\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 316, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2016 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2016/316\hfil Hardy operator in weighted Lebesgue spaces] {Connection of two nonlinear differential equations with a two-dimensional Hardy operator in weighted Lebesgue spaces \\ with mixed norms} \author[R. A. Bandaliyev \hfil EJDE-2016/316\hfilneg] {Rovshan A. Bandaliyev} \address{Rovshan A. Bandaliyev \newline Department of Mathematical Analysis, Institute of Mathematics and Mechanics of ANAS, Az 1141, B. Vahabzade str., 9, Baku, Azerbaijan} \email{bandaliyev.rovshan@math.ab.az} \thanks{Submitted February 5, 2016. Published December 12, 2016.} \subjclass[2010]{46B50, 47B38, 34L30} \keywords{Two-dimensional Hardy operator; weight functions; \hfill\break\indent Lebesgue spaces with mixed norm} \begin{abstract} In this article we study a connection between two nonlinear differential equations with a two-dimensional Hardy operator in weighted Lebesgue spaces with mixed norm. In particular, it is proved that the weight functions which are the coefficients of nonlinear differential equations are included in the estimate of the two-dimensional Hardy operator in this space. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} In the modern theory of the equations of mathematical physics the functional analysis methods, founded in the classical works of Hilbert, are widely applied. In the study of elliptic equations, embedding theorems play an important role, and been studied by several mathematicians; see for example \cite{VGM}. The study of the embedding theorems in arbitrary open sets in $\mathbb{R}^n$ requires investigation of the multidimensional Hardy operator, and there is a need an estimates of this operator in various function spaces, among which the weighted Lebesgue spaces are very important. On the other hand, the multidimensional Hardy operator has numerous applications in the spectral theory of operators, in the theory of partial differential equations, in the theory of integral equations, in the theory of function spaces and etc. see \cite{VGM,KP,EE} and the references therein). In the one-dimensional case, we refer to the well-known monograph \cite{HLP}. A systematic study of the Hardy inequality was carried out in \cite{PRB,T1,T2,Muck,JSB,KP} and references therein. Note that in \cite{PRB,T1,T2} the Hardy inequality in its equivalent differential form \begin{equation} \begin{gathered} \Big(\int_0^\infty |f(x)|^q\,w(x)\, dx\Big)^{1/q} \le C \Big(\int_0^\infty |f'(x)|^p\,u(x)\, dx\Big)^{1/p}\\ f(0)= f(+0)= 0 \end{gathered} \label{eq:1} \end{equation} was connected with the Euler-Lagrange differential equation. In particular, in \cite{T1} and \cite{T2} a necessary and sufficient condition on weight functions for validity \eqref{eq:1} was obtained in the case $p = q$. The study of the case with different parameters p and q was started in \cite{JSB} and developed in \cite{KP}, \cite{VGM} and etc. In the case $p\ne q$ the other type criterion on weight functions for validity \eqref{eq:1} was obtained in \cite{PG}. Namely, in \cite{PG} the inequality \eqref{eq:1} was connected with nonlinear ordinary differential