\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 27, pp. 1--7.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/27\hfil Exact multiplicity of solutions] {Exact multiplicity of solutions for a class of two-point boundary value problems} \author[Y. An, R. Ma\hfil EJDE-2010/27\hfilneg] {Yulian An, Ruyun Ma} % in alphabetical order \address{Yulian An \newline Department of Mathematics, Shanghai Institute of Technology, shanghai 200235, China} \email{an\_yulian@tom.com} \address{Ruyun Ma \newline Department of Mathematics, Northwest Normal University, Lanzhou 730070, China} \email{mary@nwnu.edu.cn} \thanks{Submitted September 30, 2009. Published February 16, 2010.} \thanks{Supported by grants: 10671158 from NSFC, 3ZS051-A25-016 from NSF of Gansu Province, \hfill\break\indent NWNU-KJCXGC-03-17, Z2004-1-62033 from the Spring-sun program, 20060736001 from \hfill\break\indent SRFDP, the SRF for ROCS, 2006[311] from SEM, YJ2009-16 A06/1020K096019 and \hfill\break\indent LZJTU-ZXKT-40728} \subjclass[2000]{34B15, 34A23} \keywords{Exact multiplicity; nodal solutions; bifurcation from infinity; \hfill\break\indent linear eigenvalue problem} \begin{abstract} We consider the exact multiplicity of nodal solutions of the boundary value problem \begin{gather*} u''+\lambda f(u)=0 , \quad t\in (0, 1),\\ u'(0)=0,\quad u(1)=0, \end{gather*} where $\lambda \in \mathbb{R}$ is a positive parameter. $f\in C^1(\mathbb{R}, \mathbb{R})$ satisfies $f'(u)>\frac{f(u)}{u}$, if $u\neq 0$. There exist $\theta_10$, if $us_2$; $uf(u)<0$, if $s_10$; \item[(C2)] $f'(u)>\frac{f(u)}{u}$, if $ u\neq 0$; \item[(C3)] The limit $f_\infty=\lim_{s\to \infty} \frac{f(s)}{s}\in (0, \infty)$. \end{itemize} They obtained a full description of the the first $N$ solution curves which bifurcate from the line of trivial solutions. Anuradha and Shivaji \cite{a1} gave some similar results where $f$ satisfied $f(0)<0$ and other conditions. Motivated by these works, we will consider the existence and uniqueness of one-sign and sign changing solutions of \eqref{e1.1} under the following conditions \begin{itemize} \item[(H1)] $f\in C^1(\mathbb{R}, \mathbb{R})$, $f'(u)>\frac{f(u)}{u}$, if $u\neq 0$; \item[(H2)] the limit $f_\infty=\lim_{|s|\to \infty} \frac{f(s)}{s}\in (0, \infty)$; \item[(H3)] There exist $\theta_10$, if $us_2$; $uf(u)<0$, if $s_10$. However, (H3) implies that $f'(0)<0$. Meanwhile, (C1) and (C2) implies that $f(u)u>0,$ if $u\neq 0$, but $f(u)u$ has negative parts if $f$ satisfying (H3). So it is interesting to find precise global bifurcation structure for the whole solution set of \eqref{e1.1} under the conditions (H1)-(H3). \end{remark} \begin{remark} \label{rmk1.4}\rm The uniqueness and exact multiplicity of positive solutions have been studied by many authors, see \cite{k1,s3} and the references therein. The exact multiplicity results about sign changing solutions have also been researched, see \cite{a1,t1} and the references therein. Bari and Rynne \cite{b1} consider the global structure of the nodal solutions of the problem \begin{gather*} (-1)^mu^{(2m)}(t)=\lambda g(u(t))u(t) , \quad t\in (0,1),\\ u^{(i)}(-1)=u^{(i)}(1)=0,\quad i=0, \dots, m-1, \end{gather*} where $\lambda>0$ is a parameter, the function $g\in C^1(\mathbb{R}, \mathbb{R})$ satisfying $\lim_{|\xi|\to \infty}g(\xi)=\infty$, and $g(0)> 0$, $\pm g'(\xi )> 0$, for all $\pm\xi>0$. \end{remark} \section{Preliminary results} Let $Y= C[0,1]$ with the norm $\|y\|_{\infty}=\max_{t\in [0,1]}|y(t)|$, and let $$ E=\{y\in C^1[0,1]: y'(0)=y(1)=0\}, $$ with the norm $\|y\|_E=\max\{\|y\|_{\infty}, \,\|y'\|_{\infty}\}$. Define the operate $L:\ D(L)\subset E\to Y$, by $Lu:=-u''$, $u\in D(L)$, where $$ D(L)=\{u\in C^2[0,1]: u'(0)=u(1)=0\}. $$ Then $L^{-1}:Y\to E$ is a completely continuous operator and \eqref{e1.1} is equivalent to the operator equation $$ u-\lambda L^{-1}(f(u))=0. $$ We introduce some notation to describe the nodal properties of solutions to \eqref{e1.1}. Firstly, for any $C^1$ function $u$, $x_0$ is a \emph{simple} zero of $u$ if $u(x_0) = 0$ and $u'(x_0)\neq 0$. Now, for any integer $k\ge1$ and any $\nu \in \{+,-\}$, we define sets $S_k^{\nu}\subset C^2[0, 1] $ as follows: if $u \in S_k^{\nu}$, then \begin{itemize} \item[(i)] $u'(0) = 0, \nu u(0) > 0$; \item[(ii)] $u$ has only simple zeros in $[0, 1]$ and has exact $k-1$ zeros in $(0, 1)$. \end{itemize} The sets $S_k^{\nu}$ are open in $E$ and disjoint. Let $\mathbb{E}=\mathbb{R}\times E$, under the product topology. We add the point $\{(\lambda, \infty)_p|\lambda\in \mathbb{R}\}$ into the space $\mathbb{E}$. Put $\Phi_k^\nu=\mathbb{R}\times S_k^\nu$. Clearly, $u \equiv 0 $ is a solution of \eqref{e1.1} for any $\lambda \in \mathbb{R}$. $(\lambda, 0)$ is called a trivial solution of \eqref{e1.1}. Note that (H1) ensures that the solution of the initial value problem for the differential equation in \eqref{e1.1} is unique. This fact will be used repeatedly in the following proof so, for brevity, it will be abbreviated to ``IVPU''. We first prove the following result about the nodal properties of nontrivial solutions of \eqref{e1.1}. \begin{lemma} \label{lem2.1} Suppose $(\lambda, u)$ is a nontrivial solution of \eqref{e1.1}. Then \begin{itemize} \item[(i)] $u\in S_k^{\nu}$ for some $k\in \mathbb{N}$ and $\nu \in \{+,-\}$; \item[(ii)] $\max_{t\in [0,1]} u(t)>\theta_2$ and $\min_{t\in [0,1]} u(t)<\theta_1$ if $k\ge 2$; $\max_{t\in [0,1]} u(t)>\theta_2$ if $u\in S_1^{+}$; $\min_{t\in [0,1]} u(t)<\theta_1$ if $u\in S_1^{-}$; \item[(iii)] $u(0)=\max_{t\in [0,1]} u(t),$ if $u\in S_k^{+}$, and $u(0)=\min_{t\in [0,1]} u(t),$ if $u\in S_k^{-}$. \item[(iv)] $u$ has no positive local minimum and/or negative local maximum. \end{itemize} \end{lemma} \begin{proof} (i) Since u is nontrivial, ``IVPU'' implies that all zeros of $u$ are simple. So, (i) is true. In particular, by the boundary condition in \eqref{e1.1}, we have $u(0)\neq 0$ since $u'(0) = 0$. We now describe the qualitative ``shape'' of the solution $u$. Without loss of generality, assume that $u\in S_k^{+}$ for some $k\in \mathbb{N}$ in the following