\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{mathrsfs} \usepackage{amssymb} \AtBeginDocument{{\noindent\small Ninth MSU-UAB Conference on Differential Equations and Computational Simulations. \emph{Electronic Journal of Differential Equations}, Conference 20 (2013), pp. 103--117.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2013 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \setcounter{page}{103} \title[\hfilneg EJDE-2013/Conf/20/ \hfil A Landesman-Lazer condition] {A Landesman-Lazer condition for the boundary-value problem $-u''=a u^+ - b u^- +g(u)$ with periodic boundary conditions} \author[Q. A. Morris, S. B. Robinson \hfil EJDE-2013/conf/20 \hfilneg] {Quinn A. Morris, Stephen B. Robinson} \address{Quinn A. Morris \newline Department of Mathematics and Statistics, The University of North Carolina at Greensboro 116 Petty Building, 317 College Avenue, Greensboro, NC 27412, USA} \email{qamorris@uncg.edu} \address{Stephen B. Robinson \newline Department of Mathematics, Wake Forest University, PO Box 7388, 127 Manchester Hall, Winston-Salem, NC 27109, USA} \email{sbr@wfu.edu} \thanks{Published October 31, 2013.} \subjclass[2000]{34B15} \keywords{Fu\u{c}\'{i}k spectrum; resonance; Landesman-Lazer condition; \hfill\break\indent variational approach} \begin{abstract} In this article we prove the existence of solutions for the boundary-value problem \begin{gather*} -u''=a u^+ - b u^- +g(u)\\ u(0)=u(2 \pi)\\ u'(0)=u'(2 \pi), \end{gather*} where $(a,b)\in \mathbb{R}^2$, $u^+ (x) = \max \{u(x),0\}$, $u^- (x) = \max \{-u(x),0\}$, and $g: \mathbb{R} \to \mathbb{R}$ is a bounded, continuous function. We consider both the resonance and nonresonance cases relative to the Fu\u{c}\'{i}k Spectrum. For the resonance case we assume a generalized Landesman-Lazer condition that depends upon the average values of $g$ at $\pm\infty$. Our theorems generalize the results in \cite{rumbos} by removing certain restrictions on $(a,b)$. Our proofs are also different in that they rely heavily on a variational characterization of the Fu\u{c}\'{i}k Spectrum given in \cite{castro}. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} We are interested in the boundary-value problem \begin{equation} \label{BVP} \begin{gathered} -u''=a u^+ - b u^- +g(u)\\ u(0)=u(2 \pi)\\ u'(0)=u'(2 \pi), \end{gathered} \end{equation} where $(a,b)\in \mathbb{R}^2$, $u^+ (x) = \max \{u(x),0\}$, $u^- (x) = \max \{-u(x),0\}$, and $g: \mathbb{R} \to \mathbb{R}$ is a bounded, continuous function. It has been known since the 1970s with the work of Fu\u{c}\'{i}k \cite{fucik} that the existence of solutions to \eqref{BVP} depends on the parameter values $(a,b)$. Consider the related boundary-value problem \begin{equation} \label{FEP} \begin{gathered} -u''=a u^+ - b u^-\\ u(0)=u(2 \pi)\\ u'(0)=u'(2 \pi ) , \end{gathered} \end{equation} and define the Fu\u{c}\'{i}k Spectrum \begin{equation*} \Sigma := \{ (a,b) \in \mathbb{R}^2 : \text{ There exists a nontrivial solution to \eqref{FEP} } \} . \end{equation*} The Fu\u{c}\'{i}k spectrum represents a nonlinear resonance set for our boundary-value problem, and therefore, in analogy to the Fredholm Alternative for linear operators, we expect that there should exist solutions to \eqref{BVP} without further restrictions when $(a,b) \notin \Sigma$; i.e., the \emph{nonresonance} case. However, when $(a,b) \in \Sigma$; i.e., the \emph{resonance} case, we will require the use of a generalized orthogonality condition, referred to in the literature as a Landesman-Lazer condition, in order to establish the existence of solutions. For future reference we let $\{\lambda_k\}_{k=1}^{\infty}$ represent the eigenvalues of \eqref{FEP} for the special case $a=b=\lambda$, and $\{\phi_k\}$ represents the associated $L^2$-normalized eigenfunctions. We will assume throughout that for some fixed $k$ we have $\lambda_k0 \}. \] Then, \begin{enumerate} \item either $(a,b(a)) \in \Sigma$, or $b(a) = \infty$, \item if $a \leq b < b(a)$, then $(a,b) \notin \Sigma$, and \item $b(a) \ge \lambda_{k+1}$. \end{enumerate} \end{theorem} To prove an existence theorem for the resonance case we will assume the following generalized Landesman-Lazer condition. \begin{definition}[LLD] \rm Let $(a,b) \in \Sigma$. If \[ G^{\pm} = \lim _{u \to \pm \infty} \frac{G(u)}{u}, \] and \[ \Big[ G^+ \int_{\Psi>0} \Psi \,dt + G^- \int_{\Psi<0} \Psi \,dt \Big] < 0, \] for every nontrivial eigenfunction $\Psi$ associated with $(a,b)$, then condition (LLD) is satisfied. \end{definition} \begin{remark} \label{rmk1} \rm