\documentclass[twoside]{article} \usepackage{amssymb} % used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil Existence of periodic solutions \hfil EJDE--1998/31} {EJDE--1998/31\hfil Petr Girg \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 1998}(1998), No.~31, pp. 1--10. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp 147.26.103.110 or 129.120.3.113 (login: ftp)} \vspace{\bigskipamount} \\ Existence of periodic solutions for a semilinear ordinary differential equation \thanks{ {\em 1991 Mathematics Subject Classifications:} 34B15, 34C15, 34C25, 34C99. \hfil\break\indent {\em Key words and phrases:} Ordinary differential equation, periodic solutions. \hfil\break\indent \copyright 1998 Southwest Texas State University and University of North Texas. \hfil\break\indent Submitted August 20, 1998. Published November 20, 1998.} } \date{} \author{Petr Girg} \maketitle \begin{abstract} Dancer [3] found a necessary and sufficient condition for the existence of periodic solutions to the equation $$\ddot x +g_1(\dot x) + g_0(x) = f(t)\,.$$ His condition is based on a functional that depends on the solution to the above equation with $g_0=0$. However, that solution is not always explicitly known which makes the condition unverifiable in practical situations. As an alternative, we find computable bounds for the functional that provide a sufficient condition and a necessary condition for the existence of solutions. \end{abstract} \newcommand{\esssup}{\mathop{\rm ess~sup}} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \section{Introduction} In this paper, we study the existence and the non-existence of solutions of the semilinear boundary-value problem \begin{eqnarray} &\ddot x(t) + g_1 (\dot x(t)) + g_0 (x(t)) = f(t)\,,&\label{fund_eq}\\ &x(0) = x(T),\ \dot x(0) = \dot x(T)\,.& \label{per_con} \end{eqnarray} Although a necessary and sufficient condition is already known \cite{Danc}, it can not be verified in practical situations because the condition is given by a related nonlinear boundary-value problem. In this article we give, on the one hand, a sufficient condition, and on the other hand a necessary condition, which can be verified for any continuous function $f$. In the first part of this article, we present a survey of known results and their physical interpretation. And in the second part, we present our main result, which is stated as Theorem~\ref{it_the}. Overall, we will suppose that $g_0$, $g_1$, $f$ are continuous real-valued functions, and $f$ is $T$--periodic. For a given $k\geq 0$, let \begin{eqnarray*} C^{k}_T &=& \big\{ u : u\mbox{ is $k$-times continuously differentiable on $[0,T]$, with} \\ &&\quad u(0)=u(T), u'(0)=u'(T), \dots, u^{(k)}(0)=u^{(k)}(T) \big\}\,. \end{eqnarray*} In these spaces the maximum norm will be denoted by $\|\cdot\|_{C^k_T}$, and $C^{0}_T$ will be denoted by $C_T$. The subspace consisting of functions with mean value zero will be denoted by $$\tilde C_T^k=\big\{u\in C_T^k : \int_0^Tu(t)\,dt =0\big\}\,.