\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 11, pp. 1--8.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/11\hfil Unique solvability] {Unique solvability for a second order nonlinear system via two global inversion theorems} \author[R. Dalmasso\hfil EJDE-2008/11\hfilneg] {Robert Dalmasso} \address{Robert Dalmasso \newline Laboratoire Jean Kuntzmann\\ Equipe EDP, Tour IRMA, BP 53 \\ 38041 Grenoble Cedex 9, France} \email{robert.dalmasso@imag.fr} \thanks{Submitted October 16, 2007. Published January 21, 2008.} \subjclass[2000]{34B15} \keywords{Nonlinear boundary value problem; existence; uniqueness} \begin{abstract} In this paper we use two global inversion theorems to establish the existence and uniqueness for a nonlinear second order homogeneous Dirichlet system. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \section{Introduction} Let $n \geq 1$ and let $f = (f_1, \dots ,f_n) : [0, 1]\times\mathbb{R}^n \to \mathbb{R}^n$ be a continuous function. We consider the system $$\begin{gathered} u''(x) + f(x, u(x)) = 0\,, \quad 0 \leq x \leq 1 ,\\ u(0) = u(1) = 0 \,. \end{gathered}\label{eq:eq1}$$ We first introduce some notations: \begin{displaymath} \|u\| = \max_{1 \leq j \leq n}(|u_j|), \quad u = (u_1, \dots , u_n) \in \mathbb{R}^n\,, \end{displaymath} $M(n)$ is the space of $n\times n$ matrices with real entries and $\rho(M)$ is the spectral radius of $M \in M(n)$, \begin{gather*} \|M\| = \max_{1 \leq j \leq n}\sum_{k = 1}^{n}|m_{jk}|, \quad M = (m_{jk})_{1\leq j,k\leq n} \in M(n)\,,\\ \|y\|_p =(\int_0^1{|y(t)|^{p}\, dt})^{1/p}, \quad y \in L^p(0,1)\,,\; 1 \leq p < + \infty \,, \\ \|y\|_{\infty} = \mathop{\rm ess\,sup}_{(0,1)} |y|, \quad y \in L^{\infty}(0,1)\,, \\ \|y\|_p = \max_{1 \leq j \leq n}(\|y_j\|_p), \quad y = (y_1, \dots , y_n) \in L^p((0, 1),\mathbb{R}^n)\,, \; 1 \leq p \leq + \infty, \\ \mathbb{R}^n_+ = \{x =(x_1,\dots, x_n) \in \mathbb{R}^n; \, x_j \geq 0 \,, \, j = 1, \dots , n\}\,. \end{gather*} Recently the author proved the following theorem. \begin{theorem}[\cite{da}] \label{thm1} Assume that the partial derivatives $\partial f_j/\partial u_k$ exist and are continuous on $[0,1]\times\mathbb{R}^n$ for $j, k =1, \dots ,n$. Let $\Lambda = (\lambda_{jk})_{1 \leq j, k \leq n}: \mathbb{R}_+ \to M(n)$ be a continuous map with $\lambda_{jk}$ nondecreasing and bounded for $j, k = 1, \dots , n$. Assume that \begin{gather} \bigl|\frac{\partial f_j}{\partial u_k}(x, u)\bigr| \leq \lambda_{jk}(\|u\|) \quad \forall \, (x, u) \, \in [0,1]\times\mathbb{R}^n \,, \; 1 \leq j, k \leq n \,, \label{eq:eq2} \\ \rho(\Lambda(t)) < \pi^2 \quad \forall \, t \, \geq 0 \,, \label{eq:eq3} \\ \int_0^{+ \infty}\det(\pi^2 I - \Lambda(t))\, dt = + \infty \,. \label{eq:eq4} \end{gather} Then the boundary value problem \eqref{eq:eq1} has a unique solution. \end{theorem} The purpose of this paper is to improve and complement Theorem \ref{thm1}. We have the following results. \begin{theorem} \label{thm2} Assume that the partial