlinear integral equations) were studied in [9] and [10]. In these papers, summable and essentially bounded solutions on $\mathbb{R}$ were constructed. In the present paper, we study the question of constructing a nontrivial, nonnegative, essentially bounded and integrable on $\mathbb{R}$ solution of equation (1) in the case $\nu(K)>0$ under essentially different restrictions on $G_0$, $G_1$ and $\mu$, and moreover, we do not assume the condition that the mapping $G_1(t,u)$ is contractive with respect to $u$. At the end of the paper, we present specific applied examples of the kernel $K$ and the functions $G_0, G_1$ and $\mu$. § 2. Auxiliary facts2.1. Basic conditions on kernel $K$ and on function $\mu$With respect to kernel $K$, we assume that the following basic conditions are satisfied: I) $K\in L_1(\mathbb{R})\cap M(\mathbb{R})$, $K(x)>0$, $x\in \mathbb{R}$, $\int_{-\infty}^\infty K(x)\,dx=1$; II) there exists $\nu(K)=\int_{-\infty}^\infty x K(x)\,dx>0$; III) there exists a number $\lambda_1>0$ such that
$$
\begin{equation*}
\int_{-\infty}^\infty K(x)e^{-\lambda x}\,dx<+\infty,\qquad \lambda\in[0, \lambda_1].
\end{equation*}
\notag
$$
Assume that $\mu(x)$ is a continuous function defined on $\mathbb{R}$, and
$$
\begin{equation}
0\leqslant\mu(x)\leqslant1,\qquad x\in \mathbb{R},\quad \mu\not\equiv1.
\end{equation}
\tag{2}
$$
2.2. Conditions on the function $G_1(t,u)$Let $G(u)$ be a continuous function defined on the set $\mathbb{R}^+:=[0,+\infty)$ and satisfying the following conditions: a) there exists a finite derivative $G'(0)>1$ such that b) the function $G(u)$ is upward convex on the set $\mathbb{R}^+$; c) there exists a number $\eta>0$ such that $G(u)\uparrow$ in $u$ on the set $[0,\eta]$, $G(\eta)=\eta$; d) there exist numbers $\varepsilon>0$ and $c>0$ such that
$$
\begin{equation*}
G(u)\geqslant G'(0)u- cu^{1+\varepsilon},\qquad u\in [0,\eta].
\end{equation*}
\notag
$$
Regarding the nonlinearity of $G_1(t,u)$, we assume that the following conditions are hold: 1) for any fixed $t\in\mathbb{R}$ the function $G_1(t,u)\uparrow$ in $u$ on $[0,\eta]$; 2) the double inequality is satisfied
$$
\begin{equation*}
0\leqslant G_1(t,u)\leqslant \eta-G(\eta-u),\qquad u\in [0,\eta],\quad t\in \mathbb{R};
\end{equation*}
\notag
$$
3) the nonlinearity of $G_1(t,u)$ on the set $\mathbb{R}\times [0,\eta]$ satisfies the Caratheodory condition with respect to argument $u$, i.e., for any fixed $u\in [0,\eta]$ the function $G_1(t,u)$ is measurable in $t$ on $\mathbb{R}$ and for almost all $t\in \mathbb{R}$ this function is continuous in $u$ on the segment $[0,\eta]$. 2.3. Diekmann functionBefore imposing the appropriate conditions on the function $G_0(x,u)$, we introduce the well-known Diekmann function for the kernel $K$ (see [1]):
$$
\begin{equation}
L(\lambda):=G'(0) \int_{-\infty}^\infty K(t)e^{-\lambda t}\,dt,\qquad \lambda\in[0,\lambda_1].
\end{equation}
\tag{3}
$$
Note that due to conditions a) and I) On the other hand, according to condition II) Since $L'(\lambda)$ is continuous, there exists $\lambda^*\in(0,\lambda_1]$ such that
$$
\begin{equation}
L'(\lambda)<0\quad\text{for}\quad\lambda\in[0,\lambda^*].
