\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 226, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/226\hfil Nonuniqueness and fractional index ] {Nonuniqueness and fractional index convolution complementarity problems} \author[D. E. Stewart \hfil EJDE-2014/226\hfilneg] {David E. Stewart} % in alphabetical order \address{David E. Stewart\newline Department of Mathematics, University of Iowa, Iowa City, IA 52242, USA} \email{david-e-stewart@uiowa.edu} \thanks{Submitted June 4, 2014. Published October 22, 2014.} \subjclass[2000]{90C33, 74M20, 34A08} \keywords{Convolution complementarity problem; mechanical impact; \hfill\break\indent viscoelasticity; uniqueness} \begin{abstract} Uniqueness of solutions of fractional index convolution complementarity problems (CCPs) has been shown for index $1+\alpha$ with $-1<\alpha\leq0$ under mild assumptions, but not for $0<\alpha<1$. Here a family of counterexamples is given showing that uniqueness generally fails for $0<\alpha<1$. These results show that uniqueness is expected to fail for convolution complementarity problems of the type that arise in connection with solutions of impact problems for Kelvin-Voigt viscoelastic rods. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \allowdisplaybreaks \section{Convolution complementarity problems} \label{sec:CCPs} A \emph{convolution complementarity problem} (CCP) is the task, given functions $m:[0,\infty)\to\mathbb{R}^{n\times n}$ and $q:[0,\infty)\to\mathbb{R}^{n}$, of finding a function $z:[0,\infty)\to\mathbb{R}^{n}$ where \begin{equation} K\ni z(t) \perp \int_{0}^{t}m(t-\tau)\, z(\tau)\, d\tau +q(t)\in K^{*}\quad\text{for almost all }t\geq0,\label{eq:ccp-def} \end{equation} where $K$ is a closed and convex cone ($x\in K$ and $\alpha\geq0$ implies $\alpha x\in K$) and $K^{*}$ is its dual cone: \begin{equation} K^{*}=\{ y\in\mathbb{R}^{n}\mid x^{T}y\geq0\text{ for all }x\in K\} .\label{eq:dual-cone} \end{equation} Most commonly $K=\mathbb{R}_{+}^{n}$, for which $K^{*}=\mathbb{R}_{+}^{n}=K$ . Also note that ``$a\perp b$'' means that $a$ and $b$ are orthogonal: $a^{T}b=0$. Convolution complementarity problems were introduced by this name in \cite{ste:ccpaip}, although this concept was used by Petrov and Schatzman \cite{ps:vmcs}. One reason for studying CCPs is their use in studying mechanical impact problems. In particular, Petrov and Schatzman \cite{ps:vmcs} studied the problem of a visco-elastic rod impacting a rigid obstacle: \begin{gather} \rho u_{tt} = Eu_{xx}+\beta u_{txx}+f(t,x),\quad x\in(0,L),\label{eq:visco-elastic-rod-1}\\ N(t) = -\left[Eu_{x}(t,0)+\beta u_{tx}(t,0)\right],\label{eq:visco-elastic-rod-2}\\ 0 = -\left[Eu_{x}(t,L)+\beta u_{tx}(t,L)\right],\label{eq:visco-elastic-rod-3}\\ 0\leq N(t) \perp u(t,0)\geq0.\label{eq:visco-elastic-rod-4} \end{gather} Here $u(t,x)$ is the displacement at time $t$ and position $x\in(0,L)$; \eqref{eq:visco-elastic-rod-1} is the equation for one-dimensional Kelvin--Voigt visco-elasticity; \eqref{eq:visco-elastic-rod-2} is the boundary condition for a contact force $N(t)$ applied at $x=0$; \eqref{eq:visco-elastic-rod-3} is the boundary condition for a free end at $x=L$; and finally, \eqref{eq:visco-elastic-rod-4} is the Signorini-type contact condition at $x=0$, indicating that separation ($u(t,0)>0$) implies no contact force ($N(t)=0$) while a positive contact force ($N(t)>0$) implies contact ($u(t,0)=0$). Because the system is time-invariant, $u(t,0)$ can be represented as $\widehat{u}(t,0)+\int_{0}^{t}m(t-\tau) N(\tau)\, d\tau$ where $\widehat{u}(t,x)$ is the solution of the linear system with $N(t)\equiv0$ and no contact conditions, and the kernel function $m(t)\sim m_{0}t^{1/2}$ as $t\downarrow0$ with $m_{0}>0$. While existence of solutions has been demonstrated for these problems \cite{ps:vmcs,ste:dcpficcp}, uniqueness has not. This paper shows why. The \emph{index} of a CCP is the number $\beta$ where $(d/dt)^{\beta}m(t)=m_{0}\,\delta(t)+m_{1}(t)$ with $\delta$ the Dirac-$\delta$ function, and $\int_{[0,\epsilon)}\| (d/dt)^{\beta}m_{1}(t)\| \, dt\to0$ as $\epsilon\downarrow0$, and $m_{0}$ is an invertible matrix. If we allow fractional derivatives in the sense of \cite{kir:gfca}, then $\beta$ need not be an integer. Typically, for index $\beta$ we have $m(t)\sim m_{0}\, t^{\beta-1}$ as $t\downarrow 0$. Basic results for fractional index CCPs with index $0<\beta<1$ were published in \cite{sw:ficcp}. In particular, combining the results of \cite{ste:ccpaip}, \cite{sw:ficcp}, and \cite{ste:dcpficcp} we can say that under fairly mild regularity and positivity conditions (related to the index), solutions exist for $0\leq\beta<2$ and are unique for $0\leq\beta\leq1$. These results can be extended to prove existence of solutions for index $\beta=2$. However, it is known that solutions are not unique in general for $\beta=2$. Neither existence nor uniqueness hold in general for $\beta>2$ (see \cite[\S3.2.5]{ste:diihc}). For clarity as to what exactly has been proven for $1<\beta<2$, we quote the main results of \cite[\S8]{ste:dcpficcp}: \begin{theorem} \label{thm1} If $m(t)=m_{0}t^{\beta-1}+m_{1}(t)$ for $t\geq0$ with $m_{0}>0$, $m_{1}$ Lipschitz, $1<\beta<2$, $\alpha=\beta-1$, $q'\in H^{\alpha/2}(0,T^{*})$ with $T^{*}>0$, and $q(0)\geq0$, then there is a solution $z(\cdot)\in H^{-\alpha/2}(0,T^{*})$ of \[ 0\le z(t) \perp (m*z)(t)+q(t)\geq0\quad\text{for all }t\geq0. \] \end{theorem} As yet, an open question has been whether uniqueness holds for $1<\beta<2$. This paper answers this question in the negative: there are functions $q(\cdot)$ for which there are at least two solutions for $z(\cdot)$ with $m(t)=t^{\alpha}$ for $0<\alpha<1$ where $\alpha=\beta-1$. The construction of a counter-example to uniqueness is somewhat involved. It proceeds in a similar manner to Mandelbaum's counter-example to uniqueness for certain differential complementarity problems \cite{man:dcp}: we first prove equivalence of uniqueness of solutions for \eqref{eq:ccp-def} for $n=1$ to non-existence of a non-zero function $\zeta:[0,\infty)\to\mathbb{R}$ satisfying \begin{equation} \zeta(t)(m*\zeta)(t)\leq0\quad\text{for all }t\geq0.\label{eq:mandelbaum-cond} \end{equation} Given such a $\zeta$ we are able to construct both a function $q(\cdot)$ a pair of solutions $z_{1}(\cdot)$ and $z_{2}(\cdot)$ of \eqref{eq:ccp-def}. The next task is then to construct a suitable $\zeta(\cdot)\not\equiv0$ satisfying \eqref{eq:mandelbaum-cond} for $m(t)=t^{\alpha}$. We define the \emph{floor} of a real number $z$ to be $\lfloor z\rfloor =\max\{ k\in\mathbb{Z}\mid k\leq z\} $, and the \emph{ceiling} of $z$ to be $\lceil z\rceil =\min\{ k\in\mathbb{Z}\mid k\geq z\} $. \section{Mandelbaum's condition for CCPs} \label{sec:Mandelbaum's-condition} In \cite{man:dcp}, Mandebaum