derivatives $\partial f_j/\partial u_k$ exist and are continuous on $[0,1]\times\mathbb{R}^n$ for $j, k = 1, \dots ,n$. Let $\Lambda = (\lambda_{jk})_{1 \leq j, k \leq n}: \mathbb{R}_+ \to M(n)$ be a continuous map with $\lambda_{jk}$ nondecreasing for $j, k = 1, \dots , n$. Assume that \eqref{eq:eq2} and \eqref{eq:eq3} hold and that $$\int_0^{+ \infty}\frac{dt}{\|(\pi^2 I - \Lambda(t))^{- 1}\|} = + \infty \,. \label{eq:eq5}$$ Then the boundary value problem \eqref{eq:eq1} has a unique solution. \end{theorem} \begin{theorem} \label{thm3} Assume that the partial derivatives $\partial f_j/\partial u_k$ exist and are continuous on $[0,1]\times\mathbb{R}^n$ for $j, k = 1, \dots ,n$. Let $b \in \mathbb{R}_+^n$ and let $A = (a_{jk})_{1 \leq j, k \leq n}: \mathbb{R}_+ \to M(n)$ be a continuous map with $a_{jk}$ nondecreasing for $j, k = 1, \dots , n$. Assume that $$u_{j}f_{j}(x,u) \leq \sum_{k=1}^{n}a_{jk}(\|u\|)|u_j u_k| + b_{j}|u_j| \,, \label{eq:eq6}$$ for all $(x, u) \, \in [0,1]\times\mathbb{R}^n$, $1 \leq j \leq n$, \begin{gather} \rho(A(t)) < \pi^2 \quad \forall \, t \, \geq 0 \,, \label{eq:eq7} \\ \lim_{t\to+\infty}\frac{t}{\|(\pi^2 I - A(t))^{- 1}\|} = + \infty \,. \label{eq:eq8} \end{gather} Let $\Lambda = (\lambda_{jk})_{1 \leq j, k \leq n}: \mathbb{R}_+ \to M(n)$ be a continuous map with $\lambda_{jk}$ nondecreasing for $j, k = 1, \dots , n$. Assume that \eqref{eq:eq2} and \eqref{eq:eq3} hold. Then the boundary value problem \eqref{eq:eq1} has a unique solution. \end{theorem} In Section 2 we recall some results from the theory of nonnegative matrices. We also give two global inversion theorems. We prove Theorem \ref{thm2} in Section 3 and Theorem \ref{thm3} in section 4. Finally in Section 5 we conclude with some examples. \section{Preliminaries} We begin with some results from the theory of nonnegative matrices. We refer the reader to \cite{bepl} for proofs. \begin{definition} \label{def1} \rm $A \in M(n)$ is called $\mathbb{R}^n_+$-monotone if $Ax \in \mathbb{R}^n_+$ implies $x \in \mathbb{R}^n_+$. $N = (n_{jk})_{1 \leq j, k \leq n}$ is nonnegative if $n_{jk} \geq 0$ for $j, k = 1, \dots, n$. \end{definition} \begin{theorem}[{\cite[p. 113]{bepl}}]\label{thm4} $A \in M(n)$ is $\mathbb{R}^n_+$-monotone if and only if $A$ is nonsingular and $A^{- 1}$ is nonnegative. \end{theorem} \begin{theorem}[{\cite[p. 113]{bepl}}] \label{thm5} Let $A = \alpha I - N$ where $\alpha \in \mathbb{R}$ and $N \in M(n)$ is nonnegative. Then the following are equivalent: \begin{itemize} \item[(i)] $A$ is $\mathbb{R}^n_+$-monotone; \item[(ii)] $\rho(N) < \alpha$. \end{itemize} \end{theorem} \begin{remark} \label{rmk1} \rm With the notations of Theorem \ref{thm5}, assume that (i) (or (ii)) holds. Then $\det A > 0$. \end{remark} The proof of Theorem \ref{thm2} makes use of the following global inversion theorem of Hadamard-L\'evy type established by M. R\v adulescu and S. R\v adulescu \cite[Theorem 2]{ra1}. \begin{theorem} \label{thm6} Let $(Y, N_0)$ be a Banach space and let $L : D(L) \to Y$ be a linear operator with closed graph, where $D(L)$ is a linear subspace of $Y$. Then $D(L)$ is a Banach space with respect to the norm defined by \begin{displaymath} N_1(u) = N_0(u) + N_0(Lu) \,, \quad u \in D(L)\,. \end{displaymath} Further, let $K : (Y, N_0) \to (Y, N_0)$ be a $C^1$ map and let $X$ be a linear subspace of $D(L)$ which is closed in the norm $N_1$. Consider the nonlinear map $S : (X, N_1) \to (Y, N_0)$ defined by $S = L - K$, and assume that $S$ is a local diffeomorphism. If there exists a continuous map $c : \mathbb{R}_+ \to \mathbb{R}_+^{\star}$ such that \begin{gather*} \int_0^{+\infty}{c(t)\, dt} = + \infty\,, \\ N_0(S'(u)(h)) \geq c(N_0(u))N_0(h)\quad \forall \, u\,,\, h \in X \,, \end{gather*} then $S$ is a global diffeomorphism. \end{theorem} The proof of Theorem \ref{thm3} makes use of the Banach-Mazur-Caccioppoli global inversion theorem (\cite{bm}, \cite{ca} and \cite{ra2}). \begin{theorem} \label{thm7} Let $E$ and $F$ be two Banach spaces. Then $S: E \to F$ is a global homeomorphism if and only if $S$ is a local homeomorphism and a proper map. \end{theorem} \section{Proof of Theorem \ref{thm2}} We begin with two lemmas. \begin{lemma} \label{lem1} Let $w \in C^1([0,1], \mathbb{R})$ be such that $w(0) = w(1) = 0$. Then \begin{displaymath} \|w'\|_2 \geq \pi\|w\|_2 \quad \text{and}\quad \|w'\|_2 \geq 2\|w\|_{\infty}\,. \end{displaymath} \end{lemma} The first inequality is known as the Wirtinger inequality and the second inequality is known as the Lees inequality. \begin{lemma} \label{lem2} Let \begin{displaymath} X = \{h \in C^2([0,1],\mathbb{R}^n) ;\, h(0) = h(1) = 0\} \,, \end{displaymath} and let $V = (v_{jk})_{1 \leq j,k\leq n}: [0,1] \to M(n)$ be a continuous map. Assume that there exists $N = (n_{jk})_{1 \leq j, k \leq n} \in M(n)$ such that $\rho(N) < \pi^2$ and \begin{displaymath} |v_{jk}(x)|\leq n_{jk} \quad \forall \, x \, \in [0,1] \,, \, \, 1 \leq j, k \leq n \,. \end{displaymath} If $T : X \to C([0,1],\mathbb{R}^n)$ is the operator defined by \begin{displaymath} T(h)(x) = h''(x) + V(x)h(x) \,, \quad h \in X\,, \, \, x \in [0,1] \,, \end{displaymath} then \begin{displaymath} \|T(h)\|_{\infty} \geq \frac{2}{\pi\|(\pi^2 I - N)^{- 1}\|}\|h\|_{\infty} \quad \forall \, \, h \in X \,. \end{displaymath} \end{lemma} \begin{proof} Let $h = (h_1, \dots ,h_n) \in X$ and let $j \in \{1, \dots ,n\}$. Integrating by parts we get \begin{align*} \int_0^1{h_j(x)T(h)_j(x)\, dx} & = \int_0^1{h_j(x)(h_j''(x) + \sum_{k = 1}^{n}v_{jk}(x)h_k(x))\, dx} \\ & = - \int_0^1{h'_j(x)^2\, dx} + \sum_{k=1}^{n} \int_0^1{v_{jk}(x)h_j(x)h_k(x)\, dx}\,.