\end{equation}
\tag{4}
$$
In what follows we will assume that Note also that
$$
\begin{equation}
L''(\lambda)=G'(0) \int_{-\infty}^\infty K(t)t^2 e^{-\lambda t}\,dt>0
\end{equation}
\tag{6}
$$
(may be infinity). Hence, the function $L(\lambda)$ is downward convex. It follows from listed properties of the function $L(\lambda)$ that there exists a unique $\sigma\in(0,\lambda^*)$ such that Fix a number $\sigma$ and consider the following function (see [1]):
$$
\begin{equation}
\mathcal{L}(x):=\max \{\eta e^{\sigma x}-M e^{(\delta+\sigma)x}, 0\},\qquad x\in \mathbb{R},
\end{equation}
\tag{8}
$$
where $M>0$ and $\delta\in(0, \lambda^*-\sigma)$ are numerical parameters. Note that the properties of the function $L(\lambda)$ immediately imply that for $\delta\in(0, \lambda^*-\sigma)$ 2.4. Conditions on function $G_0(x,u)$We assume that $G_0(x,u)$ satisfies the following conditions: $\mathrm{i}_1)$ for any fixed $x\in\mathbb{R}$, the function $G_0(x,u)\uparrow$ in $u$ on the segment $[0,\eta]$; $\mathrm{i}_2)$ the function $G_0(x,u)$ on the set $\mathbb{R}\times[0,\eta]$ satisfies the Caratheodory condition with respect to the argument $u$; $\mathrm{i}_3)$ there exists number $\xi\in(0,1)$ such that
$$
\begin{equation}
G_0(x, q_\xi(x))\geqslant q_\xi(x),\quad G_0(x,\eta)\leqslant q_1(x),\qquad x\in\mathbb{R},
\end{equation}
\tag{10}
$$
where
$$
\begin{equation}
q_\alpha(x):=\alpha\mathcal{L}(x)(1-\mu(x)),\qquad \alpha>0,\quad x\in\mathbb{R}.
\end{equation}
\tag{11}
$$
§ 3. The main resultThe following theorem holds. Theorem 1. Let conditions I)–III), 1)–3), $\mathrm{i}_1)$–$\mathrm{i}_3)$, (2) and (5) be satisfied. Then equation (1) has a nonnegative nontrivial bounded and summable solution $f(x)$, and $\lim_{x\to\pm\infty}f(x)=0$. Proof. Step I (a priori estimate for $\mathcal{L}(x)$). First we choose the parameters $M>0$ and $\delta\in(0, \lambda^*-\sigma)$ in such a way that the following inequality holds:
$$
\begin{equation}
\mathcal{L}(x)\leqslant \eta-\eta e^{\sigma x},\qquad x\in (-\infty,0].
\end{equation}
\tag{12}
$$
To this end, consider the following auxiliary function: Note that when $M>\eta$.
On the other hand, for $\delta<\sigma$ the following inequality holds:
$$
\begin{equation*}
\begin{aligned} \, F'(x) &=-2\eta\sigma e^{\sigma x}+M(\sigma+\delta)e^{(\delta+\sigma)x}= e^{\sigma x}\bigl(-2\eta\sigma+M(\sigma+\delta)e^{\delta x}\bigr) \\ &\leqslant e^{\sigma x}\bigl(-2\eta\sigma +M(\sigma+\delta)e^{\ln(\eta/M)}\bigr)=\eta e^{\sigma x}(\delta-\sigma)<0,\qquad x\leqslant \frac{1}{\delta}\ln\frac{\eta}{M}. \end{aligned}
\end{equation*}
\notag
$$
Therefore, $F(x)\downarrow$ on the set $(-\infty, (1/\delta)\ln(\eta/M)]$. From the listed properties of the function $F$, it immediately follows that
$$
\begin{equation}
F(x)>0,\qquad x\in \biggl(-\infty, \frac{1}{\delta}\ln\frac{\eta}{M}\biggr].
\end{equation}
\tag{15}
$$
By direct checking, we can verify that for $M>\eta$ the function $\mathcal{L}(x)$ reaches its maximum at the point
$$
\begin{equation*}
x_{\mathrm{max}}=\frac{1}{\delta}\ln\frac{\eta\sigma}{M(\sigma+\delta)}<0\quad\text{and}\quad \mathcal{L}(x)=0\quad\text{for}\quad x\geqslant\frac{1}{\delta}\ln\frac{\eta}{M}.