The classic Landesman-Lazer condition, and associated existence theorems, can be found in \cite{LL}. That condition requires $g$ to have limits at infinity. Here, however, we only require that $g$ attain a finite limit on average at infinity. Simple examples to consider are $g(t)=\arctan(u)+c\sin(u)$ for $c\in\mathbb{R}$. The classical Landesman-Lazer condition is applicable for $c=0$, but the generalized condition above is applicable for arbitrary $c$. Related generalized conditions are found in \cite{rumbos},\cite{marius} and \cite{marius2} and references therein, where the latter two references are for the case $a=b$. We note that the condition in \cite{rumbos} does not require $\frac{G(u)}{u}$ to have limits at infinity, but rather expresses its condition in terms of $\liminf$ and $\limsup$. \end{remark} \section{Properties of the functional $J$} We begin by defining $X:=\operatorname{span} \{ \phi _1,\phi _2, \ldots ,\phi _k \}$ and let $Y:=X^{\perp}$; i.e, $Y=\operatorname{span} \{ \phi _{k+1},\phi _{k+2}, \ldots ,\phi _n, \ldots \}$, and let the parameter $a\in (\lambda_k,\lambda_{k+1})$. With this splitting in hand, we prove the following lemmas. \begin{lemma} \label{ineq} Choose $\epsilon$ such that $a=(1+\epsilon) \lambda_k$, and let $s=b-a \ge 0$. Let $\delta=\min \{ \frac{\epsilon}{2} \lambda_k, \frac{\epsilon}{2} \}$ and let \[ D=\langle \nabla J (x_2+y_2) - \nabla J (x_1+y_1),x_2-x_1 \rangle , \] where $x_1,x_2 \in X$ and $y_1,y_2 \in Y$. Then \begin{equation} \label{HI} D \le -\delta \| x_2-x_1\| _H^2 + s \big( \| x_2-x_1 \| _{L^2}+\| y_2-y_1 \| _{L^2} \big) \| y_2-y_1 \| _{L^2} \end{equation} \end{lemma} \begin{proof} Observe that for $u,v\in H$, \begin{align*} \nabla J(u)\cdot v &= \int_0^{2\pi}u'v'-a\int_0^{2\pi}u^+v+b\int_0^{2\pi}u^-v \\ &= \int_0^{2\pi}u'v'-a\int_0^{2\pi}uv+s\int_0^{2\pi}u^-v, \end{align*} so \begin{align*} D & = \langle (x_2 + y_2)'-(x_1+y_1)',(x_2-x_1)'\rangle_{L^2} -a\langle (x_2+y_2)-(x_1+y_1),x_2-x_1\rangle_{L^2}\\ &\quad + s\langle (x_2+y_2)^- -(x_1 + y_1)^- , x_2 - x_1 \rangle _{L^2} \end{align*} Using the orthogonality of $X$ and $Y$ to cancel terms, and then recollecting we obain \[ D = \|x_2'-x_1'\|_{L^2}^2-a\|x_2-x_1\|_{L^2}^2 +s\langle (x_2+y_2)^--(x_1+y_1)^-,x_2-x_1\rangle_{L^2} \] To estimate the last part of this expression we make the substitution $x_2-x_1=(x_2+y_2) - (x_1+y_1) - (y_2-y_1)$, use the monotonicity of the function $f(t) = t^-$, and the fact that $|f(t_2)-f(t_1)| \le |t_2-t_1|$, to see that \begin{align*} \langle (x_2 +y_2)^- - (x_1 + y_1)^- , x_2 - x_1 \rangle _{L^2} &\leq -\langle (x_2 +y_2)^- - (x_1 + y_1)^- , y_2-y_1 \rangle _{L^2}\\ &\leq \|(x_2 +y_2)^- - (x_1 + y_1)^-\|_{L^2}\|y_2-y_1\|_{L^2}\\ &\leq \|(x_2 +y_2) - (x_1 + y_1)\|_{L^2}\|y_2-y_1\|_{L^2}\\ &\leq \big(\|x_2-x_1\|_{L^2}+\| y_2-y_1\|_{L^2}\big)\|y_2-y_1\|_{L^2} \end{align*} To estimate the other part we let $a=(1+ \epsilon ) \lambda _k$, and write \[ \|x_2'-x_1'\|_{L^2}^2-a\|x_2-x_1\|_{L^2}^2 =(1+\frac{\epsilon}{2})\|x_2'-x_1'\|^2_{L^2}- \frac{\epsilon}{2}\|x_2'-x_1'\|^2_{L^2}-(1+\epsilon)\lambda_k\|x_2-x_1\|^2_{L^2}. \] Standard estimates show that $\|x'\|^2_{L^2}\leq\lambda_k\|x\|^2_{L^2}$ for all $x\in X$, so \[ \|x_2'-x_1'\|_{L^2}^2-a\|x_2-x_1\|_{L^2}^2 \leq -\frac{\epsilon\lambda_k}{2}\|x_2-x_1\|^2_{L^2} -\frac{\epsilon}{2}\|x_2'-x_1'\|^2_{L^2}. \] The desired inequality now follows. \end{proof} \begin{remark} \label{rmk2} \rm While the necessity of the above lemma may not be immediately obvious, it was determined after repeated estimations of the same type that such an inequality would have wide application to this particular problem. \end{remark} \begin{theorem}[Reduction theorem]\quad \begin{enumerate} \item For fixed $y \in Y$, $J $ is concave and anticoercive on the set $y+X:=\left\{y+x: x \in X\right\}$ and achieves a unique maximum on this set. \item Let $r:Y\to X$ such that $J(r(y)+y)=\max_{x\in X}J(x+y)$, then $r$ is Lipschitz continuous as a function of $L^2(0,2\pi)$ into $H$, and $r(cy)=cr(y)$ for any $c\geq 0$ and $y\in Y$. \item Let $\tilde{J}:Y\to \mathbb{R}:\tilde J(y)=J(r(y)+y)$, then \begin{enumerate} \item $\tilde{J} \in C^1(Y,\mathbb{R})$, and \item $ \tilde{J} (c y )=c^2 \tilde{J} (y)$ for all $c\geq 0$ and $y \in Y$. \end{enumerate} \end{enumerate} \end{theorem} This theorem is well-known from works such as \cite{castro} and \cite{castro2}, but we reproduce it here for completeness, because the properties of $r$ and $J$ play a crucial role in subsequent proofs, and