\\ \end{align*} Then using the Cauchy-Schwarz inequality and Lemma \ref{lem1} we can write \begin{align*} \|h_j\|_2\|T(h)_j\|_2 & \geq - \int_0^1{h_j(x)T(h)_j(x)\, dx} \\ & = \|h'_j\|_2^2 - \sum_{k=1}^{n} \int_0^1{v_{jk}(x)h_j(x)h_k(x)\, dx}\\ & \geq \pi\|h_j\|_2\|h'_j\|_2 - \sum_{k=1}^{n}n_{jk} \|h_j\|_2 \|h_k\|_2 \\ & \geq \pi\|h_j\|_2\|h'_j\|_2 - \frac{1}{\pi} \sum_{k=1}^{n}n_{jk} \|h_j\|_2 \|h'_k\|_2 \,, \end{align*} from which we deduce that $$\|T(h)_j\|_2 \geq \pi\|h'_j\|_2 - \frac{1}{\pi}\sum_{k=1}^{n}n_{jk} \|h'_k\|_2 \,, \label{eq:eq9}$$ for $j = 1, \dots, n$. Let $a$, $b$ denote the vectors \begin{displaymath} a = (\|h'_j\|_2)_{1\leq j \leq n} \quad \text{and} \quad b = (\pi \|T(h)_j\|_2)_{1\leq j \leq n}\,. \end{displaymath} Inequality \eqref{eq:eq9} can be written \begin{displaymath} b - (\pi^2 I - N)a \in \mathbb{R}^n_+\,. \end{displaymath} Theorem \ref{thm5} implies that $\pi^2 I - N$ is $\mathbb{R}^n_+$-monotone. Then using Theorem \ref{thm4} we obtain $$(\pi^2 I - N)^{- 1}b - a \in \mathbb{R}^n_+\,, \label{eq:eq10}$$ which implies that \begin{displaymath} \pi \|(\pi^2 I - N)^{- 1}\|\,\|T(h)\|_2 \geq \|h'_j\|_2 \,, \end{displaymath} for $j = 1, \dots , n$. Using Lemma \ref{lem1} and the fact that $\|T(h)\|_2 \leq \|T(h)\|_{\infty}$ we deduce that \begin{displaymath} \|T(h)\|_{\infty} \geq \frac{2}{\pi \|(\pi^2 I - N)^{- 1}\|}\|h\|_{\infty} \,, \end{displaymath} and the lemma is proved. \end{proof} Now we can complete the proof of Theorem \ref{thm2}. Let $Y = C([0,1],\mathbb{R}^n)$ be equipped with the sup norm $\|.\|_{\infty}$ and let $L: D(L) \to Y$ be the linear operator defined by \begin{displaymath} Lu = u'' \,, \quad u \in D(L) \,, \end{displaymath} where $D(L) = C^2([0,1],\mathbb{R}^n)$. Since $L$ has closed graph, it follows from Theorem \ref{thm6} that $D(L)$ is a Banach space with respect to the norm $N_1$ defined by \begin{displaymath} N_1(u) = \|u\|_{\infty} + \|Lu\|_{\infty}\,, \quad u \in D(L) \,. \end{displaymath} Let $K: (Y, \|.\|_{\infty}) \to (Y, \|.\|_{\infty})$ be given by \begin{displaymath} K(u)(x) = - f(x,u(x)) \,, \quad u \in Y\,, \, x \in [0,1] \,. \end{displaymath} The regularity assumptions on $f$ imply that $K$ is of class $C^1$. The set $X = \{u \in D(L); u(0) = u(1) = 0\}$ is a closed subspace of $D(L)$ in the norm $N_1$. Let $S = L - K$. Clearly $S: (X, N_1) \to (Y, \|.\|_{\infty})$ is of class $C^1$. Let $u \in X$ be fixed and let $V = (v_{jk})_{1 \leq j, k \leq n}: [0, 1] \to M(n)$ be such that \begin{displaymath} v_{jk}(x) = \frac{\partial f_j}{\partial u_k}(x, u(x)) \,,\quad x \, \in [0,1] \,, \, \, 1 \leq j, k \leq n \,. \end{displaymath} We have \begin{displaymath} S'(u)(h)(x) = h''(x) + V(x)h(x)\,, \quad h \in X \,, \, x \in [0,1]\,. \end{displaymath} Also \eqref{eq:eq2} implies \begin{displaymath} |v_{jk}(x)| \leq \lambda_{jk}(\|u\|_{\infty}) \quad \forall \, x \, \in [0,1]\,, \, \, 1 \leq j, k \leq n \,. \end{displaymath} Then using Lemma \ref{lem2}, we get $$\|S'(u)(h)\|_{\infty} \geq \frac{2}{\pi\|(\pi^2 I - \Lambda(\|u\|_{\infty}))^{- 1}\|}\|h\|_{\infty} \quad \forall\, h \in X \,. \label{eq:eq11}$$ Let $Q: X \to Y$ be defined by \begin{displaymath} Q(h)(x) = - V(x)h(x)\,,\quad h \in X \,, \, x \in [0,1] \,. \end{displaymath} The operator $L: X \to Y$ is one to one and onto. We have $S'(u) = L - Q = L(I- L^{- 1}Q)$. By \eqref{eq:eq11} $\ker (S'(u)) = \{0\}$. Then $\ker(I - L^{- 1}Q) = \{0\}$. Since $L^{- 1}: (Y, \|.