\end{equation*}
\notag
$$
Thus, by (15) for $M>\eta$ and $\delta\in(0, \min(\lambda^*-\sigma, \sigma))$, inequality (12) holds. Geometrically, inequality (12) is shown in Fig. 1.
Step II (nonlinear auxiliary equation). In addition to equation (1), consider the following auxiliary integral equation with nonlinearity $G$:
$$
\begin{equation}
\varphi(x)=\int_{-\infty}^\infty K(x-t)G(\varphi(t))\,dt,\qquad x\in \mathbb{R},
\end{equation}
\tag{16}
$$
with respect to the required function $\varphi(x)$.
From the results of [11] it follows that when A) $M>\max\biggl\{\eta,\dfrac{c\eta^{1+\varepsilon}L(\sigma+\delta)}{G'(0)(1-L(\delta+\sigma))}\biggr\}$, B) $\delta\in(0, \min(\varepsilon\sigma, \sigma, \lambda^*-\sigma))$ equation (16) has a nonnegative nontrivial bounded monotonically increasing solution $\varphi(x)$ on $\mathbb{R}$, and
$$
\begin{equation}
\mathcal{L}(x)\leqslant\varphi(x)\leqslant\begin{cases} \eta e^{\sigma x}, &x\leqslant0, \\ \eta, &x>0, \end{cases} \qquad x\in\mathbb{R}.
\end{equation}
\tag{17}
$$
Moreover,
$$
\begin{equation}
\lim_{x\to+\infty}\varphi(x)=\eta,\qquad \eta-\varphi\in L_1(\mathbb{R}^+).
\end{equation}
\tag{18}
$$
It was also proved in [5] that if $h(x)$ is any continuous, monotonically nondecreasing nonnegative and convex upward function on $\mathbb{R}^+$, and for some $m\in\mathbb{N}$ the following integral converges then The listed properties of the function $\varphi(x)$ play an important role in our further reasoning. In what follows we will assume that the parameters $M$ and $\delta$ have the properties A) and B), respectively. Note that the following lower bound holds:
$$
\begin{equation}
\eta-\varphi(x)\geqslant\mathcal{L}(x),\qquad x\in \mathbb{R}.
\end{equation}
\tag{21}
$$
Indeed, by (8), (17) and A), the inequality (21) for $x>0$ is obviously satisfied. Let $x\leqslant0$. Then taking into account (12) and (17), we get
$$
\begin{equation*}
\eta-\varphi(x)\geqslant \eta-\eta e^{\sigma x}\geqslant\mathcal{L}(x).
\end{equation*}
\notag
$$
Step III (successive approximations for equation (1)). Consider the following successive approximations for the main equation (1):
$$
\begin{equation}
\begin{aligned} \, f_{n+1}(x) &=G_0(x,f_n(x)) \nonumber \\ &\qquad+\mu(x)\int_{-\infty}^\infty K(x-t)G_1(t,f_n(t))\,dt,\qquad x\in\mathbb{R},\quad n=0,1,2,\dots, \end{aligned}
\end{equation}
\tag{22}
$$
as the zero approximation we choose By induction on $n$, we prove that:
$\mathrm{k}_1)$ $f_n(x)\uparrow$ in $n, x\in\mathbb{R}$; $\mathrm{k}_2)$ $f_n(x)$ are measurable on $\mathbb{R}$, $n=0,1,2,\dots$; $\mathrm{k}_3)$ $f_n(x)\leqslant\min\{\eta-\varphi(x), \varphi(x)\}$, $n=0,1,2,\dots$, $x\in\mathbb{R}$. The measurability of the zero approximation immediately follows from the definition of the function $q_\xi(x)$. We have from (11), (21), (17) and (2) that
$$
\begin{equation*}
0\leqslant f_0(x)\leqslant\mathcal{L}(x)\leqslant \min\{\eta-\varphi(x), \varphi(x)\},\qquad x\in\mathbb{R}.
\end{equation*}
\notag
$$
Prove that $f_1(x)\geqslant f_0(x)$, $x\in\mathbb{R}$. Taking into account I), (2), 2) and $\mathrm{i}_3)$, we obtain from (22) that
$$
\begin{equation*}
f_1(x)\geqslant G_0(x, q_\xi(x))\geqslant q_\xi(x)=f_0(x),\qquad x\in\mathbb{R}.