considered differential complementarity problems of the form \begin{gather} \frac{dw}{dt}(t) = M z(t)+q'(t),\quad w(0)=q(0),\label{eq:dcp-1}\\ 0\leq w(t) \perp z(t)\geq0\label{eq:dcp-2} \end{gather} for all $t$. He was able to show that multiple solutions may exist even for $M=\begin{bmatrix} 2 & -1\\ 3 & 1 \end{bmatrix}$ which is positive definite, but not symmetric. The tool that Mandelbaum used was the following theorem. \begin{theorem} \label{thm:mandelbaum} The system \eqref{eq:dcp-1}, \eqref{eq:dcp-2} has a unique solution if and only if$\omega(t)\circ\zeta(t)\leq0$ and $d\omega/dt(t)=M\zeta(t)$ for $t\geq0$ and $\omega(0)=0$ implies that $\zeta(t)=0$ for all $t\geq0$. \end{theorem} Note that ``$a\circ b$'' is the Hadamard product given by $(a\circ b)_{i}=a_{i}b_{i}$ for all $i$. In the scalar case ($n=1$), the Hadamard product reduces to the ordinary product of real numbers. \begin{theorem} \label{thm:ccp-mandelbaum} The system \eqref{eq:ccp-def} with $n=1$ has unique solutions for all $q(\cdot)$ if and only if $\zeta(t)(m*\zeta)(t)\leq0$ for all $t\geq0$ implies $\zeta(t)=0$ for all $t\geq0$. \end{theorem} \begin{proof} The proof is based on Mandelbaum's proof. The sufficiency of the condition for uniqueness can be shown via the contrapositive: if the system \eqref{eq:ccp-def} has two distinct solutions $z_{1}(\cdot)$ and $z_{2}(\cdot)$ then we can set $\zeta(t)=z_{1}(t)-z_{2}(t)$ not identically zero where \begin{align*} \zeta(t)(m*\zeta)(t) & = (z_{1}(t)-z_{2}(t))(m*z_{1}+q-m*z_{2}-q)(t)\\ & = z_{1}(t)(m*z_{1}+q)(t)-z_{1}(t)(m*z_{2}+q)(t)\\ & \quad -z_{2}(t)(m*z_{1}+q)(t)+z_{2}(t)(m*z_{2}+q)(t)\\ & = -z_{1}(t)(m*z_{2}+q)(t)-z_{2}(t)(m*z_{1}+q)(t)\leq0 \end{align*} for all $t\geq0$, since $z_{1}(t),\, z_{2}(t)\geq0$, $(m*z_{1}+q)(t),(m*z_{2}+q)(t)\geq0$ and $z_{1}(t)(m*z_{1}+q)(t)=z_{2}(t)(m*z_{2}+q)(t)=0$. To show necessity, we again use the contrapositive, and suppose that there is a function $\zeta(\cdot)$ which is not everywhere zero and $\zeta(t)(m*\zeta)(t)\leq0$ for all $t\geq0$. Let $\omega=m*\zeta$. Note that $\omega(t)\zeta(t)\leq0$. We wish to find functions $q(\cdot)$, $z_{1}(\cdot)$, and $z_{2}(\cdot)$ such that $z_{1}(\cdot)$ and $z_{2}(\cdot)$ are both solutions to \eqref{eq:ccp-def}. Let $E^{+}=\{ t\geq0\mid\omega(t)>0\} $, $E^{-}=\{ t\geq0\mid\omega(t)<0\} $, and $E^{0}=\{ t\geq0\mid\omega(t)=0\} $. Let $w_{1}(t)=\max(\omega(t),0)$ and $w_{2}(t)=\max(-\omega(t),0)$. For $t\in E^{+}$ we set $z_{1}(t)=0$ and $z_{2}(t)=-\zeta(t)>0$; for $t\in E^{-}$ we set $z_{1}(t)=\zeta(t)\geq 0$ and $z_{2}(t)=0$; for $t\in E^{0}$ we set $z_{1}(t)=\max(\zeta(t),0)$ and $z_{2}(t)=\max(-\zeta(t),0)$. Then $\zeta(t)=z_{1}(t)-z_{2}(t)$ and $z_{1}(t),\, z_{2}(t),\, w_{1}(t),\, w_{2}(t)\geq0$ for all $t\geq0$. For $t\in E^{+}$, $w_{1}(t)z_{1}(t)=0$ since $z_{1}(t)=0$, and $w_{2}(t)z_{2}(t)=0$ since $w_{2}(t)=0$; for $t\in E^{-}$, $w_{1}(t)z_{1}(t)=0$ since $w_{1}(t)=0$, and $w_{2}(t)z_{2}(t)=0$ since $z_{2}(t)=0$; for $t\in E^{0}$, $w_{1}(t)z_{1}(t)=w_{2}(t)z_{2}(t)=0$ since $w_{1}(t)=w_{2}(t)=0$. Thus both $(z_{1}(\cdot),w_{1}(\cdot))$ and $(z_{2}(\cdot),w_{2}(\cdot))$ satisfy the complementarity conditions. We now check the dynamic conditions. Let $q(t)=w_{1}(t)-(m*z_{1})(t)$ for all $t\geq0$. Then, clearly, $w_{1}(t)=(m*z_{1})(t)+q(t)$. On the other hand, $w_{1}(t)-w_{2}(t)=\omega(t)$ and $z_{1}(t)-z_{2}(t)=\zeta(t)$ for