$$ For functions with domain $[0,T]$, with distributional derivatives, we define: \begin{eqnarray*} &L^{p} = \big\{u : \int_{0}^T |u(t)|^{p} dt < \infty \big\},\ 1\leq p < +\infty\,;&\\ & L^\infty =\big\{ u : \esssup_{t\in [0,T]} |u(t)|<\infty \big\}, &\\ & W^{1,2}_T = \big\{u\in L^2 : u'\in L^2,\; u(0)=u(T)\big\} \,,&\\ & W^{2,\infty}_T = \big\{ u\in L^{\infty} : u''\in L^{\infty}, \; u(0)=u(T),\; u'(0)=u'(T) \big\}\,. & \end{eqnarray*} For a subset $X$ on the space of integrable functions on $[0,T]$, we define $\tilde X = \big\{u\in X : \int_0^T u(t)dt = 0\big\}$. For integrable functions, we use the decomposition $$f = \tilde f + \bar f\,, \quad\mbox{with } \bar f = \frac{1}{T} \int_0^T f(t)\,dt\,.$$ We will assume that $f$, the right-hand side of (\ref{fund_eq}), belongs to $C_T$, and the solution $x$ belongs to $C^2_T$. Although the results in \cite{Danc} assume that $f$ is in a certain Lebesgue space and that $x$ is in a certain Sobolev space, it is not hard to get analogous results for $f$ in $C_T$ and $x$ in $C^2_T$. So when we cite results from \cite{Danc}, we do a conversion to our function spaces (except in section 2). When $g_0=0$, for each $\tilde f \in \tilde C_T$ there exists a value $s(\tilde f)$ such that $$\label{s_eq} \ddot x(t) + g_1(\dot x(t)) = \tilde f(t) + s(\tilde f)$$ has a periodic solution, \cite[Theorem~1]{Danc}. Equivalently, the range of the operator $H_1:C^2_{T} \rightarrow C_{T}$, $H_1(x)=\ddot x + g_1\circ \dot x$, can be written as $$\label{rank} {\cal R}_1 = \{\tilde f + s(\tilde f) : \tilde f\in \tilde C_T\}\,.$$ Under the assumption that $g_0$ is bounded, continuous, and satisfies $$g_0(-\infty):=\lim_{\xi \rightarrow -\infty}g_0(\xi) < \lim_{\xi \rightarrow + \infty}g_0(\xi)=: g_0(+\infty)\,,$$ Dancer \cite[Theorem~2]{Danc} showed that a function $f \in C_T$ belongs to ${\cal R}$, the range of $H:C^2_{T} \rightarrow C_{T}$, $H(x)=\ddot x + g_1\circ \dot x+ g_0\circ x$, if $$\label{danc_con} g_0(-\infty) < \bar f - s(\tilde f) < g_0(+\infty)\,.$$ Thus, (\ref{danc_con}) is a sufficient condition for (\ref{fund_eq}) to posses a periodic solution. However, if we also have $$g_0(-\infty) 0, or \dot xg_1(\dot x) < 0, which represents a self--excitation (positive damping). \medskip For general functions g_1 and g_0, with g_0 bounded as in the Landesman--Lazer case, Dancer \cite{Danc} proved that the range H(W^{2,\infty}_T) is enclosed by two manifolds parallel to the range H_1(W^{2,\infty}_T). A sufficient condition for the solvability of Problem (\ref{fund_eq})--(\ref{per_con}) is given by (\ref{danc_con}). Note that if g_0(-\infty)<0\frac{TkK}{2(k+K)}. Without lost of generality, we may assume that the maximum norm is attained at a point t_0=\frac{kT}{2(k+K)}, 0\leq t_0\leq T/2. If necessary multiply w by -1, interchange the roles of k and K, and shift w suitably in time. Then$$w(t_0)=\|w\|_{C_T} > \frac{TkK}{2(k+K)}=Kt_0\,. $$Our strategy is to prove the following two inequalities for t\in[0,\frac{T}{2}]: \begin{eqnarray} &w(t)>\min\big\{Kt, k(\frac{T}{2}-t) \big\}\,,&\label{le1}\\ &w(t+\frac{T}{2})>-\min\big\{kt,K(\frac{T}{2}-t) \big\}\,.&\label{le2} \end{eqnarray} Which lead us to the contradiction that \int_0^T w=0 and$$ \int_0^T w(t)\,dt=\int_0^{T/2}w(t)\,dt+\int_0^{T/2} w(t+\frac{T}{2})\,dt > 0\,. $$For (\ref{le1}), we consider the two cases: If 0 Kt_0+ (t_0-t)(-K)=Kt\,.$$ and if $t_0 Kt_0 + (t-t_0)(-k)= k(\frac{T}{2}-t)\,. $$For (\ref{le2}), we put u(t)=w(t+\frac{T}{2}) and notice that u(0)=w(\frac{T}{2})>0 and u(\frac{T}{2})=w(T)=w(0)>0. For t in [0,\frac{T}{2}] we have$$w\left(t+\mbox{$\frac{T}{2}$}\right)=u(t)= u(0)+\int_0^t\dot u(\tau)\,d\tau > -kt$$and$$w\left(t+\mbox{$\frac{T}{2}$}\right)=u(t)=u\left(\mbox{$\frac{T}{2}$}\right) +\int_t^{T/2}\bigl(-\dot u(\tau)\bigr)\,d\tau>-K(\frac{T}{2}-t)\,.