\end{equation*}
\notag
$$
Assume now that for some positive integer $n$ the function $f_n(x)$ is measurable, $f_n(x)\geqslant f_{n-1}(x)$, $x\in\mathbb{R}$, and $f_n(x)\leqslant \min\{\eta-\varphi(x), \varphi(x)\}$. Then, by conditions 3), $\mathrm{i}_2)$, I) and (2), the function $f_{n+1}(x)$ will also be measurable. Since the functions $G_0(x,u)$ and $G_1(t,u)$ are monotonous in $u$ and the kernel $\mu(x)K(x-t)$ is nonnegative, by the induction assumption, we have from (22) that
$$
\begin{equation*}
f_{n+1}(x)\geqslant G_0(x, f_{n-1}(x))+\mu(x)\int_{-\infty}^\infty K(x-t)G_1(t,f_{n-1}(t))\,dt=f_n(x).
\end{equation*}
\notag
$$
We now prove that $f_{n+1}(x)\leqslant \min\{\eta-\varphi(x), \varphi(x)\}$. First we prove that $f_{n+1}(x)\leqslant\varphi(x)$, $x\in\mathbb{R}$. Indeed, taking into account the inductive assumption, formulas (2), (10), (16), (17), (21), conditions 2), 1), b) and Jensen’s inequality, we obtain from (22) that
$$
\begin{equation*}
\begin{aligned} \, f_{n+1}(x) &\leqslant G_0(x, \varphi(x))+\mu(x)\int_{-\infty}^\infty K(x-t)G_1(t,\varphi(t))\,dt \\ &\leqslant G_0(x,\eta)+\mu(x)\int_{-\infty}^\infty K(x-t)\bigl(\eta-G(\eta-\varphi(t))\bigr)\,dt \\ &\leqslant q_1(x)+\mu(x)\int_{-\infty}^\infty K(x-t)\bigl(G(\eta)-G(\eta-\varphi(t))\bigr)\,dt \\ &= q_1(x)+ \mu(x)\int_{-\infty}^\infty K(x-t) \biggl(\frac{\varphi(t)}{\eta}G(\eta) +\frac{\eta-\varphi(t)}{\eta}G(0) \\ &\qquad +\frac{\eta-\varphi(t)}{\eta}G(\eta)+\frac{\varphi(t)}{\eta}G(0)-G(\eta-\varphi(t))\biggr)\,dt \\ &\leqslant q_1(x)+\mu(x)\int_{-\infty}^\infty K(x-t) \bigl(G(\varphi(t))+G(\eta-\varphi(t))-G(\eta-\varphi(t))\bigr)\,dt \\ &=q_1(x)+\mu(x)\varphi(x)\leqslant(1-\mu(x))\varphi(x)+\mu(x)\varphi(x)=\varphi(x). \end{aligned}
\end{equation*}
\notag
$$
Now we prove that $f_{n+1}(x)\leqslant \eta-\varphi(x),\,x\in\mathbb{R}$. By conditions 2), 1), $\mathrm{i}_1)$ and formulas (2), (10), taking into consideration inductive hypothesis and relations (16), (17), (21), we have from (22) that
$$
\begin{equation*}
\begin{aligned} \, f_{n+1}(x) &\leqslant G_0(x, \eta-\varphi(x))+\mu(x)\int_{-\infty}^\infty K(x-t)G_1(t,\eta-\varphi(t))\,dt \\ &\leqslant G_0(x,\eta)+\mu(x)\int_{-\infty}^\infty K(x-t)(\eta-G(\varphi(t)))\,dt \\ &\leqslant \mathcal{L}(x)(1-\mu(x))+\mu(x)(\eta-\varphi(x)) \\ &\leqslant (\eta-\varphi(x))(1-\mu(x))+\mu(x)(\eta-\varphi(x))=\eta-\varphi(x). \end{aligned}
\end{equation*}
\notag
$$
Thus, it follows from $\mathrm{k}_1)$–$\mathrm{k}_3)$ that the sequence $\{f_n(x)\}_{n=0}^\infty$ of measurable functions has a pointwise limit when $n\to\infty$: $\lim_{n\to\infty}f_n(x)=f(x)$, and the limit function satisfies the following double inequality:
$$
\begin{equation}
q_\xi(x)\leqslant f(x)\leqslant \min\{\eta-\varphi(x), \varphi(x)\},\qquad x\in\mathbb{R}.