all $t\geq0$, so \begin{align*} w_{2}(t) & = w_{1}(t)-\omega(t)\\ & = (m*z_{1})(t)+q(t)-(m*\zeta)(t)\\ & = (m*(z_{1}-\zeta))(t)+q(t)\\ & = (m*z_{2})(t)+q(t). \end{align*} Thus the dynamic conditions also hold, and we have two distinct solutions of \eqref{eq:ccp-def}, as we wanted. \end{proof} This theorem can be extended to the $n>1$ case by working componentwise. \section{Constructing the counter-example} \label{sec:Construction} Much like the examples given for related non-smooth dynamical systems \cite{bk:rdspoR,man:dcp,ste:usdcp}, there is a self-similar structure to the counter-example created here. The counter-example involves non-analytic $q(\cdot)$. The construction begins with a ``bump'' function $\theta:\mathbb{R}\to\mathbb{R}$ where $\theta(s)\geq0$ for all $s\in\mathbb{R}$, $\operatorname{supp}\theta\subseteq[-1,+1]$, $\int_{-\infty}^{+\infty}\theta(s)\, ds=1$, and $\theta$ is $C^{\infty}$. Let $\psi_{\alpha}(t)=t^{\alpha}$ for $t>0$ and $\psi_{\alpha}(t)=0$ for $t\leq0$. We will consider $0<\alpha<1$; the CCP \begin{equation} 0\leq z(t) \perp (\psi_{\alpha}*z)(t)+q(t)\geq0\label{eq:psi-alpha-ccp} \end{equation} then has index $1+\alpha$. The case $\alpha=\frac{1}{2}$ corresponds to the viscoelastic impact problem in Petrov and Schatzman \cite{ps:vmcs} where, asymptotically, $m(t)\sim m_{0}\sqrt{t}$ as $t\downarrow0$. The case $m(t)=t^{\alpha}$ has additional structure that we will exploit in the construction here. We will construct a function $\zeta(t)$ satisfying $\zeta(t)(\psi_{\alpha}*\zeta)(t)\leq0$ for all $t\geq0$ and $\zeta(t)=0$ for $t<0$. Let $\zeta_{1}(s;\eta)=\eta^{-1}\,\theta(\eta^{-1}(s-\widehat{s}))$ where $\eta>0$ and $\widehat{s}$ are parameters to be determined. We set \begin{equation} \zeta(t;\eta)=\sum_{k\in\mathbb{Z}}(-1)^{k}\mu^{-k} \zeta_{1}(\gamma^{k}t;\eta)\label{eq:construct-zeta} \end{equation} where $0<\mu$, $1<\gamma$ are to be determined. Let $\widehat{s}=\frac{1}{2}(1+\gamma)$. Note that $\zeta_{1}(s;\eta)\to\delta(s-\widehat{s})$ as $\eta\downarrow0$ in the sense of distributions where $\delta$ is the ``Dirac-$\delta$ function''. If we write \[ \widehat{\zeta}(t)=\sum_{k\in\mathbb{Z}}(-1)^{k}\mu^{-k}\gamma^{-k} \delta(t-\gamma^{-k}\widehat{s}), \] then $\zeta(\cdot;\eta)\to\widehat{\zeta}$ as $\eta\downarrow0$ in the sense of distributions, and in terms of weak{*} convergence of measures. Note that \begin{equation} \begin{aligned} \zeta(\gamma t;\eta) & = \sum_{k\in\mathbb{Z}}(-1)^{k}\mu^{-k} \zeta_{1}(\gamma^{k+1}t;\eta) \\ & = \sum_{\ell\in\mathbb{Z}}(-1)^{\ell-1}\mu^{-\ell+1}\zeta_{1}(\gamma^{\ell}t;\eta) \quad(\ell=k+1) \\ & = -\mu\sum_{\ell\in\mathbb{Z}}(-1)^{\ell}\mu^{-\ell}\zeta_{1}(\gamma^{\ell}t;\eta) =-\mu\zeta(t;\eta). \end{aligned} \label{eq:self-similarity} \end{equation} Also note that \begin{align*} (\psi_{\alpha}*f(\gamma\cdot))(t) &= \int_{0}^{t}\psi_{\alpha}(t-\tau) f(\gamma\tau)\, d\tau\\ &= \int_{0}^{\gamma t}(t-\gamma^{-1}\sigma)^{\alpha} f(\sigma)\gamma^{-1} \, d\sigma\quad(\sigma=\gamma\tau)\\ &= \gamma^{-1-\alpha}\int_{0}^{\gamma t}(\gamma t-\sigma)^{\alpha} f(\sigma)\, d\sigma\\ &= \gamma^{-1-\alpha}(\psi_{\alpha}*f)(\gamma t). \end{align*} Thus $-\mu\gamma^{1+\alpha}(\psi_{\alpha}*\zeta(\cdot;\eta))(t) =(\psi_{\alpha}*\zeta(\cdot;\eta))(\gamma t)$. From these relationships, if $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq0$ for $1\leq t\leq\gamma$, then $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq0$ for all $t>0$. The reason is that $\zeta(\gamma t;\eta)=(-\mu\zeta(t;\eta))$ and so $\zeta(\gamma t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(\gamma t)=(-\mu)(-\mu\gamma^{1+\alpha})\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)$ and therefore \[ \operatorname{sign}\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t) =\operatorname{sign}\zeta(\gamma t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(\gamma t). \] Once we know that $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq0$ for all $t\in[1,\gamma]$, it follows that $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq0$ for all $t>0$. Since $\operatorname{supp}\zeta\cap[1,\gamma]=\widehat{s}+[-\eta,+\eta]$, it is sufficient to check that $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq0$ for $t\in\widehat{s}+[-\eta,+\eta]$; since $\zeta(t;\eta)\geq0$ for $1\leq t\leq\gamma$, it suffices to check that $(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq0$ for $t\in\widehat{s}+[-\eta,+\eta]$. We will consider the limit as $\eta\downarrow0$, so it becomes a matter of ensuring simply that $(\psi_{\alpha}*\zeta(\cdot;\eta))(\widehat{s})<0$. There are some additional technical issues that must be addressed, but this will be done later. Now we compute $\psi_{\alpha}*\zeta(\cdot;\eta)$: \begin{align*} (\psi_{\alpha}*\zeta(\cdot;\eta))(t) &= \sum_{k\in\mathbb{Z}}(-1)^{k}\mu^{-k}(\psi_{\alpha}*\zeta_{1}(\gamma^{k}\cdot;\eta))(t)\\ &= \sum_{k\in\mathbb{Z}}(-1)^{k}\mu^{-k}(\gamma^{k})^{-1-\alpha}(\psi_{\alpha}*\zeta_{1} (\cdot;\eta))(\gamma^{k}t)\\ &= \sum_{k=\left\lfloor \ln t/\ln\gamma\right\rfloor }^{\infty}(-1)^{k} (\mu\gamma^{1+\alpha})^{-k}(\psi_{\alpha}*\zeta_{1}(\cdot;\eta))(\gamma^{k}t) \end{align*} since $\zeta_{1}(s;\eta)=0$ for $s\leq1$ and therefore $(\psi_{\alpha}*\zeta_{1}(\cdot;\eta))(s)=0$ for $s\leq1$. In particular, for $1\leq t\leq\gamma$, \[ (\psi_{\alpha}*\zeta(\cdot;\eta))(t) = \sum_{k=0}^{\infty}(-1)^{k}(\mu\gamma^{1+\alpha})^{-k}(\psi_{\alpha} *\zeta_{1}(\cdot;\eta))(\gamma^{k}t). \] For this sum to converge, we need $\mu\gamma>1$: asymptotically $(\psi_{\alpha}*\zeta_{1}(\cdot;\eta))(s)\sim s^{\alpha}$ as $s\to\infty$, so $(\psi_{\alpha}*\zeta_{1}(\cdot;\eta))(\gamma^{k}t) \sim(\gamma^{\alpha})^{k}t^{\alpha}$ as $k\to\infty$. Furthermore, $(\psi_{\alpha}*\zeta_{1}(\cdot;\eta))(s) \to\psi_{\alpha}(s-\widehat{s})=(s-\widehat{s})^{\alpha}$ as $\eta\downarrow0$. So for $1\leq t\leq\gamma$, \begin{align*} (\psi_{\alpha}*\zeta(\cdot;\eta))(t) & \to \sum_{k=0}^{\infty}(-1)^{k}(\mu\gamma^{1+\alpha})^{-k} (\gamma^{k}t-\widehat{s})^{\alpha}\quad\text{as }\eta\downarrow0\\ &= \sum_{k=0}^{\infty}(-1)^{k}(\mu\gamma)^{-k}(t-\gamma^{-k}\widehat{s})^{\alpha}. \end{align*} In particular, for $t=\widehat{s}$, \begin{align*} (\psi_{\alpha}*\zeta(\cdot;\eta))(\widehat{s}) & \to \sum_{k=0}^{\infty}(-1)^{k}(\mu\gamma)^{-k} (1-\gamma^{-k})^{\alpha}(\widehat{s})^{\alpha}\quad\text{as }\eta\downarrow0. \end{align*} Note that the term in the sum with $k=0$ is zero, and so can be ignored in the limit as $\eta\downarrow0$. So we now want to evaluate the sum \begin{equation} \widehat{v}(\mu,\gamma):=\sum_{k=1}^{\infty}(-1)^{k}(\mu\gamma)^{-k} (1-\gamma^{-k})^{\alpha},\label{eq:alt-sum} \end{equation} and check that the value is negative. Note that if $\mu\gamma=\rho>1$ is held fixed, then $\widehat{v}(\mu,\gamma) =\sum_{k=1}^{\infty}(-1)^{k}\rho^{-k}(1-\gamma^{-k})^{\alpha} \to\sum_{k=1}^{\infty}(-1)^{k}\rho^{-k}=-\rho^{-1}/(1+\rho^{-1})<0$ as $\gamma\to\infty$. Thus for sufficiently large $\gamma>1$ with $\mu\gamma=\rho>1$ fixed, we have $\widehat{v}(\mu,\gamma)<0$ as we want. Also, $\rho\widehat{v}(\mu,\gamma)\to-(1-\gamma^{-1})^{\alpha}$ as $\rho\to\infty$ with fixed $\gamma>1$. \subsection{Regularity of $\zeta$ and $\psi_{\alpha}*\zeta$, and choice of parameters} First we consider the question of how to ensure that $\zeta\in L^{1}(0,\gamma)$: Since $\| \zeta_{1}(\cdot;\eta)\| =1$ independently of $\eta>0$, we have \[ \| \zeta(\cdot;\eta)\| _{L^{1}(0,\gamma)}\leq\sum_{k=0}^{\infty}(\mu\gamma)^{-k} =\frac{1}{1-\rho^{-1}} \] which is finite as long as $\rho=\mu\gamma>1$. Note that this bound is independent of $\eta>0$. Also, $\psi_{\alpha}$ is uniformly H\"older continuous: $|\psi_{\alpha}(t)-\psi_{\alpha}(s)|=|t^{\alpha}-s^{\alpha}|\leq|t-s|^{\alpha}$ for any $s, t\in\mathbb{R}$ as $0<\alpha<1$. Combining these results shows that for $s, t\in[0,\gamma]$, $|(\psi_{\alpha}*\zeta(\cdot;\eta))(t)-(\psi_{\alpha}*\zeta(\cdot;\eta))(s)|\leq|t-s|^{\alpha}\| \zeta(\cdot;\eta)\| _{L^{1}(0,\gamma)}$. That is, $\left(\psi_{\alpha}*\zeta(\cdot;\eta)\right)|_{[0,\gamma]}$ is uniformly H\"older continuous, independently of $\eta>0$. Thus, provided \eqref{eq:alt-sum} is negative, for sufficiently small $\eta>0$, we have $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq0$ for all $1\leq t\leq\gamma$. To see this rigorously, recall that $\zeta(t)\neq0$ for $1\leq t\leq\gamma$ only if $|t-\widehat{s}|<\eta$. Choose $\eta>0$ sufficiently small so that $|(\psi_{\alpha}*\zeta(\cdot;\eta))(\widehat{s}) -\widehat{v}(\mu,\gamma)|\leq\frac{1}{4}|\widehat{v}(\mu,\gamma)|$. Now for$|t-\widehat{s}|\leq\eta$, \begin{align*} |(\psi_{\alpha}*\zeta(\cdot;\eta))(t)-\widehat{v}(\mu,\gamma)| & \leq |(\psi_{\alpha}*\zeta(\cdot;\eta))(t)-(\psi_{\alpha}* \zeta(\cdot;\eta))(\widehat{s})|+\frac{1}{4}|\widehat{v}(\mu,\gamma)|\\ & \leq |t-\widehat{s}|^{\alpha}\| \zeta(\cdot;\eta)\| _{L^{1}(0,\gamma)} +\frac{1}{4}|\widehat{v}(\mu,\gamma)|\\ & \leq \eta^{\alpha}\| \zeta(\cdot;\eta)\| _{L^{1}(0,\gamma)} +\frac{1}{4}|\widehat{v}(\mu,\gamma)|. \end{align*} Choose $\eta>0$ sufficiently small so that it also satisfies $\eta^{\alpha}\| \zeta(\cdot;\eta)\| _{L^{1}(0,\gamma)} \leq\frac{1}{4}|\widehat{v}(\mu,\gamma)|$. Then $\zeta(t;\eta)\neq0$ and $1\leq t\leq\gamma$ imply that $(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq\frac{1}{2}\widehat{v}(\mu,\gamma)<0$. Since $\zeta(t;\eta)\geq0$ for $1\leq t\leq\gamma$, we have $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq0$ for all $1\leq t\leq\gamma$. Consequently, from the self-similarity property \eqref{eq:self-similarity}, $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq0$ for all $t\geq0$. If we allow $\mu>1$ we can get much stronger regularity on $\zeta$. If $\mu>1$ then by the Weierstass $M$-test (see, e.g., \cite[Thm.~3.106, p.