because the approach below is somewhat different in certain details from that found in the references. \begin{proof} Let $u_1,u_2 \in y+X$. By definition of the set $y+X$, write $u_2=y+x_1$ and $u_1=y+x_1$. Consider the quantity, \begin{align*} \langle \nabla J (u_2) - \nabla J (u_1) , u_2 - u_1 \rangle & = \langle \nabla J (x_2+y) - \nabla J (x_1+y),x_2-x_1 \rangle \\ & \le - \delta \| x_2-x_1 \| _H ^2, \end{align*} by \eqref{HI}. Therefore, $J $ is strictly concave on $y+X$. That $J $ is anticoercive on $y+X$ follows from the strict concavity and the Fundamental Theorem of Calculus. The concavity of $J $ on $y+X$ implies that $J $ is weakly upper semicontinuous, and therefore, $J $ must achieve a maximum. That this maximum is unique is a consequence of the strict concavity. The argument so far shows that $r$ is well defined. To see that $r(y)$ is a continuous function, we let $x_1=r(y_1), x_2=r(y_2)$ in \eqref{HI} and note that $D=\langle \nabla J (r(y_2)+y_2) - \nabla J (r(y_1)+y_1) , r(y_2)-r(y_1) \rangle = 0$, since $r(y_1),r(y_2)$ are both critical points with respect to $X$. Making these substitutions, we can solve \eqref{HI} to get \begin{equation}\label{rcont2} \| r(y_2)-r(y_1) \| _H \le \Big( \frac{s+\sqrt{s^2+4 \delta s}}{2 \delta} \Big) \| y_2-y_1 \| _{L^2}. \end{equation} \begin{remark} \label{rmk3} \rm Note also that if $\{y_k\}_{k=1}^{\infty}$ is a bounded sequence in $H$, then $\{y_k\}_{k=1}^{\infty}$ has a convergent subsequence in $L^2(0,2\pi)$, call it $\{y_{k_i}\}_{i=1}^{\infty}$, which, by \eqref{rcont2}, gives us that $\{r(y_{k_i})\}_{i=1}^{\infty}$ converges in $H$. Hence, $r$ is a compact mapping, sending bounded sets in $L^2(0,2\pi)$ into precompact sets in $H$. \end{remark} We now show that $\tilde{J} \in C^1 (Y,\mathbb{R})$. \begin{remark} \label{rmk4} \rm While it may seem at first to be a trivial consequence of the Chain Rule, we must be careful to note that $r$ is not necessarily a $C^1$ function, and therefore, a more technical argument must be made. \end{remark} Consider the quantity $\tilde{J} ( y_2) - \tilde{J} ( y_1 )$. We note that \begin{equation} \label{switch} \begin{split} \tilde{J} ( y_2) - \tilde{J} ( y_1 ) & = J (r(y_2)+y_2)-J ( r( y_1 )+ y_1) \\ & = \left( J (r( y_2 )+y_2 )- J (r( y_2 )+y_1 ) \right) \\ &\quad + \left( J ( r( y_2 )+y_1 )-J (r( y_1 )+y_1 ) \right) \\ & \le \left( J (r( y_2 )+y_2 )- J (r( y_2 )+y_1 ) \right) , \end{split} \end{equation} since $J (r( y_1 ) +y_1 )$ maximizes $J(x+y_1)$. Then \[ \tilde{J} ( y_2) - \tilde{J} ( y_1 ) \le \nabla J (r( y_2 ) + y_1 ) \cdot ( y_2-y_1 ) + o( \| y_2-y_1 \| _H ) \] since $J \in C^1(H,\mathbb{R})$. Then \begin{align*} \tilde{J} ( y_2) - \tilde{J} ( y_1 ) & \le \Big( \nabla J ( r( y_1 )+ y_1)+ \nabla J ( r( y_2 )+y_1 ) - \nabla J (r( y_1 )+y_1 ) \Big) \cdot ( y_2-y_1 ) \\ &\quad + o( \| y_2-y_1\| _H ) \\ & = \nabla J ( r( y_1 )+ y_1) \cdot ( y_2-y_1 ) \\ &\quad + \Big( \nabla J ( r( y_2 )+y_1 ) - \nabla J (r( y_1 )+y_1 ) \Big) \cdot ( y_2-y_1 ) + o( \| y_2-y_1\| _H ) \\ & = \nabla J ( r( y_1 ) + y_1) \cdot (y_2-y_1)+ o( \| y_2-y_1\| _H ), \end{align*} since \begin{align*} & \lim _{\| y_2-y_1\| _H \to 0} \big| \frac{( \nabla J ( r( y_2 )+y_1 ) - \nabla J (r( y_1 )+y_1 ) ) \cdot ( y_2-y_1 )}{\| y_2-y_1\| _H} \big| \\ &\leq \lim _{\| y_2-y_1\| _H \to 0}\| \nabla J ( r( y_2 )+y_1 ) - \nabla J (r( y_1 )+y_1 )\|_{H} = 0, \end{align*} by the continuity of both $\nabla J $ and $r$. If, instead of adding and subtracting $J (r( y_2 )+y_1 )$ in \eqref{switch}, we had added and subtracted $J (r( y_1 )+y_2 )$, we would have concluded that \[ \tilde{J} ( y_2) - \tilde{J} ( y_1 ) \ge \nabla J ( r( y_1 )+ y_1) \cdot (y_2-y_1)+ o( \| y_2-y_1\| _H ). \] Combining these two results, we conclude that \[ \tilde{J} ( y_2) - \tilde{J} ( y_1 ) = \nabla J ( r( y_1 )+ y_1) \cdot (y_2-y_1)+ o( \| y_2-y_1\| _H ), \] and therefore $ \tilde{J} \in C^1(Y,\mathbb{R})$ and $\nabla \tilde{J} (y)= \nabla_Y J (r(y)+y)$. We finish the proof by showing two homogeneity properties, one of the function $r$ and another of the functional $\tilde{J} $. The case $c=0$ is trivial, so we assume $c>0$. For all $u \in H$, \begin{eqnarray}\label{homog1} \begin{aligned} J (cu) &= \frac{1}{2} \Big( \int _0 ^{2 \pi} \left( (cu)' \right) ^2 \,dt - a \int _0 ^{2 \pi} \left( (cu)^+ \right) ^2 \,dt - b \int _0 ^{2 \pi} \left( (cu)^- \right) ^2 \,dt \Big) \\ &= \frac{1}{2} \Big( c^2 \int _0 ^{2 \pi} \left( (u)' \right) ^2 \,dt - a c^2 \int _0 ^{2 \pi} \left( (u)^+ \right) ^2 \,dt - b c^2 \int _0 ^{2 \pi} \left( (u)^- \right) ^2 \,dt \Big) \\ &= c^2 J (u), \end{aligned} \end{eqnarray} since positive constants can be factored out of $(\cdot)^+$ and $(\cdot)^-$. Now consider \begin{equation}\label{homog2} J (x+c y) = J \left( c \left( \frac{x}{c}+ y \right) \right) =c^2 J \left( \frac{x}{c}+y\right) \end{equation} Since $J (x+c y)$ is uniquely maximized at $x=r(c y)$ and $c^2 J (\frac{x}{c}+y)$ will be uniquely maximized at $\frac{x}{c}=r(y)$, then $r(c y)=cr(y)$. Finally, combining \eqref{homog1} and \eqref{homog2} we see that \[ \tilde{J} (c y) = J (r(cy)+c y) = J (c r(y)+c y) =c^2 J (r(y)+ y) = c^2 \tilde{J} (y). \] \end{proof} \begin{remark} \label{rmk5} \rm Note that some of the arguments above can be simplified using the fact that $X$ is finite dimensional, and so, for example, the $L^2$ and $H$ norms are equivalent on $X$. We avoid that simplification to preserve some generality in the arguments that is of interest in other situations. For example, for other choices of boundary conditions we often have two primary curves emanating from the point $(\lambda_{k+1},\lambda_{k+1})$. The lower curve is characterized as we have described here, and the upper curve can be characterized via a similar process where the roles of $X$ and $Y$ are switched in the reduction of $J$. Thus the arguments for $X$, above, must be adapted to the infinite dimensional space $Y$. \end{remark} The properties of $\tilde{J}$ above lead to the proof of Theorem \ref{varchar}. The main idea is to obtain a critical value of $\tilde{J}$ on the set $\{y\in Y:\|y\|_{L^2}=1\}$ by minimization. One then observes that a critical point of $\tilde{J}$ is only a critical point of $J$ if the associated critical value is $0$. We leave the remaining details of this argument for the reader to find in \cite{castro2}. However, we do state the following lemma, which will be helpful in the next section. \begin{lemma} \label{Jest} \rm Assume that $a0$ such that $\tilde{J}(y)\geq \epsilon$ for all $y\in Y$ such that $\|y\|_{L^2}=1$. \item[(ii)] If $b=b(a)$, then $\tilde{J}(y)\geq 0$ for all $y\in Y$ such that $\|y\|_{L^2}=1$, and $\tilde{J}(y)=0$ if and only if f $y$ is an eigenfunction associated with $(a,b)\in\Sigma$. \end{itemize} \end{lemma} Now that we know something about the geometry of $J$, we can say something about the geometry of $E$. \section{The geometry of the functional $E$} As noted previously, the geometry of $E$ is dominated to a large extent by the geometry of $J$. For the nonresonance case this leads to relatively straight forward arguments to prove a saddle geometry, and later the (PS) condition. For the resonance case we will see that the nonresonance arguments are sufficient to reduce both questions to analyzing what happens to the functional in the direction of an eigenfunction. In that case the (LLD) condition will provide a sufficient tool for finishing the analysis. \begin{lemma} \label{lem3} Assume that $b>0$ sufficiently large such that \[ \sup _{x\in X,\| x \|=R} E( x ) < \inf _{u \in \mathscr{Y}} E(u). \] \end{lemma} \begin{proof} Consider the functional $E$ restricted to the subspace $X$. If we assume that $g$ is bounded, then, since $J$ is anticoercive on $X$, and in fact satisfies a quadratic estimate, we can conclude that \begin{equation} \label{eanti} E(x) \le -\eta \| x \|_H^2+ M \| x \|_{L^2}, \end{equation} for appropriate $\eta>0$ and $M>0$, and therefore $E$ is anticoercive on $X$. Now consider $E$ restricted to the set $\mathscr{Y}$. Note first that for $y\neq 0$ \begin{equation} \label{jrange} \tilde{J}(y)=\tilde{J} \Big( \| y \|_{L^2} \frac{y}{\| y \|_{L^2}}\Big) =\| y \|_{L^2}^2 \tilde{J} (\frac{y}{\| y \|_{L^2}}), \end{equation} and so if $\inf _{\| y \|_{L^2}=1} \tilde{J}(y) \ge \epsilon$, as in Theorem \ref{Jest}, we conclude that $\tilde{J}(y) \ge \epsilon \| y \|_{L^2}^2$. Recall that $r(y)$ is Lipschitz continuous, so $\| r(y) \|_H \le M' \| y \|_{L^2}$, for some $M'>0$, and we see that \begin{flalign} \label{mtrange} E (r( y ) + y) &\ge \epsilon \| y \|_{L^2}^2 - M \| r(y)+y \|_{L^2} \\ &\ge \epsilon \| y \|_{L^2}^2 - M \left( \| r(y) \|_{L^2} +\| y \|_{L^2} \right) \\ &\ge \epsilon \| y \|_{L^2}^2 - M ( M'+1) \| y \|_{L^2}. \end{flalign} Thus $E$ is bounded below on $\mathscr{Y}$ (In fact, $E$ is coercive, but that is not necessary here). It follows that there exists some $R$ sufficiently large such that, \[ \sup _{\| x \|=R} E( x ) < \inf _{u \in \mathscr{Y}} E(u). \] \end{proof} A similar estimate is possible in the resonance case, but