proof. When $u\in S_k^{-}$, the proof is similar. It follows from the fact that $f$ is independent of $t$ and ``IVPU'' that the graph of $u$ consists of a sequence of positive and negative bumps, together with a half bump at the left end of the interval $[0, 1]$, with the following properties (ignoring the half bump): \begin{itemize} \item[(a)] all the positive (resp. negative) bumps have the same shape (the shapes of the positive and negative bumps may be different); \item[(b)] all the positive (resp. negative) bumps attain the same maximum (resp. minimum) value. \item[(c)] if $\xi \in (\alpha,\beta)\subset(0,1)$ is a critical point of $u$ and $\alpha,\beta$ are two consecutive zeros of $u$, then the graph of $u$ is symmetric about $t=\xi$ on the interval $(\alpha,\beta)$. \end{itemize} Armed with these properties on the shape of $u$ we can continue the proof of the Lemma. (ii) On the contrary, suppose $\max_{t\in [0,1]} u(t)\le \theta_2$. Let $u(0)=c$, then $00$ when $t> 0$ is small. Suppose $t_1$ is the first zero of $u$, then $u(t)>0$ on $[0,t_1)$ and $u(t_1)=0$, $u'(t_1)<0$. Note that $(\lambda, u)$ satisfies the equation \begin{equation} u''=-\lambda f(u). \label{e2.1} \end{equation} Multiplying both sides of \eqref{e2.1} by $u'$, \begin{equation} u''(t)u'(t)=-\lambda f(u(t))u'(t). \label{e2.2} \end{equation} Integrating \eqref{e2.2} from 0 to $t_1$, \begin{equation} \int_0^{t_1}u''(t)u'(t)dt=-\lambda \int_0^{t_1} f(u(t))u'(t)dt.\label{e2.3} \end{equation} It follows from \eqref{e2.3} and (H3) that \begin{equation} \frac{1}{2}(u'(t_1))^2 =-\lambda \int_0^{t_1}f(u(t))du(t) =-\lambda \int_c^0f(u)du =\lambda \int_0^cf(u)du\le 0. \label{e2.4} \end{equation} since $c\le \theta_2$. However, the left side of \eqref{e2.4} is positive. This is a contradiction. If $k=1$, then the proof is completed. If $k\ge 2$, suppose $\min_{t\in [0,1]} u(t)\ge \theta_1$. Denote $t_2$ is the second zero of $u$, then $u(t)<0$ on $(t_1, t_2)$ and $u(t_1)=u(t_2)=0,\ u'(t_1)<0$. From (a) and (b) in (i), there exists a $\xi_1 \in (t_1, t_2)$ such that $u'(\xi_1)=0$ and $u(\xi_1)=\min_{t\in [0,1]} u(t)\ge \theta_1$. Integrating \eqref{e2.2} from $t_1$ to $\xi_1$, \begin{equation} \int_{t_1}^{\xi_1}u''(t)u'(t)dt=-\lambda \int_{t_1}^{\xi_1} f(u(t))u'(t)dt.\label{e2.5} \end{equation} It follows from \eqref{e2.5} and (H3) that \begin{equation} -\frac{1}{2}(u'(t_1))^2 =-\lambda \int_{t_1}^{\xi_1}f(u(t))du(t) =-\lambda \int_0^{u(\xi_1)}f(u)du =\lambda \int_{u(\xi_1)}^0f(u)du\ge 0. \label{e2.6} \end{equation} since $u(\xi_1)\ge \theta_1$. However, the left side of \eqref{e2.6} is negative. This is a contradiction. Thus, (ii) is true. Statements (iii) and (iv) follow from (c) in (i). \end{proof} \begin{remark} \label{rmk2.2} (iv) implies that the zeros of $u$ and the zeros of $u'$ are separated, that is each bump of $u$ contains a single zero of $u'$, and there is exact one zero of $u$ between consecutive zeros of $u'$. Moreover, if $u \in S_k^{\nu}$, then $u$ has $k-1$ zeros in $(0, 1)$ and $u'$ has exact $k-1$ zeros in $(0, 