\|_{\infty}) \to (X, \|.\|_{\infty})$ is compact, $L^{- 1}Q$ is compact too. By the Fredholm alternative we obtain that $I - L^{- 1}Q$ is onto. Therefore $S'(u): (X, N_1) \to (Y, \|.\|_{\infty})$ is an invertible operator. By the local inversion theorem we have that $S$ is a local diffeomorphism. Now let \begin{displaymath} c(t) = \frac{2}{\pi\|(\pi^2 I - \Lambda(t))^{- 1}\|}\,, \quad t \geq 0 \,. \end{displaymath} This function satisfies the hypotheses of Theorem \ref{thm6}. Therefore $S$ is a global diffeomorphism and consequently the equation $Su = 0$ has a unique solution $u \in X$. This is also the unique solution of the boundary value problem \eqref{eq:eq1}. \section{Proof of Theorem \ref{thm3}} We keep the notations introduced in Section 3. In the same way we show that $S: (X, N_1) \to (Y, \|\cdot\|_{\infty})$ is a local diffeomorphism. Now let $u = (u_1, \dots ,u_n) \in X$ and let $j\in \{1, \dots ,n\}$. Integrating by parts we get \begin{align*} \int_0^1{u_j(x)S(u)_j(x)\, dx} & = \int_0^1{u_j(x)(u_j''(x) + f_j(x,u(x))\, dx}\\ & = - \int_0^1{u'_j(x)^2\, dx} + \int_0^1{u_j(x)f_j(x,u(x))\, dx}\,.\\ \end{align*} Then using the Cauchy-Schwarz inequality, \eqref{eq:eq6} and Lemma \ref{lem1} we can write \begin{align*} \|u_j\|_2\|S(u)_j\|_2 & \geq - \int_0^1{u_j(x)S(u)_j(x)\, dx} \\ & \geq \|u'_j\|_2^2 - \sum_{k=1}^{n} \int_0^1{a_{jk}(\|u(x)\|)|u_j(x)u_k(x)|\, dx} - b_j\int_0^1 |u_j(x)|\, dx \\ & \geq \|u'_j\|_2^2 - \sum_{k=1}^{n}a_{jk}(\|u\|_{\infty}) \|u_j\|_2 \|u_k\|_2 - b_j\|u_j\|_2 \\ & \geq \pi\|u_j\|_2\|u'_j\|_2 - \frac{1}{\pi} \sum_{k=1}^{n}a_{jk}(\|u\|_{\infty})\|u_j\|_2 \|u'_k\|_2 - b_j\|u_j\|_2 \,, \end{align*} from which we deduce that $$\|S(u)_j\|_2 \geq \pi\|u'_j\|_2 - \frac{1}{\pi}\sum_{k=1}^{n}a_{jk}(\|u\|_{\infty}) \|u'_k\|_2 - b_j \,, \label{eq:eq12}$$ for $j = 1, \dots, n$. Let $r$, $s$ and $b$ denote the vectors \begin{displaymath} r = (\|u'_j\|_2)_{1\leq j \leq n}\,, \quad s = (\pi \|S(u)_j\|_2)_{1\leq j \leq n}, \quad b = (\pi b_j)_{1\leq j \leq n}. \end{displaymath} Inequality \eqref{eq:eq12} can be written as \begin{displaymath} s - (\pi^2 I - A(\|u\|_{\infty}))r + b \in \mathbb{R}^n_+\,. \end{displaymath} Theorem \ref{thm5} implies that $\pi^2 I - A(\|u\|_{\infty})$ is $\mathbb{R}^n_+$-monotone. Then using Theorem \ref{thm4} we obtain $$(\pi^2 I - A(\|u\|_{\infty}))^{- 1}(s + b) - r \in \mathbb{R}^n_+\,, \label{eq:eq13}$$ which implies that \begin{displaymath} \pi\|(\pi^2 I - A(\|u\|_{\infty}))^{- 1}\|(\|S(u)\|_2 + \|b\|) \geq \|u'_j\|_2 \,, \end{displaymath} for $j = 1, \dots , n$. Using Lemma \ref{lem1} and the fact that $\|S(u)\|_2 \leq \|S(u)\|_{\infty}$ we deduce that $$\|S(u)\|_{\infty} \geq \frac{2\|u\|_{\infty}}{\pi\|(\pi^2 I - A(\|u\|_{\infty}))^{- 1}\|} - \|b\|\,. \label{eq:eq14}$$ We shall prove that \eqref{eq:eq14} implies that $S: (X, N_1) \to (Y, \|.