the proof requires the (LLD) condition. \begin{lemma} \label{lem4} Assume that $b=b(a)$ and that {\rm (LLD)} is satisfied. There exists some $R>>0$ sufficiently large such that \[ \sup _{x\in X,\| x \|=R} E( x ) < \inf _{u \in \mathscr{Y}} E(u). \] \end{lemma} \begin{proof} The argument that $E$ is anticoercive on $X$ remains the same as above. The analysis of $E$ restricted to $\mathscr{Y}$ requires more care. Once again we will show that $E$ restricted to $\mathscr{Y}$ is bounded below, but this time we use an argument by contradiction. Recall that for $u=r(y)+y\in \mathscr{Y}$ we have \[ E(u)=\tilde{J}(y)-\int_0^{2\pi}G(u)\,. \] Let $\{u_n\}\subset \mathscr{Y}$ be a minimizing sequence for $E$, with $u_n=r(y_n)+y_n$. If $E(u_n)$ is bounded below, then we are done, so we assume, without loss of generality, that $E(u_n)\downarrow -\infty$. We know that $\tilde{J}(y_n)\geq 0$. If $\{y_n\}$ were $L^2$ bounded, then $\{u_n\}$ would be bounded, and the integral of $G(u_n)$ would be bounded and we would be done, so, without loss of generality, $\|y_n\|_{L^2}\to\infty$. Moreover, if there is an $\epsilon>0$ such that $\tilde{J}(\frac{y_n}{\|y_n\|_{L^2}})\geq \epsilon$, then we have $E(u_n)\to\infty$ by the estimates in the previous proof. Thus $\tilde{J}(\frac{y_n}{\|y_n\|_{L^2}})\downarrow 0$. Let $v_n=y_n/\|y_n\|_{L^2}$ and let $w_n=u_n/\|y_n\|_{L^2}=r(v_n)+v_n$. It is clear that $\{v_n\}$ is $L^2$ bounded. It follows that $\{r(v_n)\}$ is bounded in $H$, and thus that $\{w_n\}$ is $L^2$ bounded. Since the functional values $J(w_n)=\tilde{J}(v_n)$ are also bounded, it follows that $\{w_n\}$ is $H$ bounded. Thus, without loss of generality, we have $w_n\rightharpoonup w,v_n\rightharpoonup v$ in $H$, and $w_n\to w,v_n\to v$ in $L^2(0,2\pi)$. By continuity, $r(v_n)\to r(v)$ in $H$, so $w=r(v)+v\in\mathscr{Y}$. Moreover, $v$ must be a unit vector in $L^2(0,2\pi)$ so $w$ is nontrivial. By weak lower semicontinuity, we have \[ \int_0^{2\pi}(w')^2\leq \liminf \Big( \int_0^{2\pi}(w_n')^2\Big), \] but $J(w_n)=\tilde{J}(v_n)\to 0$, so \[ \lim\Big( \int_0^{2\pi}(w_n')^2\Big) =\lim \Big(a\int_0^{2\pi}(w_n^+)^2+b\int_0^{2\pi}(w_n^-)^2 \Big) =a\int_0^{2\pi}(w^+)^2+b\int_0^{2\pi}(w^-)^2. \] Hence \[ \int_0^{2\pi}(w')^2\leq a\int_0^{2\pi}(w^+)^2+b\int_0^{2\pi}(w^-)^2, \] i.e. $J(w)=\tilde{J}(v)\leq 0$. But we already know that $\tilde{J}(v)\geq 0$, so it must be that we have equality. Hence $w$ is a nontrivial eigenfunction associated with $(a,b)$. Moreover, it must be that \[ \lim\Big( \int_0^{2\pi}(w_n')^2\Big)=\int_0^{2\pi}(w')^2, \] so $w_n\to w$ in $H$. Applying (LLD) we obtain \[ \frac{1}{\|y_n\|_{L^2}}\int_0^{2\pi}G(u_n) =\int_0^{2\pi}\Big( \frac{G(u_n}{u_n}\Big) w_n \to G^+\int_0^{2\pi}w^+-G^-\int_0^{2\pi}w^-<0. \] Finally, this leads to \[ E(u_n)=\tilde{J}(y_n)-\int_0^{2\pi}G(u_n)\geq -\int_0^{2\pi}G(u_n)\to \infty, \] which is a contradiction. The proof is complete. \end{proof} We now fix $R\gg 0$ such that \[ \sup _{x\in X,\| x \|=R} E( x ) < \inf _{u \in \mathscr{Y}} E(u). \] The final element in establishing the geometry of $E$ is the following \emph{linking} lemma. \begin{lemma}\label{link} Suppose that either $a\leq b \sup _{x \in \partial B_R} E(x). \] \end{lemma} \begin{proof} Let $\gamma:B_R \subseteq X \to H$ be a continuous function such that $\gamma (\partial B_R)= \{x+y:y=0, \| x \|_{L^2}=R\}$. Let $\gamma ( x )=\gamma_X ( x ) + \gamma_Y ( x )$ where $\gamma_X ( x ) \in X$ and $\gamma_Y ( x ) \in Y$. To show that $\gamma (B_R ) \cap \mathscr{Y} \neq \emptyset$, we wish to find $x \in B_R$ so that $\gamma_X ( x )=r(\gamma_Y (x))$. Let $F(x)=\gamma_X ( x )-r(\gamma_Y (x))$. Now, let $h(x,t)=t F(x) + (1-t) x$. Note first that if $x \in \partial B_R $, then $F(x)=x \neq 0$, so $h(x,t)=t x + (1-t) x=x$. Applying Brouwer degree, we see that $\deg (F, B_R ,0 )=\deg (I, B_R ,0 )=1$, and hence, \[ \inf _{\gamma \in \Gamma} \sup _{x \in B_R} E(\gamma ( x)) \ge \inf _{u \in \mathscr{Y}} E(u) > \sup _{x \in \partial B_R} E(x) \] \end{proof} \section{The Palais-Smale condition} In this section it is simpler to prove (PS) for both the nonresonance and resonance cases in one theorem. \begin{theorem} \label{PS2} If $(a,b) \notin \Sigma$, or if $(a,b) \in \Sigma$ and (LLD) is satisfied, then the functional $E$ satisfies (PS). \end{theorem} \begin{proof} First, suppose that $\{ u_k \}_{k=1}^{\infty}$ is a sequence such that $\{E(u_k)\}_{k=1}^{\infty}$ is bounded and $\nabla E( u_k ) \to 0$ in $H$. We must show that $\{u_k\}$ has a converging subsequence in $H$. The crucial step is to show that some subsequence is $L^{\infty}$ bounded. Suppose to the contrary that $\| u_k \| _{\infty} \to \infty$. Then let $v_k=u_k/\| u_k \| _{\infty}$. Note that if we divide the energy functional through by $\| u_k \| _{\infty} ^2$, we obtain \[ \frac{E(u_k)}{\| u_k \| _{\infty} ^2} = \frac{1}{2}\int _0 ^{2 \pi} (v_k')^2 \,dt - \frac{a}{2} \int _0 ^{2 \pi} (v_k^+)^2 \,dt - \frac{b}{2} \int _0 ^{2 \pi} (v_k ^-)^2 \,dt - \int _0 ^{2 \pi} \frac{G(u_k)}{\| u_k \| _{\infty} ^2} \,dt \] If we take a limit, the term $ E(u_k)/\| u_k \| _{\infty} ^2 \to 0$, since $\{E(u_k)\}_{k=1}^{\infty}$ is bounded, and \[ \int _0 ^{2 \pi} \frac{G(u_k)}{\| u_k \| _{\infty} ^2} \,dt \to 0, \] since $G'=g$ is a bounded function, and thus $| G(u_k)| \le C | u_k |$, where $|g(u_k)| \le C$ $\forall u_k$. Also note that $\| v_k^{\pm} \| _{\infty} \le 1$, so $\int _0 ^{2 \pi} (v^{\pm})^2 \,dt$ is likewise bounded. Therefore, we may conclude that \[ \frac{1}{2}\int _0 ^{2 \pi} (v_k')^2 \,dt , \] is bounded and therefore $\| v_k \| _H $ is bounded. Thus, without loss of generality, there exists $\Psi \in H$ such that $v_k \rightharpoonup \Psi$ in $H$ and $v_k \to \Psi$ in $L^2 (0,2 \pi)$ and $C[0,2 \pi]$, by Alaoglu's theorem and a standard compact embedding theorem. We know that $\| \Psi \| _{\infty} =1$ since $\| v_k \| _{\infty} =1$ $\forall k$ and convergence is uniform, so $\Psi$ is nontrivial. Using this convergence, we can now show that, for any $w\in H$, \begin{align*} 0&= \lim_{k \to \infty} \frac{\nabla E(u_k)}{\| u_k \| _{\infty} } \cdot w \\ &= \lim_{k \to \infty} \Big[ \int _0 ^{2 \pi} v_k' w' \,dt - a \int _0 ^{2 \pi} v_k^+ w \,dt +b \int _0 ^{2 \pi} v_k ^- w \,dt - \int _0 ^{2 \pi} \frac{g(u_k)}{\| u_k \| _{\infty} } w \,dt \Big] \\ &= \int _0 ^{2 \pi} \Psi' w' \,dt - a \int _0 ^{2 \pi} \Psi ^+ w \,dt + b \int _0 ^{2 \pi} \Psi ^- w \,dt \end{align*} Thus $\Psi$ is a weak solution of the Fu\u{c}\'{i}k eigenvalue problem, \eqref{FEP}, and hence $\Psi$ is a nontrivial Fu\u{c}\'{i}k eigenfunction. If $(a,b) \notin \Sigma$, then this is a contradiction and $\| u_k \| _{\infty} $ is bounded as claimed. If $(a,b) \in \Sigma$, then consider the quantity, \begin{equation} \label{gG} \frac{2 E(u_k) - \nabla E(u_k) \cdot u_k}{\| u_k \| _{\infty} } =-2 \int _0 ^{2 \pi} \frac{G(u_k)}{\| u_k \| _{\infty} } + \int _0 ^{2 \pi} g(u_k) \frac{u_k}{\| u_k \| _{\infty} } \end{equation} Note first that, by assumption, \[ \lim_{k \to \infty} \frac{2 E(u_k) - \nabla E(u_k) \cdot u_k}{\| u_k \| _{\infty} } = 0. \] We can rewrite the first term on the right hand side of \eqref{gG} so that \begin{equation}\label{G} \begin{split} \lim_{k \to \infty} \int _0 ^{2 \pi} \frac{G(u_k)}{\| u_k \| _{\infty} } \,dt &=\lim_{k \to \infty} \int _0 ^{2 \pi} \frac{G(u_k)}{u_k} \frac{u_k}{\| u_k \| _{\infty} } \,dt \\ &=\lim_{k \to \infty} \int_{\Psi<0} \frac{G(u_k)}{u_k} \frac{u_k}{\| u_k \| _{\infty} } \,dt + \int_{\Psi>0} \frac{G(u_k)}{u_k} \frac{u_k}{\| u_k \| _{\infty} } \,dt \\ &= G^- \int_{\Psi<0} v \,dt + G^+ \int_{\Psi>0} v \,dt, \end{split} \end{equation} where we have used the fact that $\Psi$ is only $0$ on a finite set, and that the integrands converge uniformly to their limits. Now, we need only to determine what the last integral in \eqref{gG} converges to in order to reach a contradiction, which will show that $\| u_k \| _{\infty} $ is bounded. The following two lemmas establish the convergence properties of the parts of the integrand. \begin{lemma} \label{lem6} $\frac{u_k}{\| u_k \| _{\infty} }$ has a convergent subsequence in $H$. \end{lemma} \begin{proof} Let \begin{gather*} P(u) \cdot v:= \langle u,v \rangle _H\\ S(u) \cdot v:=-(a+1) \int _0 ^{2 \pi} u^+ v \,dt + (b+1) \int _0 ^{2 \pi} u^- v \,dt\\ T(u) \cdot v:=-\int _0 ^{2 \pi} g(u) v \,dt \end{gather*} so that \[ \nabla E(u) \cdot v = (P(u)+S(u)+T(u)) \cdot v. \] First, let us consider $S(u)$. Since $ u_k/\| u_k \| _{\infty} \overset{L^2}{\to} \Psi$, it follows that $\big(u_k/\| u_k \| _{\infty} \big) ^+ \overset{L^2}{\to} \Psi ^+$ and $\big( u_k/\| u_k \| _{\infty} \big) ^- \overset{L^2}{\to} \Psi ^-$ by the Lebesgue Dominated Convergence Theorem. Noting that \begin{align*} \frac{S(u_k)}{\| u_k \| _{\infty} } \cdot v & = S \Big( \frac{u_k}{\| u_k \| _{\infty} } \Big) \cdot v \\ & = -(a+1) \int _0 ^{2 \pi} \Big( \frac{u_k}{\| u_k \| _{\infty} } \Big) ^+ v \,dt + (b+1) \int _0 ^{2 \pi} \Big( \frac{u_k}{\| u_k \| _{\infty} } \Big) ^- v \,dt, \end{align*} we conclude that $S \big( u_k/\| u_k \| _{\infty} \big) \cdot v \to S ( \Psi) \cdot v$ for all $v \in H$. Since \begin{align*} &\Big| \Big( S \big( \frac{u_k}{\| u_k \| _{\infty} } \big) - S ( \Psi ) \Big) \cdot v \Big| \\ &= \Big| -(a+1) \int _0 ^{2 \pi} \Big( \big( \frac{u_k}{\| u_k \| _{\infty} } \big) ^+-\Psi ^+ \Big) v \,dt + (b+1) \int _0 ^{2 \pi} \Big(\big( \frac{u_k}{\| u_k \| _{\infty} } \big) ^- -\Psi ^- \Big) v \,dt \Big| \\ & \le (a+1) \big\| \big( \frac{u_k}{\| u_k \| _{\infty} } \big) ^+ -\Psi ^+ \big\| _{L^2} +(b+1) \| \big( \frac{u_k}{\| u_k \| _{\infty} } \big) ^--\Psi ^- \| _{L^2}, \end{align*} for $\| v \| _{L^2} \le 1$, then $S ( u_k/\| u_k \| _{\infty}) \to S ( \Psi)$ in $H^*$. Now, considering $T(u)$, we see that \[ T(u) \cdot v = - \int _0 ^{2 \pi} g(u) v \,dt, \] so $\{T(u_k)\}$ is bounded in $H^*$ since \[ \| T(u)\| _{H^*} \le \| g(u) \| _{L^2} \le C. \] So $\| T(u_k)/\| u_k \| _{\infty} \| _{H^*} \to 0$ as $k \to \infty$. Finally we consider $P$. By the Riesz Representation Theorem, there is an isomorphism, $i: H^* \to H$ such that $i \circ P (u) = u$ for all $u \in H$. So, $P$ is an invertible linear operator with continuous inverse. Recalling that $\nabla E(u) = P(u) + S(u) + T(u)$ and that, by a hypothesis of the Palais-Smale condition, $\nabla E(u_k) \to 0$ in $H^*$ as $k \to \infty$, we see that \[ \frac{\nabla E(u_k)}{\| u_k \| _{\infty} } =P\Big( \frac{u_k}{\| u_k \| _{\infty} } \Big) + S\Big( \frac{u_k}{\| u_k \| _{\infty} } \Big) + \frac{T(u_k)}{\| u_k \| _{\infty} } \] can be rewritten as \[ \frac{u_k}{\| u_k \| _{\infty} } =P^{-1} \Big( \frac{\nabla E(u_k)}{\| u_k \| _{\infty} } - S\big( \frac{u_k}{\| u_k \| _{\infty} } \big) - \frac{T(u_k)}{\| u_k \| _{\infty} } \Big). \] Therefore, invoking the continuity of $P^{-1}$ and taking a limit as $k \to \infty$, we conclude that \[ \frac{u_k}{\| u_k \| _{\infty} } \overset{H}{\to} P^{-1} \big( 0 - S( \Psi ) - 0 \big) = P^{-1}(-S(\Psi ))=\Psi. \] \end{proof} \begin{lemma} \label{lem7} \[ g(u_k) \rightharpoonup G^+ \chi _{\Psi > 0} + G^- \chi _{\Psi < 0} \] \end{lemma} \begin{proof} By Alaoglu's Theorem, we know that $\left\{g(u_k)\right\}_{k=1}^{\infty}$ has a weakly convergent subsequence since $\{g(u_k)\}_{k=1}^{\infty}$ is bounded in $L^2 [0,2 \pi]$. Let $g(u_k) \rightharpoonup \mathscr{G}$. Now we need only to show that \[ \mathscr{G} = G^+ \chi _{\Psi > 0} + G^- \chi _{\Psi < 0} \] It will be helpful to recall some standard properties of Fu\u{c}\'{i}k eigenfunctions, $\Psi$, namely that they are continuously differentiable and have a finite number of critical points. For a proof of such properties and an explicit formulation for such $\Psi$, see \cite{rumbos}. Let $v=\chi _{[ c,d ]}$ be the characteristic function of some closed interval where $0 \le c < d \le 2 \pi$ and $[c,d] \subset \{ x:\Psi (x) >0, \Psi ' (x) >0 \}$. Then we may write \begin{equation} \label{weakconv} \begin{split} \int _0 ^{2 \pi} g(u_k) \chi _{[ c,d ]} &= \int_c^d g(u_k) \\ &= \int_c^d g(u_k) \Big( 1-\frac{\frac{u_k'}{\| u_k \| _{\infty} }} {\Psi ' (e)} \Big) +\int_c^d g(u_k) \Big(\frac{\frac{u_k'}{\| u_k \| _{\infty} }}{\Psi ' (e)} \Big), \end{split} \end{equation} where $c0$ and define $M:=\| \mathscr{G} \| _{\infty}$. The fact that $M$ exists is a consequence of the boundedness of $g$. Choose $c_i, d_i,e_i$ such that \[ \cup_{i=i}^n [c_i, d_i]=[c,d],\text{ }|d-c| =\sum_{i=1}^n |d_i-c_i|,\text{ and } |1 - \frac{\Psi ' (x)}{\Psi ' (e_i)} | < \frac{\epsilon}{M}\quad \forall x \in [c_i, d_i]. \] Then \begin{align*} \int _0 ^{2 \pi} g(u_k) \chi _{[c,d ]} &= \sum_{i=1}^n\int_{c_i}^{d_i} g(u_k) \\ &= \sum_{i=1}^n\int_{c_i}^{d_i} g(u_k) \Big( 1-\frac{\frac{u_k'}{\| u_k \| _{\infty} }}{\Psi ' (e_i)} \Big) +\sum_{i=1}^n\int_{c_i}^{d_i} g(u_k) \Big(\frac{\frac{u_k'}{\| u_k \| _{\infty} }}{\Psi ' (e_i)} \Big). \end{align*} We see that \[ \sum_{i=1}^n\int_{c_i}^{d_i} g(u_k) \Big(\frac{\frac{u_k'}{\| u_k \| _{\infty} }}{\Psi ' (e_i)} \Big) \to \sum_{i=1}^n\int_{c_i}^{d_i} G^+\chi_{[c_i,d_i]}=\int_c^d G^+\chi_{[c,d]}, \] while \[ \sum_{i=1}^n\int_{c_i}^{d_i} g(u_k) \Big( 1-\frac{\frac{u_k'}{\| u_k \| _{\infty} }}{\Psi ' (e_i)} \Big) \to \sum_{i=1}^n \int_{c_i}^{d_i} \mathscr{G} \Big( 1 - \frac{\Psi ' (x)}{\Psi ' (e_i)} \Big) \] and \[ \big|\sum_{i=1}^n \int_{c_i}^{d_i} \mathscr{G} \big( 1 - \frac{\Psi ' (x)}{\Psi ' (e_i)} \big)\big| \leq \sum_{i=1}^n \int_{c_i}^{d_i}\big| \mathscr{G} \big( 1 - \frac{\Psi ' (x)}{\Psi ' (e_i)} \big)\big| \le \sum_{i=1}^n \epsilon |d_i-c_i| = \epsilon (d-c) \] Since $\epsilon$ was chosen arbitrarily, we may let $\epsilon \to 0$, and hence \begin{equation} \label{equals} \lim_{k \to \infty} \int _0 ^{2 \pi} g(u_k) \chi _{[c,d]} = \int _0 ^{2 \pi} G^+ \chi _{[c,d]} \; \forall [c,d] \subset \{ x:\Psi (x) >0, \Psi ' (x) >0 \}. \end{equation} The exact same calculations will show that, given $[c,d] \subset \{ x:\Psi (x) >0, \Psi ' (x) <0\}$, we get the same conclusion as in \eqref{equals}. For $[c,d] \subset \{ x:\Psi (x) < 0, \Psi ' (x) > 0 \}$ and $[c,d] \subset \{ x:\Psi (x) < 0, \Psi ' (x) < 0 \}$, we can complete the same calculations, but will this time find that \[ \lim_{k \to \infty} \int _0 ^{2 \pi} g(u_k) \chi _{[c,d]} = \int _0 ^{2 \pi} G^- \chi _{[c,d]} . \] Hence, we may recombine the integrals to see that \begin{equation}\label{g} \lim_{k \to \infty} \int _0 ^{2 \pi} g(u_k) \chi _{[c,d]} = \int _0 ^{2 \pi} \big( G^+ \chi _{\Psi >0} + G^- {\chi}_{\Psi < 0} \big) {\chi}_{[c,d]}. \end{equation} We proceed, via standard arguments, to replace $\chi _{[c,d]}$ in \eqref{g} by arbitrary $v\in L^2(0,2\pi)$. So far the closed intervals above avoid critical points of $\psi$. If an interval does include critical points, however, we may delete an arbitrarily small neighborhood of each of the finitely many critical points so that the total change in the integral is less than some $\epsilon$. Hence \eqref{g} holds for arbitrary $[c,d]$ up to an arbitrary $\epsilon$. Let $\epsilon$ go to zero and we have \eqref{g} for all closed subintervals of $[0,2\pi]$. We can immediately generalize to step functions, and then to arbitrary $v\in L^2(0,2\pi)$ by taking limits of approximating step functions. Thus we have \[ \lim_{k \to \infty} \int _0 ^{2 \pi} g(u_k) v = \int _0 ^{2 \pi} \left( G^+ \chi _{\Psi >0} + G^- {\chi}_{\Psi < 0} \right) v , \] which proves the lemma. \end{proof} As a consequence of this lemma we have \begin{equation} \int _0 ^{2 \pi} g(u_k) \frac{u_k}{\| u_k \| _{\infty} }\to G^+ \int_{\Psi>0} \Psi + G^- \int_{\Psi<0} \Psi. \label{g2} \end{equation} Combining \eqref{gG}, \eqref{G}, and \eqref{g2}, we now find that \[ 0= - \Big[G^+ \int_{\Psi>0} \Psi + G^- \int_{\Psi<0} \Psi \Big] , \] a contradiction of (LLD). Hence, $\{ u_k \}_{k=1}^{\infty}$ is a bounded sequence in $L^{\infty}$. We note that $\{E(u_k)\}_{k=1}^{\infty}$ is bounded by hypothesis and all the integral terms, except the one involving $u_k'$, are bounded by virtue of $\{ u_k \}_{k=1}^{\infty}$ being bounded in $L^{\infty}$. Hence, $\{ u_k \}_{k=1}^{\infty}$ is a bounded sequence in $H$. Now, as before, consider $\nabla E(u_k)=P(u_k)+S(u_k)+T(u_k)$. Since $\{ u_k \}_{k=1}^{\infty}$ is bounded in $H$, then there exists a subsequence such that $u_k \overset{H}{\rightharpoonup} u$ and $u_k \overset{L^2,\, C}{\to} u$. Now, taking a limit, we see that \[ 0=\lim_{k \to \infty} \nabla E(u_k)= \lim_{k \to \infty} \left ( P(u_k)+S(u_k)+T(u_k)\right ) \] and since $P$ is invertible, $S(u_k)\to S(u)$, and $T(u_k)\to T(u)$, we may rearrange the equation to see that \[ u_k \overset{H}{\to} u=P^{-1} ( -S(u) - T(u)). \] Hence $\{ u_k \}_{k=1}^{\infty}$ has a subsequence which converges in $H$, and therefore we have satisfied (PS). \end{proof} \section{Main Result} \begin{theorem} \label{ethm2} Assume that $g:\mathbb{R}\to\mathbb{R}$ is bounded and continuous. If $(a,b) \notin \Sigma$ or if $(a,b) \in \Sigma$ and (LLD) is satisfied, then there exists at least one weak solution to \eqref{BVP}. \end{theorem} \begin{proof} Recall that the functional $E$ satisfies (PS) due to Theorem \ref{PS2}. Also, if \[ \Gamma:=\{\gamma:B_R \subseteq X \to H : \gamma \big|_{\partial B_R}(x)=x, \gamma \mbox{ cont. }\}, \] then \[ \inf _{\gamma \in \Gamma} \sup _{x \in B_R} E(\gamma ( x)) > \sup _{x \in \partial B_R} E(x), \] due to Lemma \ref{link}. Hence, by Theorem \ref{SPT}, \[ c:= \inf_{\gamma \in \Gamma} \sup_{x \in X} E(\gamma (x) ) \] is a critical value, and so \eqref{BVP} has a weak solution. \end{proof} \subsection*{Acknowledgements} We would like to acknowledge the many significant contributions that Alfonso Castro has made to Nonlinear Analysis in general, and to the inspiration for this paper in particular. \begin{thebibliography}{99} \bibitem{rumbos} D. Bliss, J. Buerger and A. Rumbos; \emph{Periodic Boundary Value Problems and the Dancer-Fu\u{c}\'{i}k Spectrum under Conditions of Resonance}. Electron. J. Differential Equations. 112 (2011), 1-34. \bibitem{castro2} A. Castro; \emph{Hammerstein Integral Equations with Indefinite Kernel}. Math. and Math. 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