\end{equation}
\tag{24}
$$
From Levi’s (see [12]) and Krasnoselsky’s (see [13]) limit theorems it follows that $f(x)$ satisfies equation (1) almost everywhere on $\mathbb{R}$. Since $\varphi\in L_1(\mathbb{R}\setminus \mathbb{R}^+)$, $ \eta- \varphi\in L_1(\mathbb{R}^+)$, $\varphi(-\infty)=0$, $\varphi(+\infty)=\eta$, then (24) yields that $f(\pm\infty)=0$ and $f\in L_1(\mathbb{R})\cap M(\mathbb{R})$. Theorem 1 is completely proved. Remark 1. By inequalities (24) and (17), the proof of the main result implies a stronger limit relation for solution $f(x)$ when $x\to-\infty$. Namely, for any number $\rho\in(0,\sigma)$ Moreover, $e^{-\rho x}f(x)\in L_1(-\infty,0)$. In Fig. 2, the preliminary zone for finding the constructed solution of equation (1) is shaded. Remark 2. In the proof of the theorem at the second step, it was noted that under condition (19), the integral (20) converges. It immediately follows from the estimate (24) that under additional condition (19), the solution of equation (1) possesses the following property: (for the properties of the function $h$ see the proof of Theorem 1, Step II). § 4. Examples of function $K$, $\mu$, $G_0$ and $G_1$In applications (see [1], [3], [5]) the following specific kernel representations $K(x)$ and function $\mu(x)$ occur: $\mathrm{p}_1)$ $K(x)=(1/\sqrt{\pi}\,)e^{-(x-l)^2}$, $x\in\mathbb{R}$, where $l>0$ is an arbitrary parameter; $\mathrm{p}_2)$ $K(x)=\int_a^b e^{-|x-l|s}Q(s)\, ds$, $x\in\mathbb{R}$, where $l>0$ is a parameter, and $Q\in C[a,b)$, $Q(s)>0$, $0<a\leqslant s\leqslant b\leqslant+\infty$, and $\mathrm{\beta}_1)$ $\mu(x)=1-e^{-|x|}$, $x\in\mathbb{R}$; $\mathrm{\beta}_2)$ $\mu(x)=e^{-x^2}$, $x\in\mathbb{R}$. First, we provide several examples of the function $G(u)$, then, using the construction of the numbers $\eta$ and $\sigma$, we will give the corresponding examples of the nonlinearities $G_0$ and $G_1$. In the mathematical theory of the spatio-temporal spread of an epidemic, the following particular examples of the nonlinearity of $G(u)$ occur (see [1]–[4]): $\mathrm{g}_1)$ $G(u)=\gamma(1-e^{-u})$, $u\in\mathbb{R}^+$, $\gamma>1$ is a numerical parameter; $\mathrm{g}_2)$ $G(u)=2u-u^2$, $u\in\mathbb{R}^+$. The parameter $\gamma>1$ has a specific biological meaning. In the above theory, the inequality $\gamma>1$ is called the threshold condition. The latter is the critical number of infected persons, beyond which the epidemic cannot be stopped. It is easy to check that for $\mathrm{g}_1)$ all conditions a)–d) are automatically satisfied. Take a closer look at the example $\mathrm{g}_2)$. Note that in this case $\eta=1$, $G'(0)=2$, $G\uparrow$ on the interval $[0,1]$, $G(0)=0$. We now construct the number $\sigma$ for the concrete kernel of the form $\mathrm{p}_1)$. In this case, the Diekmann function has the following form:
$$
\begin{equation*}
L(\lambda)=\frac{2}{\sqrt{\pi}} \int_{-\infty}^\infty e^{-(x-l)^2}e^{-\lambda x}\,dx,\qquad \lambda\geqslant0,
\end{equation*}
\notag
$$
or after certain calculations
$$
\begin{equation*}
L(\lambda)=2e^{\lambda^2/4-l\lambda},\qquad \lambda\geqslant0.