\|_{\infty})$ is a proper map. Let $(u_n)_{n \in \mathbb{N}}$ be a sequence in $X$ and $v \in Y$ such that $S(u_n) \to v$ as $n \to + \infty$. \eqref{eq:eq8} and \eqref{eq:eq14} imply that there exists a constant $M > 0$ such that $\|u_n\|_{\infty} \leq M$ for every $n \in \mathbb{N}$. Since $K: (X, N_1) \to (Y, \|.\|_{\infty})$ is a compact operator, it follows that the sequence $(K(u_n))_{n \in \mathbb{N}}$ contains a convergent subsequence. Without loss of generality we may assume that $(K(u_n))_{n \in \mathbb{N}}$ is convergent to $w \in Y$. Letting $n \to + \infty$ in the equality \begin{displaymath} u_n = L^{- 1}S(u_n) + L^{- 1}K(u_n) \,, \end{displaymath} we obtain $$\lim_{n \to +\infty}\|u_n - L^{- 1}(v) - L^{- 1}(w)\|_{\infty} = 0\,. \label{eq:eq15}$$ Then we have \begin{align*} &\lim_{n \to +\infty}\|L(u_n) - L(L^{- 1}(v) + L^{-1}(w))\|_{\infty}\\ &= \lim_{n \to +\infty}\|(S(u_n) - v) + (K(u_n) - w)\|_{\infty} = 0\,. %\label{eq:eq16} \end{align*} From this equality and \eqref{eq:eq15}, we deduce that \begin{displaymath} \lim_{n \to +\infty}N_1(u_n - L^{- 1}(v + w)) = 0\,. \end{displaymath} Therefore $S: (X, N_1) \to (Y, \|\cdot\|_{\infty})$ is a proper map. Using Theorem \ref{thm7} we conclude that $S$ is a global homeomorphism and consequently the equation $Su = 0$ has a unique solution $u \in X$. This is also the unique solution of the boundary value problem \eqref{eq:eq1}. \section{Examples} In this section we give two examples to illustrate Theorems \ref{thm2} and \ref{thm3}. Define $a$, $h: \mathbb{R} \to \mathbb{R}_+$ by \begin{displaymath} a(t) = \begin{cases} 0 & \text{if } t \leq 1\,,\\ 1 - \frac{1}{t^{\alpha}} & \text{if } t \geq 1 \end{cases}\quad \text{and} \quad h(t) = \int_1^t{a(s)\, ds}\,, \quad t \in \mathbb{R} \,, \end{displaymath} where $\alpha > 0$. \begin{example} \label{exa1} \rm Let $n =2$. We set \begin{displaymath} f_1(x,u) = \pi^2 h(u_1) + g_1(x),\quad f_2(x,u) = |u_1|^{\beta} + \pi^2h(u_2) + g_2(x)\,, \end{displaymath} for $(x,u) \in [0,1]\times\mathbb{R}^2$. $\beta > 1$ is a constant and $g_1$, $g_2 \in C([0,1],\mathbb{R})$. Then we can take \begin{gather*} a_{11} = a_{22} = \pi^2 a \,, \quad a_{12} = 0 \,, \quad a_{21}(t) = t^{\beta - 1} \,, \quad t \geq 0 \,,\\ b_j = \|g_j\|_{\infty} \,,\quad j = 1, \, 2 \,, \\ \lambda_{11} = \lambda_{22} = \pi^2 a \,, \quad \lambda_{12} = 0 \,, \quad \lambda_{21}(t) = \beta t^{\beta - 1} ,\quad t \geq 0\,. \end{gather*} We easily verify that $\rho(A(t)) = \rho(\Lambda(t)) = \pi^2 a(t) < \pi^2$ for $t\geq 0$, \begin{gather*} \frac{t}{\|(\pi^2 I - A(t))^{-1}\|} = \frac{\pi^4 t^{1-\alpha}}{\pi^2 + t^{\alpha + \beta - 1}} \quad \textrm{for} \quad t \geq 1 \,, \\ \|(\pi^2 I - \Lambda(t))^{- 1}\| = \frac{t^{\alpha}}{\pi^2} + \frac{\beta}{\pi^4}t^{2\alpha + \beta - 1} \quad \textrm{for} \; t\geq 1\,. \end{gather*} Note that $a_{21}$ and $\lambda_{21}$ are unbounded. If $2\alpha + \beta < 2$ we can use either Theorem \ref{thm2} or Theorem \ref{thm3}. Now