$$Hence$$ w(t+\mbox{$\frac{T}{2}$})>\max\big\{-kt, -K(\frac{T}{2}-t)\big\}= -\min\big\{kt, K(\frac{T}{2}-t)\big\}. $$Which concludes the present proof. \hfill \diamondsuit \begin{lemma}\label{lm_3} Let w be a solution to Problem (\ref{fst_eq})-(\ref{w_cond}), with \|w\|_{C_T}\leq b and \tilde f\not\equiv 0. Then$$ -k\leq \dot w(t) \leq K\,,$$where k and K are the positive constants: -k= \min_{t\in [0,T]} \tilde f(t)+m and K=\max_{t\in [0,T]} \tilde f(t) +M, where \begin{eqnarray*} &m = \sup_{c\in{\mathbb R}}\min_{|\xi|\leq b}(g_1(\xi)-c\xi)-\max_{|\xi|\leq b} g_1(\xi)\,,&\\ &M = \inf_{c\in{\mathbb R}}\max_{|\xi|\leq b}(g_1(\xi)-c\xi)-\min_{|\xi|\leq b} g_1(\xi)\,.& \end{eqnarray*} \end{lemma} \paragraph{Proof} From (\ref{fst_eq}) we obtain$$ \min_{t\in [ 0,T ]}\left(\tilde f(t) + s(\tilde f) - g_1(w(t))\right) \leq \dot w(t) \leq \min_{t\in [ 0,T ]}\left(\tilde f(t) + s(\tilde f) - g_1(w(t))\right)\,. $$Using the estimates for s(\tilde f) in (\ref{bas_est}), we obtain the desired inequality. Notice that because g_1 is continuous and the extrema is computed on a bounded interval, then \begin{eqnarray*} &-\infty<\min_{|\xi|\leq b}(g_1(\xi)-0\cdot\xi)-\max_{|\xi|\leq b} g_1(\xi) \leq m \,,&\\ & M \leq \max_{|\xi|\leq b}(g_1(\xi)-0\cdot\xi)-\min_{|\xi|\leq b}g_1(\xi)<\infty\,.& \end{eqnarray*} It is left only to check that k and K are positive. This follows from the fact that -k<\dot w(t)0, such that \|w\|_{C_T}\leq b_0 (for instance: b_0=\|\tilde f\|_2\sqrt{T/12} due to Lemma \ref{lm_0}). Then for n=0,1,2,\dots, let k_n, K_n be the constants obtained in Lemma~\ref{lm_3} with b=b_n, and let$$b_{n+1}= \frac{Tk_nK_n}{2(k_n+K_n)}\,.$$\begin{lemma} Let b_n, k_n, K_n be defined as above. If b_1\leq b_0, then b_{n+1}\leq b_n for all n\geq 1. \end{lemma} \paragraph{Proof} We proceed by induction. First notice that b_1\leq b_0 is one of the hypotheses. Now assume that b_n\leq b_{n-1}. Then in the statement of Lemma~\ref{lm_3} we see that$$ 0\geq m_n\geq m_{n-1}\quad\mbox{and}\quad 0\leq M_n \leq M_{n-1}\,.$$Thus, k_n\leq k_{n-1} and K_n\leq K_{n-1}. Since \frac{TkK}{2(k+K)} is a decreasing function of k, and of K, we have b_{n+1}\leq b_n. \hfill\diamondsuit\medskip From the above lemma, iterations can be repeated indefinitely. However, in practice the process should stop when the decrement in b_n is less than a predetermined value. Now, we define the lower and upper bounds for s(\tilde f). \begin{theorem} \label{it_the} Let b_n be as defined above. Put b=\inf\{b_0,b_1,\dots\}, and \begin{eqnarray*} &a(\tilde f)= \sup_{c\in{\mathbb R}}\min_{|\xi|\leq b} \left (g_1(\xi)-c\xi\right )\,,&\\ &A(\tilde f) = \inf_{c\in{\mathbb R}}\max_{|\xi|\leq b} \left(g_1(\xi)-c\xi\right )\,.& \end{eqnarray*} Then the functional s(\tilde f) satisfies a(\tilde f) \leq s(\tilde f) \leq A(\tilde f). \end{theorem} \paragraph{Proof} Notice that by Lemma~\ref{lm_2}, \|w\|_\infty\leq b_n for all n. Therefore, from the basic estimate (\ref{bas_est}), the statement of this theorem follows. Notice that even if A(\tilde f) is not the absolute infimum over c, the equality in this Theorem is still valid. The same statement applies for a(\tilde f). \hfill\diamondsuit\medskip Computational experiments show that the iteration method refines estimates if the ratio -\max(\tilde f)/\min(\tilde f) is much larger than one, or very close to zero. To illustrate this case, we study the following boundary-value problem \paragraph{Example 1} Consider \dot w(t) + g_1(w(t)) = \tilde f(t)+s(\tilde f), where$$ \tilde f(t) =\left\{ \begin{array}{ll} -\sin(t)/20 & \mbox{if } 0\leq t\leq \pi\\ \sin(20t) & \mbox{if } \pi0$, consider the equation $$\dot w(t) + \arctan(w(t)) = \alpha\sin(t)+s(\tilde f)\,.