\end{equation*}
\notag
$$
Since $L'(\lambda)=(\lambda-2l)e^{\lambda^2/4-l\lambda}$, then $L(\lambda)\downarrow$ on the interval $[0, 2l]$. We have
$$
\begin{equation*}
L''(\lambda)=\biggl(1+\frac{1}{2}(\lambda-2l)^2 \biggr)e^{\lambda^2/4-l\lambda}>0,
\end{equation*}
\notag
$$
which implies, in particular, that $L(\lambda)$ is downward convex. It is obvious that $L(0)\,{=}\,2$, $L(2l)=2e^{-l^2}<1$ for $l>\sqrt{\ln2}$. Therefore, when $l\in(\sqrt{\ln2},+\infty)$, there is a unique number $\sigma\in(0,2l)$ such that $L(\sigma)=1$. For example, in the case when $l=1$, after appropriate calculations, we find $\sigma\approx0.893$. Finally, we provide concrete examples of the functions $G_0(x,u)$ and $G_1(x,u)$. Examples. As $G_0(x,u)$ the following functions we can be considered: $\mathrm{a}_1)$ $G_0(x,u)=q_1(x)u/(u+q_{\varepsilon_0}(x))$, $u\geqslant0$, $x\in \mathbb{R}$, where $\varepsilon_0\in (0, 1-\xi]$ is an arbitrary number; $\mathrm{a}_2)$ $G_0(x,u)=q_1(x)u/(u+q_{\varepsilon_0}(x))+a(x)u^p$, $u\geqslant0$, $x\in \mathbb{R}$, $p>1$ is a parameter, $\varepsilon_0\in (0, 1-\xi]$, $a(x)$ is an arbitrary nonnegative continuous on $\mathbb{R}$ function and
$$
\begin{equation*}
a(x)\leqslant \frac{q_1(x)q_{\varepsilon_0}(x)}{\eta^{p+1}+\eta^pq_{\varepsilon_0}(x)},\qquad x\in \mathbb{R}.
\end{equation*}
\notag
$$
It is easy to prove that the above examples $\mathrm{a}_1)$ and $\mathrm{a}_2)$ satisfy conditions $\mathrm{i}_1)$–$\mathrm{i}_3)$. As $G_1(t,u)$, we can consider the following functions: $\mathrm{b}_1)$ $G_1(t,u)=b(t)(\eta-G(\eta-u))$, $u\geqslant0$, $t\in\mathbb{R}$, where $b(t)$ is any continuous function on $\mathbb{R}$ and $\mathrm{b}_2)$ $G_1(t,u)=(u/(u+S(t)))(\eta-G(\eta-u))$, $u\geqslant0$, $t\in\mathbb{R}$, where $S(t)\geqslant0$, $t\in\mathbb{R}$ and $S\in C(\mathbb{R})$. Direct checking will prove that conditions 1)–3) hold for examples $b_1)$ and $b_2)$. For example, when $G(u)=2u-u^2$, as $G_1(t,u)$ we can choose the function $G_1(t,u)=b(t)u^2$, $u\geqslant0$, $t\in\mathbb{R}$. § 5. On the uniqueness of the solution of equation (1) in an important particular caseAssume that nonlinearity $G_0(x,u)$ admits a representation $a_1)$, where $\varepsilon_0=1- \xi$, $G_1(t,u)$ is defined according to $\mathrm{b}_1)$, and $\mu(x)$, in addition to (2), satisfies the following additional restriction: In this case, equation (1) takes the following form:
$$
\begin{equation}
f(x)=\frac{q_1(x)f(x)}{f(x)+q_{1-\xi}(x)}+\mu(x) \int_{-\infty}^\infty K(x-t)b(t)(\eta-G(\eta-f(t)))\,dt,\qquad x\in\mathbb{R}.