let $2\alpha < 1$ and $\beta = 2(1 - \alpha)$. Then Theorem \ref{thm2} applies but Theorem \ref{thm3} does not apply. \end{example} \begin{example} \label{exa2} \rm Let $n =2$. We set \begin{displaymath} f_1(x,u) = \pi^2h(u_1) + g_1(x), \quad f_2(x,u) = \cos|u_1|^{\beta} + \pi^2h(u_2) + g_2(x)\,, \end{displaymath} for $(x,u) \in [0,1]\times\mathbb{R}^2$. $\beta > 1$ is a constant and $g_1$, $g_2 \in C([0,1],\mathbb{R})$. Then we can take \begin{gather*} a_{11} = a_{22} = \pi^2 a \,, \quad a_{12} = a_{21} = 0 \,, \quad b_1 = \|g_1\|_{\infty} \,, \quad b_2 = 1 + \|g_2\|_{\infty} \,, \\ \lambda_{11} = \lambda_{22} = \pi^2 a \,, \quad \lambda_{12} = 0 \,, \quad \lambda_{21}(t) = \beta t^{\beta - 1} ,\quad t \geq 0\,. \end{gather*} We easily verify that $\rho(A(t)) = \rho(\Lambda(t)) = \pi^2 a(t) <\pi^2$ for $t\geq 0$, \begin{gather*} \frac{t}{\|(\pi^2 I - A(t))^{-1}\|} = \frac{ t^{1-\alpha}}{\pi^2} \quad \text{for } t \geq 1 \,, \\ \|(\pi^2 I - \Lambda(t))^{- 1}\| = \frac{t^{\alpha}}{\pi^2} + \frac{\beta}{\pi^4}t^{2\alpha + \beta - 1} \quad \textrm{for} \; t\geq 1\,. \end{gather*} Notice that $\lambda_{21}$ is unbounded. If $2\alpha + \beta \leq 2$, then Theorem \ref{thm2} and Theorem \ref{thm3} apply. If $2\alpha + \beta > 2$ and $\alpha < 1$, Theorem \ref{thm3} still applies but not Theorem \ref{thm2}. \end{example} We conclude this paper with the following remark. \begin{remark} \label{rmk2} \rm With the notations of Theorems \ref{thm2} and \ref{thm3}, assume that $\lambda_{jk}$ are bounded for $j$, $k = 1, \dots, n$. Then \eqref{eq:eq4} implies \eqref{eq:eq5}. Indeed we have \begin{displaymath} (\pi^2 I - \Lambda(t))^{- 1} = \frac{1}{\det(\pi^2 I - \Lambda(t))}B(t) ,\quad t \geq 0 \,, \end{displaymath} where $B(t) \in M(n)$ is nonnegative and $\det(\pi^2 I - \Lambda(t))> 0$ (see Remark \ref{rmk1}). Since $\lambda_{jk}$ are bounded for $j$, $k = 1, \dots, n$, there exists a constant $d > 0$ such that $\|B(t)\| \leq d$ for all $t \geq 0$. Then we can write \begin{displaymath} \frac{1}{\|(\pi^2 I - \Lambda(t))^{- 1}\|} = \frac{\det(\pi^2 I - \Lambda(t))}{\|B(t)\|} \geq \frac{1}{d} \det(\pi^2 I - \Lambda(t))\,, \quad t \geq 0 \,, \end{displaymath} and our claim follows. \end{remark} It is easily seen that \eqref{eq:eq5} does not imply \eqref{eq:eq4} in general. Indeed let $\lambda_{11}(t) = \lambda_{22}(t) = \pi^2(1 - \frac{1}{t}) \,, \quad t \geq 1$ and $\lambda_{12} = \lambda_{21} = 0$. Then we have \begin{displaymath} \frac{1}{\|(\pi^2 I - \Lambda(t))^{- 1}\|} = \frac{\pi^2}{t} \quad \textrm{and} \quad \det(\pi^2 I - \Lambda(t)) = \frac{\pi^4}{t^2}\,, \quad t \geq 1 \,. \end{displaymath} \begin{thebibliography}{0} \bibitem{bepl} A. Berman and J. Plemmons; \emph{Nonnegative matrices in the mathematical sciences}, Academic Press, New York, 1979. \bibitem{bm} S. Banach and S. Mazur; \emph{\"Uber mehrdeutige stetige Abbildungen}, Studia Math. 5 51934, 174-178. \bibitem{ca} R. 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