$$ Notice that $\max\tilde f=-\min\tilde f=1$, the period is $T=2\pi$, $\|\tilde f\|_2=\alpha\sqrt\pi$, and the estimate for $\|w\|_\infty$ is $b_0=\alpha\pi/\sqrt{6}$. The following table shows the estimates obtained for several values of $\alpha$. \smallskip \begin{tabular}{|c|c|c|c|} \hline & $\alpha=0.01$ & $\alpha=0.1$ & $\alpha=1$ \\ \hline min-max $g$ & $|s|\leq$ 1.2824e-2 & $|s|\leq$ 0.12756 & $|s|\leq$ 0.90856\\ \hline basic est. & $|s|\leq$ 2.7064e-7 & $|s|\leq$ 2.6716e-4 & $|s|\leq$ 0.11593\\ \hline iterated & $|s|\leq$ 2.7064e-7 & $|s|\leq$ 2.6716e-4 & $|s|\leq$ 0.11593\\ \hline \end{tabular} \paragraph{Remark} For all functions $g_1$ and all $\alpha\neq 0$ in $\dot w(t) + g_1(w(t)) = \alpha\sin(t)+s(\tilde f)$ the iterated method fails to improve the basic estimate. To prove this statement, notice that $\max\tilde f=-\min\tilde f=|\alpha|$, the period is $T=2\pi$, $\|\tilde f\|_2=|\alpha|\sqrt\pi$, and the estimate for $\|w\|_\infty$ is $b_0=|\alpha|\pi/\sqrt{6}$. As in Lemma~\ref{lm_3}, $m_0$ and $M_0$ are non-negative quantities; thus, $k_0\geq |\alpha|$ and $K_0\geq |\alpha|$. Since $b_1$ is an increasing function of $k_0$ and of $K_0$, it follows that $$b_1 \geq \frac \pi 2|\alpha| > \frac \pi {\sqrt 6}|\alpha|=b_0\,.$$ Which indicates that the iteration method is unsuccessful in this case. \paragraph{Example 3} Consider $\dot w(t)+g_1(w(t))=\tilde f(t)+s(\tilde f)$ with $$g_1(\xi)=2\bigl(\arctan\bigl(10000(\xi+0.12)\bigr)+ \arctan\bigl(10000(\xi-0.12)\bigr)\bigr)$$ and $\tilde f$ defined as in Example 1. Note that $g_1$ varies significantly only in the neighbourhood of several points (namely $-0.12$ and $0.12$). In such a case it is better no to apply the iteration method directly, but apply the iteration method with $b_0=\|\tilde f\|_2\sqrt{T/12}$ to $$\dot v(t) + d(v(t))=\tilde f(t)+s_d(\tilde f)\,,$$ where $d(\xi)=g_1(\xi) \mbox{ for } |\xi|<\delta$ and $d(\xi)= g_1(\delta\mbox{sgn}(\xi))$ otherwise with some $0<\delta\leq b_0$. If $b=\inf\{b_0,b_1,\dots\}\leq \delta$ then considering $\|v\|_{\infty}\leq b$ and $d(\xi)=g_1(\xi)$ for $|\xi|\leq\delta$ we get $w=v$. Thus $\|w\|_{\infty} \leq b$ and $s(\tilde f)=s_d(\tilde f)$. The following table shows the estimates obtained for direct application of iteration method and different values of $\delta$.\smallskip \begin{tabular}{|c|c|c|c|} \hline & direct application & $\delta=0.11$ & $\delta=0.1$ \\ \hline $b$ & 1.5056e-1 & 1.1946e-1$>$0.11& 9.0409e-2$<$0.1 \\ \hline min-max $g$ & $|s|\leq$ 6.2759 & $|s|\leq$ 6.2759 & $|s|\leq$ 9.0908e-3\\ \hline basic est. & $|s|\leq$ 4.8202 & $|s|\leq$ 4.8202 & $|s|\leq$ 1.4010e-3\\ \hline iterated & $|s|\leq$ 4.8202 & $|s|\leq$ 4.8202 & $|s|\leq$ 1.6342e-3\\ \hline \end{tabular} \paragraph{Remark} Note that for $\delta=0.1$ the basic estimate yields better result than iterated although $b=$ 9.0409e-2 \$