\end{equation}
\tag{26}
$$
Below we will show that the following statement holds. Theorem 2. Under conditions a)–d), (2) and (25), if $\xi\in(M_0,1)$, then equation (26) cannot have more than one solution in the following class of measurable and bounded functions on $\mathbb{R}$: where Proof. Assume the opposite: equation (26) has two solutions $f$ and $\widetilde{f}$ from the class $\mathfrak{M}$. Below we verify that
$$
\begin{equation}
\varkappa:=\sup_{x\in\mathbb{R}}e^{-\sigma x}|f(x)-\widetilde{f}(x)|<+\infty,
\end{equation}
\tag{28}
$$
where the number $\sigma$ is determined from (7). Consider the following possibilities:
$\mathrm{r}_1)$ $x\geqslant(1/\delta)\ln(\eta/M)$; $\mathrm{r}_2)$ $x<(1/\delta)\ln(\eta/M)$, where $\delta\in(0, \min\{\sigma\varepsilon, \sigma, \lambda^*-\sigma\})$. In the case $\mathrm{r}_1)$, by the triangle inequality, we have
$$
\begin{equation*}
e^{-\sigma x}|f(x)-\widetilde{f}(x)|\leqslant 2\eta e^{-\sigma x}\leqslant 2\eta e^{-(\sigma/\delta)\ln(\eta/M)}=2\eta\biggl(\frac{M}{\eta}\biggr)^{\sigma/\delta}.
\end{equation*}
\notag
$$
Now consider $\mathrm{r}_2)$ $x<(1/\delta)\ln(\eta/M)$. Since $f, \widetilde{f}\in \mathfrak{M}$ and taking into account the definition of the function $\mathcal{L}(x)$, by (25), we obtain the following inequality:
$$
\begin{equation*}
\begin{aligned} \, &-\eta\bigl(1-\xi(1-\mu(x))\bigr)e^{\sigma x} - \xi M(1-\mu(x)) e^{(\delta+\sigma) x}\leqslant f(x)-\widetilde{f}(x) \\ &\qquad \leqslant\eta\bigl(1-\xi(1-\mu(x))\bigr)e^{\sigma x} + \xi M(1-\mu(x)) e^{(\delta+\sigma) x}, \end{aligned}
\end{equation*}
\notag
$$
which implies that
$$
\begin{equation*}
e^{-\sigma x}|f(x)-\widetilde{f}(x)|\leqslant \xi M+\eta(1-\xi+\xi M_0), \qquad x<\frac{1}{\delta}\ln\frac{\eta}{M}.
\end{equation*}
\notag
$$
Thus
$$
\begin{equation}
\varkappa\leqslant \max\biggl\{2\eta\biggl(\frac{M}{\eta}\biggr)^{\sigma/\delta},\, \xi M+\eta(1-\xi+\xi M_0)\biggr\}.
\end{equation}
\tag{29}
$$
From properties a)–d) of the function $G(u)$, it easily follows that for arbitrary $u_1, u_2\in[0,\eta]$ the following inequality holds: Taking into account (30), the definition of the function $L(\lambda)$ and the fact that $f, \widetilde{f}\in \mathfrak{M}$, we have from (26) that
$$
\begin{equation*}
\begin{aligned} \, e^{-\sigma x}|f(x)-\widetilde{f}(x)| &\leqslant \frac{q_1(x)q_{1-\xi}(x)e^{-\sigma x}}{(q_\xi(x)+q_{1-\xi}(x))^2}|f(x)-\widetilde{f}(x)| \\ &\qquad+ e^{-\sigma x}\mu(x)G'(0)\int_{-\infty}^\infty K(x-t)|f(t)-\widetilde{f}(t)|\,dt \\ &\leqslant (1-\xi)\varkappa+ e^{-\sigma x}G'(0)\varkappa M_0\int_{-\infty}^\infty K(x-t)e^{\sigma t}\,dt \\ &= (1-\xi)\varkappa+\varkappa G'(0)e^{-\sigma x}M_0\int_{-\infty}^\infty K(y)e^{-\sigma (y-x)}\, dy \\ &=(1-\xi+M_0)\varkappa, \end{aligned}
\end{equation*}
\notag
$$
and therefore, we get or $\varkappa(\xi-M_0)\leqslant0$. Since $\xi\in(M_0, 1)$, the latter inequality yields that $\varkappa=0$. The latter means that $f(x)=\widetilde{f}(x)$ almost everywhere on $\mathbb{R}$. Theorem 2 is proved. Remark 4. Unfortunately, the problem of the uniqueness of a solution of equation (1) in $\mathfrak{M}$ under general restrictions on $G_0$ and $G_1$ is still an open problem.
Citation in format AMSBIB
\Bibitem{KhaPet22} |
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