\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 $0<t<\delta$ implies $|r(t)|<\epsilon$.
Thus for $j\geq-\ln(\delta/\widehat{s})/\ln\gamma$, 
$|r((1-\gamma^{-\ell})\gamma^{-j}\widehat{s})|<\epsilon$,
and so
\[
\big|\sum_{\ell=1}^{\infty}(-1)^{\ell}(\mu\gamma)^{-\ell}
(1-\gamma^{-\ell})^{\alpha}r((1-\gamma^{-\ell})\gamma^{-j}\widehat{s})\big|
\leq\frac{\epsilon}{1-(\mu\gamma)^{-1}}.
\]
Since $\gamma^{-j}\leq t\leq\gamma^{-j+1}$ and $\zeta(t)\neq0$ implies
$|t-\gamma^{-j}\widehat{s}|\leq\gamma^{-j}\eta$, we can
use the bound $|(m*\zeta)(t)-(m*\zeta)(\gamma^{-j}\widehat{s})|
\leq M(\eta\gamma^{-j})^{\alpha}\| \zeta\| _{L^{1}(0,\gamma^{-j+1})}
\leq M\eta^{\alpha}\gamma^{-\alpha j}(\mu\gamma)^{-j}/(1-(\mu\gamma)^{-1})$
for $|t-\gamma^{-j}\widehat{s}|\leq\gamma^{-j}\eta$. Thus
for $\gamma^{-j}\leq t\leq\gamma^{-j+1}$ and $\zeta(t)\neq0$,
\begin{align*}
 &  |(m*\zeta)(t)-(-1)^{j}\widehat{s}^{\alpha}(\mu\gamma^{1+\alpha})^{-j}\widehat{v}(\mu,\gamma)|\\
 & \leq\frac{M\eta^{\alpha}(\mu\gamma^{1+\alpha})^{-j}}{1-(\mu\gamma)^{-1}}+\frac{(\mu\gamma^{1+\alpha})^{-j}M\eta^{\alpha}}{1-(\mu\gamma^{1+\alpha})^{-1}}+\frac{\widehat{s}^{\alpha}(\mu\gamma^{1+\alpha})^{-j}\epsilon}{1-(\mu\gamma)^{-1}}\\
 & \leq(\mu\gamma^{1+\alpha})^{-j}
\big[\frac{M\eta^{\alpha}}{1-(\mu\gamma)^{-1}}
 +\frac{M\eta^{\alpha}}{1-(\mu\gamma^{1+\alpha})^{-1}}
 +\frac{\widehat{s}^{\alpha}\epsilon}{1-(\mu\gamma)^{-1}}\big].
\end{align*}
Note that $\gamma>1$ 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 $0<t\leq\gamma^{-J}$.
By setting $\widehat{\zeta}(t)=\zeta(t)$ for $0\leq t\leq\gamma^{-J}$
and $\widehat{\zeta}(t)=0$ for $t\geq\gamma^{-J}$ (noting that $\zeta(t)=0$
in a neighborhood of $\gamma^{-k}$ for any $k\in\mathbb{Z}$), we see that
$\widehat{\zeta}(t)(m*\widehat{\zeta})(t)\leq 0 $ for all $t\geq0$, and
thus we have non-uniqueness of solutions for \eqref{eq:ccp-def} where
$m(t)\sim m_{0}t^{\alpha}$ and $m'(t)\sim m_{0}\alpha t^{\alpha-1}$
as $t\downarrow0$ provided $m_{0}>0$ 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). How this can be resolved
is a subject for future investigation.

\begin{thebibliography}{10}

\bibitem{bk:rdspoR}
Alain Bernard and Ahmed el~Kharroubi.
\newblock R\'egulations d\'eterministes et stochastiques dans le premier
  ``orthant'' de {${\bf R}\sp n$}.
\newblock {\em Stochastics Stochastics Rep.}, 34(3-4):149--167, 1991.

\bibitem{kir:gfca}
Virginia Kiryakova.
\newblock {\em Generalized fractional calculus and applications}, volume 301 of
  {\em Pitman Research Notes in Mathematics Series}.
\newblock Longman Scientific \& Technical, Harlow, 1994.

\bibitem{man:dcp}
Avishai Mandelbaum.
\newblock The dynamic complementarity problem.
\newblock Unpublished manuscript, 1989.

\bibitem{ps:vmcs}
Adrien Petrov and Michelle Schatzman.
\newblock Visco\'elastodynamique monodimensionnelle avec conditions de
  {Signorini}.
\newblock {\em Comptes Rendus Acad. Sci., S\'er. I}, 334:983--988, 2002.

\bibitem{ste:ccpaip}
David~E. Stewart.
\newblock Convolution complementarity problems with application to impact
  problems.
\newblock {\em IMA J. Applied Math.}, 71(1):92--119, 2006.

\bibitem{ste:dcpficcp}
David~E. Stewart.
\newblock Differentiating complementarity problems and fractional index
  convolution complementarity problems.
\newblock {\em Houston J. Mathematics}, 33(1):301--322, 2006.

\bibitem{ste:usdcp}
David~E. Stewart.
\newblock Uniqueness for solutions of differential complementarity problems.
\newblock {\em Math. Program.}, 118(2, Ser. A):327--345, 2009.

\bibitem{ste:diihc}
David~E. Stewart.
\newblock {\em Dynamics with Inequalities: impacts and hard constraints}.
\newblock Number 123 in Applied Mathematics Series. SIAM Publ., Philadelphia,
  PA, July 2011.

\bibitem{sw:ficcp}
David~E. Stewart and Theodore~J. Wendt.
\newblock Fractional index convolution complementarity problems.
\newblock {\em Nonlinear Anal. Hybrid Syst.}, 1(1):124--134, 2007.

\bibitem{str:icra}
K.~R. Stromberg.
\newblock {\em An Introduction to Classical Real Analysis}.
\newblock Wadsworth, Belmont, CA, 1981.

\end{thebibliography}

\end{document}