1)$. \end{remark} For a nontrivial solution of \eqref{e1.1}, $(\lambda, u)$ is \emph{degenerate} if the problem \begin{equation} \begin{gathered} w''+\lambda f'(u)w=0 ,\ \quad t\in (0,1),\\ w'(0)=0,\quad w(1)=0 \end{gathered}\label{e2.7} \end{equation} has a nontrivial solution, otherwise it is \emph{nondegenerate}. Now, we consider the initial value problem \begin{equation}\begin{gathered} \phi''+\lambda f'(u)\phi=0 , \quad t\in (0,\ 1),\\ \phi'(0)=0,\quad \phi(0)=1. \end{gathered}\label{e2.8} \end{equation} It plays very important role to study the exact multiplicity of solutions of \eqref{e1.1}. Note that if $\phi$ is the unique solution of \eqref{e2.8}, then any solution of \eqref{e2.7} can be written $w=c\phi$, where $c\in {\mathbb R}$ is a constant. \begin{lemma} \label{lem2.3} If $(\lambda, u)\in \Phi_k^{\nu}$ is a nontrivial solution of \eqref{e1.1}. Then $(\lambda, u)$ is nondegenerate. \end{lemma} \begin{proof} Suppose $(\lambda, w), (\lambda, \phi)$ is the solutions of \eqref{e2.7}, \eqref{e2.8}, respectively. We claim that \begin{equation} \phi(1)\neq 0.\label{e2.9} \end{equation} From this claim, we obtain immediately that \eqref{e2.7} has only trivial solution since $w(1)=c\phi(1)=0$ if and only if $c=0$. So $(\lambda,u)$ is nondegenerate. Therefore, we only need to prove that \eqref{e2.9} holds. Since $u\in S_k^{\nu}$, then all zeros of $u$ are simple. By Lemma \ref{lem2.1} and Remark \ref{rmk2.2}, $u$ has exact $k-1$ zeros in $(0, 1)$, and especially, $u'$ has also exact $k-1$ zeros in $(0, 1)$. The function $u$ satisfies \begin{equation} u''+\lambda f(u)=0,\quad t\in (0,1).\label{e2.10} \end{equation} Define the function $$ p(t)=\begin{cases} \frac{f(u(t))}{u(t)}, &u(t)\neq 0,\\ f'(0),& u(t)= 0. \end{cases} $$ Then \eqref{e2.10} is equivalent to \begin{equation} u''+\lambda p(t)u=0.\label{e2.11} \end{equation} On the other hand, note that $\phi$ and $u'$ satisfy the following equations respectively: \begin{gather} \phi''+\lambda f'(u)\phi=0,\label{e2.12} \\ (u')''+\lambda f'(u)u'=0.\label{e2.13} \end{gather} By (H1), (H2), (H3), we have $p(t)\le f'(u(t))$ for all $t\in (0,1)$. Applying the Sturm comparison lemma to \eqref{e2.11} and \eqref{e2.12}, we obtain, there exists at least one zero of $\phi$ between any two consecutive zeros of $u$. We extend evenly $u, \phi$ to $[-1, 0)$, then $u$ has exact $2(k-1)$ zeros in $(-1, 1)$, that is, $u$ has exact $2k$ zeros in $[-1, 1]$. This implies that $\phi$ has at least $2k-1$ zeros in $(-1, 1)$. Note that $\phi$ is a even function in $[-1, 1]$, and $\phi(0)\neq 0$, then $\phi$ has at least $2k$ zeros in $(-1, 1)$. Therefore, $\phi$ has at least $k$ zeros in $(0, 1)$. On the other hand, between any two consecutive zeros of $\phi$, there exists at least one zero of $u'$. Suppose \eqref{e2.9} does not hold, i.e., $\phi(1)=0$. Then $\phi$ has at least $k+1$ zeros in $(0, 1]$. Moreover, $u'$ has at least $k$ zeros in $(0, 1)$. It is impossible! Thus, $\phi(1)\neq 0$. \end{proof} The following Lemma shows that every solution of \eqref{e1.1} which