equation in weighted Lebesgue spaces. In this paper we prove a connection between two nonlinear differential equations with a two-dimensional Hardy operator in weighted Lebesgue spaces with mixed norm. In other words, we prove that the weight function involved in the definition of the weighted Lebesgue spaces with mixed norm connects these differential equations with a two-dimensional Hardy operator in this space. This article is organized as follows. Section 2 contains some preliminaries along with the standard ingredients used in the proofs. The main results are stated and proved in Section 3. \section{Preliminaries} \label{section2} Let $1< p_1, p_2 < \infty$ and $\rho_1(t)$, $\rho_2(t)$ be weight functions, that are, Lebesgue measurable, positive and a.e. finite functions on $(0, \infty)$. Let $p'_i= \frac{p_i}{p_i- 1}$ and $q'_i= \frac{q_i}{q_i- 1}$, where $i= 1,2$. Throughout this article we assume that the functions are Lebesgue measurable. Now we define the weighted Lebesgue spaces with mixed norm. Let $(0,\infty)^2= (0, \infty)\times (0, \infty)$ and $f$ is an arbitrary real function on $(0,\infty)^2$. \begin{definition}[\cite{BP}] \rm We denote by $L_{(p_1, p_2, \rho_1, \rho_2)}[(0,\infty)^2]$ the space of all measurable functions $f$ on $(0,\infty)^2$ such that the mixed norm $$ \|f\|_{L_{(p_1, p_2, \rho_1, \rho_2)}[(0,\infty)^2]} = \Big(\int_0^\infty (\int_0^ \infty |f(x_1, x_2)|^{p_1} \rho_1(x_1) dx_1 )^{p_2/p_1} \rho_2(x_2) dx_2\Big)^{1p_2} $$ is finite. \end{definition} The variable Lebesgue space with mixed norm was introduced and studied in \cite{AB}. We denote by $C^{1}(0, \infty)$ the space of continuously differentiable functions on $(0, \infty)$. The set of all locally absolutely continuous functions on $(0, \infty)$ is denoted by $AC_{\rm loc} (0,\infty)$. Let $v_i(t), \omega_i(t)$ be weight functions defined on $(0, \infty)$, $v_i\in C^{1}(0, \infty)$ and $\lambda_i> 0$, where $i= 1,2$. We consider two nonlinear differential equations \begin{gather} \lambda_1\frac{d}{dt}\Big([v_1(t)]^{q_1/p_1} [y'_1(t)]^{q_1/p'_1}\Big) + \omega_1(t) [y_1(t) ]^{q_1/p'_1}= 0, \label{e2.1}\\ \lambda_2\frac{d}{dt}\Big([v_2(t)]^{q_2/p_2} [y'_2(t) ]^{q_2/p'_2}\Big) + \omega_2(t) [y_2(t) ]^{q_2/p'_2}= 0, \label{e2.2} \end{gather} where \begin{equation} y_i(t)> 0,\quad y'_i(t)> 0\quad ( t> 0),\quad y'_i\in AC_{\rm loc}(0,\infty),\quad i= 1,2. \label{e2.3} \end{equation} We say that a pair of functions $(y_1(t),y_2(t))$ is a solution of problem \eqref{e2.1}-\eqref{e2.3}, if these functions a.e. on $(0, \infty)$ satisfy the equations \eqref{e2.1}, \eqref{e2.2} and condition \eqref{e2.3}. We denote $y_i(0)= \lim_{t\to +0} y_i(t)$, $i= 1,2$. First we prove the following theorem. \begin{theorem} \label{thm2.3} Let $1< p_i\le q_i< \infty$ and $v_i(t), \omega_i(t)$ be weight functions on $(0, \infty)$, $v_i\in C^{1}(0, \infty)$, $\lambda_i> 0$, where $i= 1,2$. If the problem \eqref{e2.1}-\eqref{e2.3} has a solution $(y_1(t),y_2(t))$ then the inequality \begin{equation} \label{e2.4} \|u\|_{L_{(q_1, q_2, \omega_1, \omega_2)}[(0,\infty)^2]} \le \lambda_1^{1/q_1}\lambda_2^{1/q_2} \big\|\frac{\partial^2 u}{\partial x_1 \partial x_2} \big\|_{L_{(p_1, p_2, v_1, v_2)}[(0,\infty)^2]} \end{equation} holds, where $u:(0,\infty)^2\mapsto R$ is an arbitrary absolutely continuous function of two variables satisfying condition \begin{equation} \label{e2.5} \begin{gathered} u(x_1,0)= \lim_{t_2\to +0} u(x_1,t_2)= 0\\ u(0, x_2)= \lim_{t_1\to +0} u(t_1, x_2)= 0. \end{gathered} \end{equation} \end{theorem} \begin{proof} It is well known that for any absolutely continuous function of two variables the representation \begin{equation} \label{e2.6} \begin{aligned} u(x_1, x_2) &= u(0, 0) + \int_0^{x_1} \frac{\partial u(\alpha_1, 0)}{\partial \alpha_1} d\alpha_1 + \int_0^{x_2} \frac{\partial u(0, \alpha_2)}{\partial \alpha_2} d\alpha_2 \\ &\quad + \int_0^{x_1} \int_0^{x_2} \frac{\partial^2 u(\alpha_1,\alpha_2)}{\partial \alpha_1 \partial \alpha_2} d\alpha_1 d\alpha_2 \end{aligned} \end{equation} is valid (see \cite{JM}). Obviously, condition \eqref{e2.5} implies $u(0, 0)= \lim_{\stackrel{t_1\to +0}{t_2\to +0}} u(t_1, t_2)= 0$. Therefore, by \eqref{e2.6} and \eqref{e2.5}, we have \[ u(x_1, x_2)= \int_0^{x_1} \int_0^{x_2} \frac{\partial^2 u(\alpha_1,\alpha_2)} {\partial\alpha_1 \partial \alpha_2} d\alpha_1 d\alpha_2. \] Note that the last integral representation defines the two-dimensional Hardy operator. Let a vector function $y(x_1, x_2)= (y_1(x_1), y_2(x_2))$ be a solution of \eqref{e2.1}-\eqref{e2.3}. Then by H\"{o}lder's inequality, we have \begin{align*} &\Big|u(x_1, x_2)\Big|^{q_1}\omega_1(x_1) \\ &= \Big|\int_0^{x_1} \int_0^{x_2} \frac{\partial^2 u}{\partial t_1 \partial t_2} dt_1 dt_2\Big|^{q_1}\omega_1(x_1) \\ &\le \Big(\int_0^{x_1} \int_0^{x_2} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big| dt_1 dt_2\Big)^{q_1} \omega_1(x_1 ) \\ &= \Big(\int_0^{x_1} \int_0^{x_2} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2} \big|[y'_1(t_1)]^{1/p_1'} [y'_1(t_1)]^{-1/p_1'} dt_1\, dt_2\Big)^{q_1} \omega_1 (x_1 ) \\ &=\Big[\int_0^{x_1} \Big(\int_0^{x_2} \big|\frac{\partial^2 u} {\partial t_1 \partial t_2}\big| [y'_1(t_1)]^{-1/p_1'} dt_2\Big) [y'_1(t_1)]^{1/p_1'} dt_1\Big]^{q_1}\omega_1 (x_1 ) \\ &\le \Big(\int_0^{x_1} y'_1(t_1) dt_1\Big)^{q_1/p_1'} \Big(\int_0^{x_1} \Big(\int_0^{x_2} \big| \frac{\partial^2 u}{\partial t_1 \partial t_2}\big|[y'_1(t_1) ]^{-1/p_1'} dt_2\Big)^{p_1} dt_1\Big)^{q_1/p_1}\omega_1(x_1 ) \\ &\le \omega_1(x_1 )\,(y_1(x_1))^{q_1/p_1'} \Big(\int_0^{x_1} (\int_0^{x_2} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2} \big|[y'_1(t_1) ]^{-1/p_1'} dt_2)^{p_1} dt_1\Big)^{q_1/p_1} \\ &=-\lambda_1\frac{d}{dx_1}\Big([v_1(x_1)]^{q_1/p_1}[y'_1(x_1)]^{ \frac{q_1}{p'_1}}\Big) \\ &\quad\times \Big(\int_0^{x_1} \Big(\int_0^{x_2} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big|[y'_1(t_1) ]^{-1/p_1'} dt_2\Big)^{p_1} dt_1\Big)^{q_1/p_1} \\ &= \Big( \int_0^{x_1} \Big[-\lambda_1\frac{d}{dx_1} \Big( [v_1(x_1)]^ {\frac{q_1}{p_1}} [y'_1(x_1)]^{ \frac{q_1}{p'_1}}\Big) \Big]^{p_1/q_1} \\ &\quad\times \Big(\int_0^{x_2} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big| [y'_1(t_1) ]^{-1/p_1'} dt_2 \Big)^{p_1} dt_1 \Big)^{q_1/p_1} . \end{align*} %\label{e2.7} Integrating both with respect to $x_1$ and applying the generalized Minkowskii inequality, we obtain \begin{align*} &\Big(\int_0^\infty\left|u(x_1,x_2)\right|^{q_1}\omega_1(x_1)dx_1 \Big)^{p_1/q_1} \\ &\le \Big\{ \int_0^\infty \Big(\int_0^{x_1} \Big[-\lambda_1\frac{d}{dx_1} ( [v_1(x_1)]^{q_1/p_1} [y'_1(x_1)]^{q_1/p'_1} ) \Big]^{p_1/q_1} \\ &\quad\times \Big(\int_0^{x_2} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big| [y'_1(t_1) ]^{-1/p_1'} dt_2 \Big)^{p_1} dt_1 \Big)^{q_1/p_1} dx_1 \Big\}^{p_1/q_1} \\ &\le \int_0^\infty \Big(\int_{t_1}^{\infty} -\lambda_1\frac{d}{dx_1} \big([v_1(x_1)]^{q_1/p_1}[y'_1(x_1)]^{q_1/p'_1}\big)\\ &\quad \times \Big(\int_0^{x_2} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big| [y'_1(t_1) ]^{-1/p_1'} dt_2 \Big)^{q_1} dx_1\Big)^{p_1/q_1} dt_1 \\ &= \int_0^\infty \Big(\int_0^{x_2} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big| [y'_1(t_1) ]^{-1/p_1'} dt_2)^{p_1} \\ &\quad\times \Big(\int_{t_1}^{\infty} -\lambda_1\frac{d}{dx_1} ([v_1(x_1)]^{q_1/p_1}[y'_1(x_1)]^{q_1/p'_1}) dx_1 \Big)^{p_1/q_1} dt_1 \\ &\le \int_0^\infty \Big(\int_0^{x_2} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big| [y'_1(t_1) ]^{-1/p_1'} dt_2\Big)^{p_1} [\lambda_1 ([v_1(t_1)]^{q_1/p_1}[y'_1(t_1)]^{ \frac{q_1}{p'_1}})]^{p_1/q_1} dt_1 \\ &= \lambda_1^{p_1/q_1} \int_0^\infty \Big(\int_0^{x_2} \left|\frac{\partial^2 u}{\partial t_1 \partial t_2}\right| dt_2\Big)^{p_1} [y'_1(t_1) ]^{-p_1/p_1'} v_1(t_1) [y'_1(t_1)]^{p_1/p'_1} dt_1 \\ &= \lambda_1^{p_1/q_1} \int_0^\infty \Big(\int_0^{x_2} \left|\frac{\partial^2 u}{\partial t_1 \partial t_2}\right| dt_2\Big)^{p_1} v_1(t_1) dt_1. \end{align*} We have $$ \Big(\int_0^\infty|u(x_1,x_2)|^{q_1}\omega_1(x_1)dx_1 \Big)^{1/q_1} \le \lambda_1^{1/q_1} \Big(\int_0^\infty (\int_0^{x_2} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big| dt_2)^{p_1} v_1(t_1) dt_1\Big)^{1/p_1}. $$ Again applying the generalized Minkowskii inequality, we obtain \begin{align*} &\Big(\int_0^\infty \Big(\int_0^{x_2} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big| dt_2\Big)^{p_1} v_1(t_1) dt_1\Big)^{1/p_1} \\ &\le \int_0^{x_2}\Big(\int_0^\infty \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big|^{p_1} v_1(t_1) dt_1\Big)^{1/p_1} dt_2. \end{align*} Thus, $$ \Big(\int_0^\infty|u(x_1,x_2)|^{q_1}\omega_1(x_1)dx_1\Big)^{1/q_1} \le \lambda_1^{1/q_1} \int_0^{x_2}\big(\int_0^\infty \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big|^{p_1} v_1(t_1) dt_1\Big)^{1/p_1} dt_2. $$ From the above estimate we obtain \begin{align*} &\Big(\int_0^\infty|u(x_1,x_2)|^{q_1}\omega_1(x_1)dx_1\Big)^{q_2/q_1} \omega_2(x_2) \\ &\le \lambda_1^{q_2/q_1} \omega_2(x_2) \Big(\int_0^{x_2} (\int_0^\infty \big|\frac{\partial^2 u}{\partial t_1 \partial t_2} \big|^{p_1} v_1(t_1) dt_1)^{1/p_1} dt_2\Big)^{q_2}. \end{align*} Analogously, we have the estimate \begin{align*} &\omega_2(x_2) \Big(\int_0^{x_2} \Big(\int_0^\infty \big|\frac{\partial^2 u}{\partial t_1 \partial t_2}\big|^{p_1} v_1(t_1) dt_1\Big)^{1/p_1} dt_2\Big)^{q_2} \\ &\le \Big( \int_0^{x_2} \Big[-\lambda_2\frac{d}{dx_2} \Big( [v_2(x_2)]^{q_2/p_2} [y'_2(x_2)]^{q_2/p'_2} \Big) \Big]^{p_2/q_2} \\ &\quad\times \Big(\int_0^{\infty} \big|\frac{\partial^2 u}{\partial t_1 \partial t_2} \big|^{p_1} v_1(t_1) [y'_2(t_2) ]^{-1/p_2'} dt_1 \Big)^{p_2/p_1} dt_2 \Big)^{q_2/p_2} . \end{align*} Therefore, \begin{align*} &\Big(\int_0^\infty\left|u(x_1,x_2)\right|^{q_1}\omega_1(x_1)dx_1\Big)^{q_2/q_1} \omega_2(x_2) \\ &\le \lambda_1^{q_2/q_1} \Big( \int_0^{x_2} \Big[-\lambda_2\frac{d}{dx_2} \Big( [v_2(x_2)]^{q_2/p_2} [y'_2(x_2)]^{ \frac{q_2}{p'_2}} \Big) \Big]^{p_2/q_2} \\ &\quad\times \Big( \int_0^{\infty}\big| \frac{\partial^2 u}{\partial t_1 \partial t_2} \big|^{p_1} v_1(t_1) [y'_2(t_2) ]^{-\frac {1}{p_2'}} dt_1 \Big)^{p_2/p_1} dt_2 \Big)^{q_2/p_2} . \end{align*} Integrating both sides of the estimate with respect to $x_2$ and applying the generalized Minkowskii inequality, we finally obtain \eqref{e2.4}. This completes the proof. \end{proof} We put \begin{equation} \label{e2.8} M_i= \frac{p'_i}{q_i}\inf_g\sup_{t> 0} \frac{1}{g_i(t)- \int_0^{t} (v_i(s))^{1- p'_i} ds} \int_0^{t} \omega_i(s) (g_i(s))^{\frac{q_i}{p'_i}+ 1} ds, \end{equation} where the infimum is taken over the class of all measurable functions $g$ such that $$ g_i(t)> \int_0^{t} (v_i(s))^{1- p'_i} ds,\quad \text{for all $t> 0$ and $i= 1,2$}. $$ The following lemma establishes a connection between problem \eqref{e2.1}-\eqref{e2.3} and $M_i$ $(i=1, 2)$. \begin{lemma} \label{lem2.4} Let $\lambda_i> 0$ be given numbers satisfying the conditions of Theorem \ref{thm2.3} and the expressions $M_i$ be defined by \eqref{e2.8}, where $i= 1,2$. Suppose $v_i$ and $\omega_i$ are weight functions on $(0, \infty)$ and the derivative $v'_i(t)$ exists for all $\;t\in (0, \infty)$. Then the following statements are equivalent: \begin{itemize} \item[(a)] if the problem \eqref{e2.1}-\eqref{e2.3} has a solution with a locally absolutely continuous first derivative, then $\lambda_i \ge M_i$; \item[(b)] if $M_i< +\infty$, then the problem \eqref{e2.1}-\eqref{e2.3} has a solution for every $\lambda_i > M_i$, where $i= 1,2$. \end{itemize} \end{lemma} \begin{proof} Assume that (a) holds. Let $y(t)= (y_1(t), y_2(t))$ be a solution of \eqref{e2.1}-\eqref{e2.3}. Let us take $w_i= \frac{y_i}{y'_i} v_i^{1- p'_i}$. Then the functions $w_1(t), w_2(t )$ are positive solutions of the equations \begin{gather} \label{e2.9} w'_1(t)= \frac{p'_1}{q_1\lambda_1} \omega_1(t)(w_1(t))^{\frac {q_1}{p'_1}+ 1} + [v_1(t)]^{1- p'_1}, \\ \label{e2.10} w'_2(t)= \frac{p'_2}{q_2\lambda_2}\,\omega_2(t)(w_2(t))^{\frac {q_2}{p'_2}+ 1} + [v_2(t)]^{1- p'_2}. \end{gather} By \eqref{e2.9} and \eqref{e2.10}, we obtain \begin{equation} \label{e2.11} w_i(t) \ge \int_0^{t} w_i'(s)ds = \frac{p'_i}{q_i\lambda_i} \int_0^{t} \omega_i(s)(w_i(s))^{\frac {q_i}{p'_i}+ 1} ds + \int_0^{t} [v_i(s)]^{1- p'_i} ds, \end{equation} for $i= 1,2$. This implies $w_i(t)\ge \int_0^{t} [v_i(s) ]^{1- p'_i} ds$ and \begin{equation} \label{e2.12} \lambda_i\ge \frac{p'_i}{q_i}\frac{1}{w_i(t) - \int_0^{t} (v_i(s))^{1- p'_i} ds} \int_0^{t} \omega_i(s) (w_i(s))^{\frac{q_i}{p'_i}+ 1} ds. \end{equation} From this inequality and \eqref{e2.8} we obtain that $\lambda_i\ge M_i$ and the proof of (a) implies (b) is complete. Now let (b) hold. Let us fix $\lambda_i> M_i$, where $i= 1,2$. By the definition of $M_i$ there exists a measurable functions $g_i(t)$ such that \begin{equation} \label{e2.13} g_i(t)\ge \int_0^{t} (v_i(s))^{1- p'_i} ds + \frac{p'_i}{q_i\,\lambda_i} \int_0^{t} \omega_i(s) (g_i(s))^{\frac{q_i}{p'_i}+ 1} ds. \end{equation} We define a sequence of functions $w_{n,i}(t)$ ($i= 1,2$) by setting \begin{equation} \label{e2.14} \begin{gathered} w_{0,i}(t)= g_i(t), \\ w_{n+1, i}(t)= \int_0^{t} (v_i(s))^{1- p'_i} ds + \frac{p'_i}{q_i\lambda_i} \int_0^{t} \omega_i(s) (w_{n,i}(s))^{\frac{q_i}{p'_i}+ 1} ds, \end{gathered} \end{equation} where $n= 0, 1, 2,\ldots$. Also \eqref{e2.13} implies $w_{0, i}(t)\ge w_{1,i}(t)$. We put $w_{n- 1,i}(t)\ge w_{n,i}(t)$. Now we prove that the sequences $\{w_{n,1}(t)\}$ and $\{w_{n,2}(t)\}$ are non-increasing with respect to $n$ on $t\in (0, \infty)$. We have $$ w_{n,i}(t)- w_{n+1,i}(t)= \frac{p'_i}{q_i\lambda_i} \int_0^{t} \omega_i(s) [(w_{n-1,i}(s))^{\frac{q_i}{p'_i}+ 1} - (w_{n,i}(s) )^{\frac{q_i}{p'_i}+ 1}] ds\ge 0. $$ Since $w_{n,i}(t)\ge 0$, the sequences in \eqref{e2.14} converge. For $i= 1, 2$, we denote $w_i(t):=\lim_{n\to \infty} w_{n, i}(t)$. By the Levi monotone convergence theorem it follows that $w_i$ is a nonnegative solution of the equation $$ w_{i}(t)= \int_0^{t} (v_i(s))^{1- p'_i} ds + \frac{p'_i}{q_i\,\lambda_i}\, \int_0^{t} \omega_i(s) (w_{i}(s))^{\frac{q_i}{p'_i}+ 1} ds, $$ where $i= 1,2$. Hence, $w_i$ is absolutely continuous and satisfies the differential equation $$ w_i'(t)= (v_i(t))^{1- p'_i} + \frac{p'_i}{q_i\lambda_i} \omega_i(t) (w_{i}(t))^{\frac{q_i}{p'_i}+ 1}. $$ Therefore, for any fixed numbers $a_i> 0$ $(i= 1,2)$ the functions $$ y_i(t)= \exp\big(\int_{a_i}^{t} [w_i(s)]^{-1} (v_i(s))^{1- p'_i}\,ds\Big) $$ satisfy problem \eqref{e2.1}-\eqref{e2.3}. This completes the proof. \end{proof} \begin{theorem} \label{thm2.5} Let $1< p_i\le q_i< \infty$, $M_i< \infty$, $v_i(t)$ and $\omega_i(t)$ be weight functions on $(0, \infty)$, where $i= 1,2$. Suppose $C> 0$ is the best possible constant, such that \begin{equation} \label{e2.15} \|u\|_{L_{(q_1, q_2, \omega_1, \omega_2)}[(0,\infty)^2]} \le C \|\frac{\partial^2 u}{\partial x_1 \partial x_2}\|_{L_{(p_1, p_2, v_1, v_2)} [(0,\infty)^2]}\,, \end{equation} where $u:(0,\infty)^2\mapsto R$ is an arbitrary absolutely continuous function of two variables satisfying condition \eqref{e2.5}. Then $$ C\le M_1^{1/q_1} M_2^{1/q_2}. $$ \end{theorem} \begin{proof} We denote $\frac{\partial^2 u}{\partial x_1 \partial x_2}= h(x_1, x_2)$. Under condition \eqref{e2.5} we have \[ u(x_1, x_2)= \int_0^{x_1}\int_0^ {x_2} h(t_1, t_2) dt_1 dt_2. \] Obviously, \begin{equation} \label{e2.16} C= \sup\Big(\int_0^\infty\Big(\int_0^\infty \Big|\int_0^{x_1} \int_0^{x_2} h(t_1, t_2) dt_1 dt_2\Big|^{q_1}\omega_1(x_1)dx_1\Big)^ {\frac{q_2}{q_1}} \omega_2(x_2) dx_2 \Big)^{1/q_2}, \end{equation} where the supremum is taken over the class of all measurable functions $h$ such that \begin{gather} \int_0^\infty\big(\int_0^\infty |h(x_1, x_2)|^{p_1} v_1(x_1) dx_1 \Big)^{p_2/p_1} v_2(x_2) dx_2= 1. \end{gather} Assume the contrary. Let $C> M_1^{1/q_1} M_2^{1/q_2}$. Then there exists a number $\mu> 0$ such that $C> \mu> M_1^{1/q_1} M_2^{1/q_2}$. Since $M_i< \infty$, by Lemma \ref{lem2.4}, problem \eqref{e2.1}-\eqref{e2.3} has a solution. Therefore, by Theorem \ref{thm2.3}, the inequality $$ \|u\|_{L_{(q_1, q_2, \omega_1, \omega_2)}[(0,\infty)^2]} \le \mu\big\|\frac{\partial^2 u}{\partial x_1 \partial x_2} \big\|_{L_{(p_1, p_2, v_1, v_2)}[(0,\infty)^2]} $$ holds for all absolutely continuous function $u(x_1, x_2)$ satisfying \eqref{e2.5}. Hence $C$ is not the best possible constant in \eqref{e2.15}. This contradiction completes the proof. \end{proof} \begin{corollary} \label{coro2.6} Let \[ B_i= \sup_{x_i> 0} \int_{x_i}^{\infty} \omega_i(s)ds \big[\int_0^{x_i} (v_i(s))^{1- p'_i} ds\Big]^{q_i/p'_i}, \quad i=1,2. \] Then \begin{gather} \label{e2.18} B_1^{\frac {1}{q_1}} B_2^{1/q_2}\le C\le M_1^{1/q_1}M_2^{1/q_2} \le \prod_{i= 1}^2 (q_i^{1/q_i} (q'_i)^{1/p'_i}) B_1^{1/q_1} B_2^{1/q_2}, \\ \label{e2.19} M_1^{1/q_1} M_2^{1/q_2} \le \prod_{i= 1}^2 \Big(q_i^{1/q_i} (q'_i)^{1/p'_i}\Big) C. \end{gather} \end{corollary} \begin{proof} Let $(\xi_1, \xi_2) \in (0, \infty)^2$ and $P_{\xi_1\xi_2} = \{(x_1, x_2) \in (0, \infty)^2: 0 < x_1 < \xi_1, 0 < x_2 < \xi_2\} $. We put $$ s(x_1, x_2)= \begin{cases} \prod_{i=1}^2\Big(\int_0^{\xi_i} [v_i(t_i)]^{1-p'_i} dt_i\Big)^{-1/p_i} [v_i(x_i)]^{1-p'_i},& \text{for } (x_1, x_2)\in P_{\xi_1\xi_2}\\ 0,& \text{for } (x_1, x_2)\notin P_{\xi_1\xi_2}, \end{cases} $$ Obviously, \begin{align*} &\int_0^\infty\Big(\int_0^\infty (s(x_1, x_2))^{p_1} v_1(x_1) dx_1 \Big)^{p_2/p_1} v_2(x_2) dx_2 \\ &=\int_0^{\xi_2}\Big(\int_0^{\xi_1} \Big[\prod_{i=1}^2\Big(\int_0^{\xi_i} [v_i(t_i)]^{1-p'_i} dt_i\Big)^{-1/p_i} [v_i(x_i)]^{1-p'_i}\Big]^{p_1} v_1(x_1) dx_1 \Big)^{p_2/p_1} \\ &\quad\times v_2(x_2) dx_2= 1. \end{align*} Then, by \eqref{e2.16}, we have \begin{align*} C&\ge \Big(\int_0^\infty\Big(\int_0^\infty \Big(\int_0^{x_1} \int_0^{x_2} s(t_1, t_2) dt_1 dt_2\Big)^{q_1}\omega_1(x_1)dx_1\Big)^{q_2/q_1} \omega_2(x_2)dx_2 \Big)^{1/q_2} \\ &\ge \Big(\int_{\xi_2}^\infty\Big(\int_{\xi_1}^\infty (\int_0^{x_1} \int_0^{x_2} s(t_1, t_2) dt_1 dt_2)^{q_1}\omega_1(x_1)dx_1\Big)^{q_2/q_1} \omega_2 (x_2)dx_2\Big)^{1/q_2} \\ &= \Big(\int_{\xi_2}^\infty \Big( \int_{\xi_1}^\infty \Big( \int_0^{\xi_1} \int_0^{\xi_2} \prod_{i=1}^2 \Big(\int_0^{\xi_i} [v_i(y_i)]^{1-p'_i} dy_i \Big)^{-1/p_i} [v_i(t_i)]^{1-p'_i} dt_1 dt_2 \Big)^{ q_1} \\ &\quad\times \omega_1(x_1) dx_1 \Big)^{q_2/q_1} \omega_2(x_2)dx_2 \Big)^ {1/q_2} \\ &= \Big( \int_0^{\xi_1} [v_1(y_1)]^{1-p'_1} dy_1 \Big)^{1/p'_1} \Big(\int_{\xi_1}^ \infty \omega_1(x_1) dx_1 \Big)^{1/q_1} \Big(\int_0^{\xi_2} [v_2(y_2)]^ {1-p'_2} dy_2 \Big)^{1/p'_2} \\ &\quad\times \Big(\int_{\xi_2}^\infty \omega_2(x_2) dx_2 \Big)^{1/q_2} . \end{align*} Passing to the supremum in the last inequality (with respect to $\xi_1$ and $\xi_2$), we obtain $C\ge B_1^{1/q_1} B_2^{1/q_2}$. The inequality $C\le M_1^{1/q_1} M_2^{1/q_2}$ follows from Theorem \ref{thm2.5}. Now we prove that $M_i\le q_i (q'_i)^{q_i/p'_i} B_i$, where $i= 1,2$. First, by the definition of $B_i$ we obtain that \[ B_i^{p_i'/q_i}\Big(\int_{x_i}^{\infty} \omega_i(y_i) dy_i\Big) ^{-p_i'/q_i}> \int_0^{x_i} [v_i(t_i)]^{1-p'_i} dt_i. \] Putting \[ g_i(x_i)= \frac{q'_i B_i^{p_i'/q_i}}{(\int_{x_i}^{\infty} \omega_i(y_i) dy_i)^{p_i'/q_i}} \] for $i= 1,2$, in \eqref{e2.8}, we obtain \begin{align*} M_i&= \frac{p'_i}{q_i} \inf\sup_{x_i> 0} \frac{1}{q'_i B_i^{p_i'/q_i}\big(\int_{x_i}^{\infty} \omega_i(t_i) dt_i\big)^{-p_i'/q_i}- \int_0^{x_i} [v_i(t_i)]^{1-p'_i} dt_i} \\ &\quad \times \int_0^{x_i} \omega_i(t_i) \Big(q'_i B_i^{p_i'/q_i} \Big(\int_{t_i}^{\infty} \omega_i(y_i) dy_i\Big)^{-p_i'/q_i} \Big)^{\frac{q_i}{p'_i}+ 1} dt_i \\ &\le \sup_{x_i> 