~141]{str:icra}), $\zeta(\cdot;\eta)$ is continuous. Furthermore, if $\mu\gamma^{-p}>1$, $\zeta$ is $p$-times continuously differentiable for $p=1,\,2,\,\ldots$, again by the Weierstrass $M$-test but applied to $\zeta^{(p)}(\cdot;\eta)$. This is equivalent to the condition that $\rho\gamma^{-p-1}>1$. If we set $\rho=2\gamma^{mp+1}$, then \begin{align*} \gamma^{p+1}\,\widehat{v}(\mu,\gamma) &= \gamma^{p+1}\sum_{k=1}^{\infty}(-1)^{k}\rho^{-k}(1-\gamma^{-k})^{\alpha}\\ &= \gamma^{p+1}\sum_{k=1}^{\infty}(-1)^{k}(2\gamma^{m+1})^{-k}(1-\gamma^{-k})^{\alpha}\\ &= \sum_{k=1}^{\infty}(-1)^{k}\frac{1}{2}(2\gamma^{p+1})^{-k+1}(1-\gamma^{-k})^{\alpha}\\ & \to -\frac{1}{2}\quad\text{as }\gamma\to\infty. \end{align*} So for sufficiently large $\gamma>1$, $\widehat{v}(\mu,\gamma)<0$. Then $\mu\gamma=\rho=2\gamma^{p+1}$, so we set $\mu=2\gamma^{p}$. We then choose $\eta>0$ sufficiently small so that $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq0$ for $1\leq t\leq\gamma$. Since $\zeta(\gamma^{-k}t;\eta)=(-\mu)^{-k}\zeta(t;\eta)$ and $(\psi_{\alpha}*\zeta(\cdot;\eta))(\gamma^{-k}t) =(-\mu\gamma^{1+\alpha})^{-k}(\psi_{\alpha}*\zeta(\cdot;\eta))(t)$, we have $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq 0$ for $\gamma^{-k}\leq t\leq\gamma^{-k+1}$ for any $k\in\mathbb{Z}$; thus $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)=0$ for any $t>0$. In addition, $(\psi_{\alpha}*\zeta(\cdot;\eta))(0)=0$, so $\zeta(t;\eta)(\psi_{\alpha}*\zeta(\cdot;\eta))(t)\leq0$ for all $t\geq0$, and there is a counter-example to uniqueness as we wanted. Furthermore, the counter-example is in $C^{p}$. \section{Extension to general $m(t)\sim m_{0}t^{\alpha}$} \label{sec:Extension} Here we assume not only that $0<\alpha<1$ but also that $m_{0}>0$. If $m_{0}<0$ so that $m(t)<0$ for $0\leq t\leq T_{1}$ with $T_{1}>0$ and $z_{1}(t)$ is a positive smooth function of $t$, then for $q_{1}(t)=-(m*z_{1})(t)$ not only is $z(t)=z_{1}(t)$ for $t\geq0$ a solution to \[ 0\leq z(t) \perp (m*z)(t)+q_{1}(t)\geq0\quad\text{for all }t\geq0, \] but $z(t)=0$ for $0\leq t\leq T_{1}$ is also a solution as $q_{1}(t)>0$ for $0\leq t\leq T_{1}$. The assumptions made on $m$ are that $m(t)\sim m_{0}t^{\alpha}$, $m'(t)\sim m_{0}\alpha t^{\alpha-1}$ as $t\downarrow0$, and $m'(t)$ is continuous in $t$ away from $t=0$. This implies that on bounded sets, $m(\cdot)$ is uniformly H\"older continuous: given a bounded interval $[a,b]$, there is an $M$ where $|m(t)-m(s)|\leq M\,|t-s|^{\alpha}$ for all $s, t\in[a,b]$. Note that dividing $m(t)$ by $m_{0}>0$ does not affect the existence of multiple solutions as \eqref{eq:ccp-def} is equivalent to \[ 0\leq z(t)\perp((m/m_{0})*z)(t)+q(t)/m_{0}\geq0\quad\text{for all }t\geq0. \] So we consider without loss of generality the case where $m(t)\sim t^{\alpha}$. As in Section~\ref{sec:Mandelbaum's-condition} we look for a non-zero function $\zeta:[0,\infty)\to\mathbb{R}$ where $\zeta(t)(m*\zeta)(t)\leq0$ for all $t\geq0$. The constructed $\zeta$ from the previous Section will also work here with some small modifications. Let $r(t)=(m(t)/\psi_{\alpha}(t))-1$. Note that $r(t)\to0$ as $t\downarrow0$. Using \eqref{eq:construct-zeta} to define $\zeta(\cdot)$, \[ \zeta(t)=\sum_{k\in\mathbb{Z}}(-1)^{k}\mu^{-k}\zeta_{1}(\gamma^{k}t;\eta), \] we can show that for $\gamma^{-j}\leq t<\frac{1}{2}\gamma^{-j}(1+\gamma)$, \begin{align*} (m*\zeta)(t) &= \sum_{k=j}^{\infty}(-1)^{k}\mu^{-k}(m*\zeta_{1}(\gamma^{k}\cdot;\eta))(t)\\ & \to \sum_{k=j+1}^{\infty}(-1)^{k}\mu^{-k}\gamma^{-k}m(t-\gamma^{-k+j}\widehat{s}) \quad\text{as }\eta\downarrow0, \end{align*} using $(m*\zeta_{1}(\cdot;\eta))(s)\to m(s-\widehat{s})$ as $\eta\downarrow0$, and $m(0)=0$. We need to distinguish between the value and the limit. First, note that if $\operatorname{supp} g\subseteq[\widehat{s}-\rho,\widehat{s}+\rho]$ and $g$ is non-negative, then for continuous $f$, \[ \Big|\int_{-\infty}^{+\infty}f(s)\, g(s)\, ds -f(\widehat{s})\int_{\widehat{s}-\rho}^{\widehat{s}+\rho}g(s)\, ds\Big| \leq\max_{s:|s-\widehat{s}|\leq\rho}|f(s)-f(\widehat{s})| \int_{\widehat{s}-\rho}^{\widehat{s}+\rho}g(s)\, ds. \] Then \[ |(m*\zeta_{1}(\gamma^{k}\cdot;\eta))(t)-\gamma^{-k}m(t-\gamma^{-k}\widehat{s})|\leq M(\gamma^{-k}\eta)^{\alpha}\gamma^{-k}=M\eta^{\alpha}(\gamma^{1+\alpha})^{-k}. \] So, for $t=\gamma^{-j}\widehat{s}$, \begin{align*} & \Big|(m*\zeta)(\gamma^{-j}\widehat{s})-\sum_{k=j}^{\infty}(-1)^{k} \mu^{-k}\gamma^{-k}m((1-\gamma^{-k+j})\gamma^{-j}\widehat{s})\Big|\\ & \leq\sum_{k=j}^{\infty}\mu^{-k}(\gamma^{1+\alpha})^{-k}\, M\eta^{\alpha} =\frac{(\mu\gamma^{1+\alpha})^{-j}M\eta^{\alpha}}{1-(\mu\gamma^{1+\alpha})^{-1}}. \end{align*} Note that \begin{align*} & \sum_{k=j+1}^{\infty}(-1)^{k}\mu^{-k}\gamma^{-k}m((1-\gamma^{-k+j})\gamma^{-j}\widehat{s})\\ & =(-1)^{j}(\mu\gamma)^{-j}\sum_{\ell=1}^{\infty}(-1)^{\ell}(\mu\gamma)^{-\ell}m((1-\gamma^{-\ell})\gamma^{-j}\widehat{s})\\ & =(-1)^{j}(\mu\gamma)^{-j}\sum_{\ell=1}^{\infty}(-1)^{\ell} (\mu\gamma)^{-\ell}((1-\gamma^{-\ell})\gamma^{-j}\widehat{s})^{\alpha} [1+r((1-\gamma^{-\ell})\gamma^{-j}\widehat{s})] \\ & =(-1)^{j}(\mu\gamma^{1+\alpha})^{-j}\widehat{s}^{\alpha} \sum_{\ell=1}^{\infty}(-1)^{\ell}(\mu\gamma)^{-\ell}(1-\gamma^{-\ell})^{\alpha} [1+r((1-\gamma^{-\ell})\gamma^{-j}\widehat{s})]. \end{align*} Since $r(t)\to0$ as $t\downarrow0$, for every $\epsilon>0$ there is a $\delta>0$ where $01$ so that $\mu\gamma^{1+\alpha}>\mu\gamma>1$. By choosing $\eta>0$ and $\epsilon>0$ sufficiently small, we can guarantee that the sign of $(m*\zeta)(t)$ for $\gamma^{-j}\leq t\leq\gamma^{-j+1}$ and $\zeta(t)\neq0$ is the sign of $(-1)^{j}\widehat{v}(\mu,\gamma)$. After choosing $\eta>0$ and $\epsilon>0$ so that this holds, we can ensure that $\zeta(t)(m*\zeta)(t)\leq0$ for $\gamma^{-j}\leq t\leq\gamma^{-j+1}$ where $j\geq J:=\left\lceil -\ln(\delta/\widehat{s})/\ln\gamma\right\rceil $. Thus $\zeta(t)(m*\zeta)(t)\leq0$ for all $00$ and $0<\alpha<1$. \section{Conclusions} Non-uniqueness of convolution complementarity problems of the form \eqref{eq:ccp-def} with convolution kernel $m(t)\sim m_{0}t^{\alpha}$ and $m'(t)\sim m_{0}\alpha t^{\alpha-1}$ with $m_{0}>0$ and $0<\alpha<1$ has been demonstrated via a generalization of a result of Mandelbaum. Note that the counter-examples can belong to any space $C^{p}$, $p=1,\,2,\,3,\,\ldots$. Counter-examples must have infinitely many oscillations in a finite time interval, and so cannot be analytic. The main non-uniqueness result is of particular interest for questions of contact mechanics, as the perpendicular impact of a Kelvin--Voigt viscoelastic rod on a rigid obstacle can be model by such a CCP (see \cite{ps:vmcs}). Note that this non-uniqueness holds in spite of the existence of an energy balance for this situation \cite{ps:vmcs}. By contrast, the perpendicular impact of a purely elastic rod on a rigid obstacle does have uniqueness of solutions, by using CCP formulations but with $\alpha=0$ \cite{ste:ccpaip}. Multidimensional contact problems then either have a problem of existence (for purely elastic bodies) or with uniqueness (for Kelvin--Voigt viscoelastic bodies). 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