belongs to $\Phi_k^+$ (resp. $\Phi_k^-$) can be parameterized by its maximum (resp. minimum). \begin{lemma} \label{lem2.4} Given $k\in \mathbb{N}$ for each $d>0$(resp. $d<0$), there exists at most one $\lambda>0$ such that \eqref{e1.1} has at most a solution $u$ which belongs to $S_k^{+}$(resp. $S_k^{-}$) and satisfies $u(0)=d$. \end{lemma} The proof of the above lemma can be found in \cite{s2}. \section{The main result and its proof} Our main result reads as follows. \begin{theorem} \label{thm3.1} Let {\rm (H1)-(H3)} hold. Then for every $k\in \mathbb{N}$ and $\nu\in \{+,-\}$, we have: \begin{itemize} \item[(i)] Equation \eqref{e1.1} has no degenerate solutions. All solutions of \eqref{e1.1} that belong to $\Phi_k^{\nu}$ lie on a unique continuous curve $D_k^{\nu}$. This curve starts from $(\frac{\lambda_k}{f_\infty}, \infty)_p\in \mathbb{E}$, and extends for increasing $\lambda$ such that $\mathop{\rm Proj}_{\mathbb{R}}D_k^{\nu}=(\frac{\lambda_k}{f_\infty}, \infty)\subset \mathbb{R}^+$. \item[(ii)] For every given parameter $\lambda\in (\frac{\lambda_k}{f_\infty}, \infty)\subset \mathbb{R}^+$, there exists exactly one solution of \eqref{e1.1} which belongs to $S_k^{\nu}$; for every given parameter $\lambda\in (0, \frac{\lambda_k}{f_\infty}]$, there exists no solution of \eqref{e1.1} which belongs to $S_k^{\nu}$, where $\lambda_k$ is the $k$th eigenvalue of the linear problem \begin{equation} \begin{gathered} \varphi''+\lambda \varphi=0 , \quad t\in (0, 1),\\ \varphi'(0)=0,\quad \varphi(1)=0. \end{gathered}\label{e3.1} \end{equation} \end{itemize} \end{theorem} \begin{remark} \label{rmk3.2}\rm It is well-known that the eigenvalues of \eqref{e3.1} satisfy $$ 0<\lambda_1<\lambda_2<\dots<\lambda_k<\lambda_{k+1}<\dots, \quad \lim_{k\to \infty} \lambda_k=\infty, $$ for each $\lambda_k$ is simple and the corresponding eigenfunction $\varphi_k$ has exactly $k-1$ zeros in $(0, 1)$. \end{remark} From Theorem \ref{thm3.1}, we obtain immediately the following corollary. \begin{corollary} \label{coro3.3} Let {\rm (H1)-(H3)} hold. Then for every $k\in \mathbb{N}$ and $\lambda>0$: \eqref{e1.1} has no nontrivial solution when $\lambda\in (0, \frac{\lambda_1}{f_\infty}]$; has exactly two nontrivial solutions, one positive and one negative, when $\lambda\in (\frac{\lambda_1}{f_\infty},\frac{\lambda_2}{f_\infty}]$; has exactly four nontrivial solutions when $\lambda\in (\frac{\lambda_2}{f_\infty},\frac{\lambda_3}{f_\infty}]$, a positive solution, a negative solution, a solution which has one zero on $(0,1)$ and $u(0)>0$ and a solution which has one zero on $(0,1)$ and $u(0)<0$. In general, when $\lambda\in (\lambda_k/f_\infty, \lambda_{k+1}/f_\infty]$, \eqref{e1.1} has exactly $2k$ nontrivial solutions, where \[ u_1\in S_1^+,\quad u_2\in S_1^- ,\quad u_3\in S_2^+,\quad u_4\in S_2^- ,\quad \dots \quad u_{2k-1}\in S_k^+,\quad u_{2k}\in S_k^-. \] \end{corollary} Let $\zeta \in C(\mathbb{R},\mathbb{R})$ be such that \begin{equation} f(u)=f_\infty u+\zeta(u). \label{e3.2} \end{equation} Clearly, \begin{equation} \lim_{|u|\to \infty} \frac {\zeta (u)}u=0. \label{e3.3} \end{equation} Let us consider \begin{equation} Lu-\lambda f_\infty u=\lambda \zeta(u) \label{e3.4} \end{equation} as a bifurcation problem from infinity. We note that \eqref{e3.4} is equivalent to \eqref{e1.1}. The results from Rabinowitz \cite{r1} for \eqref{e3.4} can be stated as follows: \begin{lemma} \label{lem3.4} For each integer $k\geq 1$, $\nu\in \{+, \ -\}$, all nontrivial solutions of \eqref{e1.1} near $\big(\frac {\lambda_k}{f_\infty}, \infty\big)_p$ lie on a smooth local curve $\mathcal{D}_{k}^\nu$, and $\mathcal{D}_{k}^\nu\setminus \{\bigl (\frac {\lambda_k}{f_\infty}, \infty\bigr)_p\}\subset \Phi_k^\nu$. \end{lemma} \begin{proof}[Proof of Theorem \ref{thm3.1}] (i) From Lemma \ref{lem2.3}, \eqref{e1.1} has no degenerate solution. We give the proof only for $u(0)>0$. When $u(0)<0$, the proof is similar. By Lemma \ref{lem3.4}, all solutions of \eqref{e1.1} near the point $(\frac{\lambda_k}{f_{\infty}},\infty)_P$ and $u(0)>0$ lie on a unique continuous local curve $\mathcal{D}_{k}^+$ which bifurcating from $\bigl (\frac {\lambda_k}{f_\infty}, \infty\bigr)_p$, and $\mathcal{D}_{k}^+\setminus \{\bigl (\frac {\lambda_k}{f_\infty}, \infty\bigr)_p\}\subset \Phi_k^+$. By Lemma \ref{lem2.3} and the implicit function theorem, we can continue this local curve to a maximal interval of definition over the $\lambda$-axis. We still denote the curve $\mathcal{D}_{k}^+$. If we extend $\mathcal{D}_{k}^+$ for decreasing $\lambda$, then this curve will intersect with the hyperplane $\{0\}\times E$ at some point $(\tilde{u}, 0)$ with $\tilde{u}(0)>\theta_2$. This contradicts $u\equiv 0$ if $\lambda=0$, since $f(0)=0$. So, we must extend $\mathcal{D}_{k}^+$ for increasing $\lambda$. By Lemma \ref{lem2.3} and the implicit function theorem, it cannot stop at a point such as $(\lambda_0 , u_0)$ where $\frac{\lambda_k}{f_\infty}<\lambda_0 <\infty$ and $u_0(0)<\infty$. On the other hand, by Lemma \ref{lem2.4}, it also can not blow up at some point $(\lambda_*, \infty)_p$ with $\frac{\lambda_k}{f_\infty}<\lambda_* <\infty$. Therefore, this curve must continue for increasing $\lambda$ such that $\mathop{\rm Proj}_\mathbb{R}D^+_k =(\frac{\lambda_k}{f_\infty},\infty)\subset \mathbb{R}$. Moreover, if $(\lambda,u)\in \mathcal{D}_{k}^+$ and $\lambda\to \infty$, then there must be a constant $M\ge \theta_2$ such that $\|u\|_\infty\to M$. Finally, we claim that all solutions of \eqref{e1.1} which belong to $\Phi_k^+$ must lie on $D_k^+$. If $M=\theta_2$, by Lemma \ref{lem2.4}, the above claim is naturally right. If $M>\theta_2$, on the contrary, we suppose there is a solution $(\lambda_0,u_0)$ of \eqref{e1.1} and $(\lambda_0,u_0)\in \Phi_k^+$, but $(\lambda_0,u_0)\not\in \mathcal{D}_{k}^+$. By Lemma \ref{lem2.3} and the implicit function theorem, all solutions of \eqref{e1.1} near $(\lambda_0,u_0)$ must lie on a unique local curve which through $(\lambda_0,u_0)$. We denote this local curve $\Gamma_0$. Then for any $(\lambda,u)\in \Gamma_0$, we have $\theta_2< \|u\|_\infty