0} \frac{(q'_i)^{\frac{q_i}{p'_i}+ 1} B^{1+ \frac {p_i'}{q_i}}} {q'_i B_i^{p_i'/q_i}\big(\int_{x_i}^{\infty} \omega_i(t_i) dt_i\big)^{-p_i'/q_i} - \int_0^{x_i} [v_i(t_i)]^{1-p'_i} dt_i} \\ &\quad\times \int_0^{x_i} \frac {d}{dt_i} \Big[\Big(\int_{t_i}^{\infty} \omega_i(y_i) dy_i\Big)^{-p_i'/q_i}\Big] \\ &\le (q'_i)^{\frac{q_i}{p'_i}+ 1} B^{1+ \frac {p_i'}{q_i}} \sup_{x_i> 0} \frac{\big(\int_{x_i}^{\infty} \omega_i(t_i) dt_i\big)^{-p_i'/q_i}} {q'_i B_i^{p_i'/q_i}\big(\int_{x_i}^{\infty} \omega_i(t_i) dt_i\big)^{-p_i'/q_i} - \int_0^{x_i} [v_i(t_i)]^{1-p'_i} dt_i}. \end{align*} The definition of $B_i$ implies that \[ \int_0^{x_i} [v_i(y_i)]^{1-p'_i} dy_i \le B^{p_i'/q_i} \Big(\int_{x_i}^{\infty} \omega_i(y_i) dy_i\Big)^{-p_i'/q_i}. \] Hence \begin{align*} &q'_i B_i^{p_i'/q_i}\Big(\int_{x_i}^{\infty} \omega_i(t_i) dt_i\Big)^{-p_i'/q_i} - \int_0^{x_i} [v_i(t_i)]^{1-p'_i} dt_i \\ &\ge (q'_i- 1) B_i^{p_i'/q_i} \Big(\int_{x_i}^{\infty} \omega_i(t_i) dt_i\Big)^{-p_i'/q_i}. \end{align*} Thus \[ M_i\le \frac{(q'_i)^{\frac{q_i}{p'_i}+ 1} B_i^{1+ \frac {p_i'}{q_i}}}{(q'_i- 1) B_i^{p_i'/q_i}}= q_i (q'_i)^{q_i/p'_i} B_i, \] where $i= 1, 2$. The inequality \eqref{e2.18} is proved. Clearly, \eqref{e2.18} implies \eqref{e2.19}. The proof is complete. \end{proof} It follows from Corollary \ref{coro2.6} that \eqref{e2.15} holds if and only if $B_i< +\infty$, where $i= 1, 2$ (see \cite{AK}). \section{Main result} \label{section3} \begin{theorem} \label{thm3.1} Let $1< p_i\le q_i< \infty$ and $v_i(t)$ and $\omega_i(t)$ be weight functions on $(0, \infty)$, $v_i\in C^{1}(0, \infty)$, where $i= 1,2$. Then for the solvability of problem \eqref{e2.1}-\eqref{e2.3}, it is necessary and sufficient that there exists a constant $C_0> 0$ such that the inequality \begin{equation} \|u\|_{L_{(q_1, q_2, \omega_1, \omega_2)}[(0,\infty)^2]} \le C_0\big\|\frac{\partial^2 u}{\partial x_1 \partial x_2} \big\|_{L_{(p_1, p_2, v_1, v_2)}[(0,\infty)^2]}, \label{e3.1} \end{equation} holds, where $u:(0,\infty)^2\mapsto R$ is an arbitrary absolutely continuous function of two variables satisfying condition \eqref{e2.5}. \end{theorem} \begin{proof} The sufficiency part of the statement follows from Theorem \ref{thm2.5}. We shall prove only the necessity part. Let $u(x_1, x_2)$ be an absolutely continuous function satisfying condition \eqref{e2.5} and let the inequality \eqref{e3.1} holds. Then $C\le C_0< \infty$, where $C$ is the constant in \eqref{e2.15}. \eqref{e2.18} implies that $M_i< \infty$, where $i= 1, 2$. Then, by Lemma \ref{lem2.4}, problem \eqref{e2.1}-\eqref{e2.3} has a solution for any $\lambda_i> M_i$, where $i= 1, 2$. This completes the proof. \end{proof} \begin{corollary} \label{coro3.2} Let $p_1= p_2= p$, $q_1= q_2= q$ and $v(x_1, x_2)= v_1(x_1) v_2(x_2)$, $\omega(x_1, x_2)= \omega_1(x_1) \omega_2(x_2)$ be weight functions on $(0, \infty)^2$ satisfy all conditions of Theorem \ref{thm3.1}. Then \eqref{e3.1} holds for any absolutely continuous function of two variables satisfying \eqref{e2.5}. \end{corollary} \begin{example} \rm Let $1< p_i\le q_i< \infty$, $v_i(x_i)= x_i^{\alpha_i}$ and $\omega_i (x_i)= x_i^{\frac{q_i}{p_i}(p_i- 1- \alpha_i)- 1}$, where $i= 1, 2$. 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