\newpage
\section*{List of Figures}
\begin{description}
\item[Figure 1:]
Radiolocations and home range boundaries of 5 adult and yearling timber
wolves radiotracked via aircraft, 27 May -- 22 October 1971 in
northeastern Minnesota. Redrawn from Fig.\ 4 in
\citeN{VanBallenberghe:WM-43-1}
with permission from The Wildlife Society (copyright holder).
\item[Figure 2:]
(a) Scent mark densities $p(u(x),v(x))$ and $q(u(x),v(x))$ are determined
uniquely by (\ref{eq:p})--(\ref{eq:q}) $m_u(q)$ and $m_u(q)$ are
convex functions. (b) If $m_u(q)$ and $m_u(q)$ exhibit switching then
$p(u(x),v(x))$ and $q(u(x),v(x))$ may not be determined uniquely if
$u(x)$ and $v(x)$ are sufficiently small.
\item[Figure 3:]
(a) Three regions where solutions to equation for wolf pack 1 (\ref{eq:uODE})
must lie. These are given by (\ref{eq:upos})--(\ref{eq:uzero}).
(b) The maximum and minimum values for $u(0)$. The area under
each curve integrates to unity and thus satisfies (\ref{eq:icon}).
\item[Figure 4:]
Functional forms for the movement behavior function $c(q)$ and the
marking behavior function $m(q)$: (a) convex $c(q)$,
(b) convex $m(q)$, (c) switching in $c(q)$ and (d) switching
in $m(q)$.
\item[Figure 5:]
Time-independent solution to (\ref{eq:w1a})--(\ref{eq:w2a}),
(\ref{eq:w1_zf})--(\ref{eq:w2_zf}), (\ref{eq:p1a})--(\ref{eq:q1a})
with $m_u=m_v=0$ and $c_u(q)=\gamma q$, $c_v(p)=\gamma p$,
$\gamma=1$, (Figure 4(a)) and $d_u=d_v=0.1$.
Shown is the large-time numerical solution with the
expected density of wolves from pack 1 ($u(x)$) and the expected
density of wolves from pack 2 ($v(x)$). This corresponds
to the analytical solution to the system of logistic equations
(\ref{eq:logistic}). Note that $u(x)+v(x)=2$ pointwise across the
domain so there is no buffer zone. The expected scent mark densities are
identical to the expected wolf densities.
Numerical solution used the Method of Lines and Gear's Method.
\item[Figure 6:]
Time-independent solution to (\ref{eq:w1a})--(\ref{eq:w2a}),
(\ref{eq:w1_zf})--(\ref{eq:w2_zf}), (\ref{eq:p1a})--(\ref{eq:q1a})
with $c_u(p)=\gamma p$, $c_v(q)=\gamma q$ (Figure 4(a)) and
$m_u(q)=\mu q$, $m_v(p)=\mu p$, (Figure 4(b))
and $\gamma=1$, $d_u=d_v=0.333$, $\mu=1.1$.
(a) Shown is the large-time numerical solution with the
expected density of wolves from pack 1 ($u(x)$), the expected
density of wolves from pack 2 ($v(x)$), the expected
scent mark density from pack 1 ($p(x)$) and the expected
scent mark density from pack 2 ($q(x)$). This corresponds
to the analytical solution to equation (\ref{eq:E}) and the
similar equation for $u$. Note the bowl shaped scent mark densities.
Intersection of the line $u=1/\mu$ with $u(x)$ yields the location
of the maximum value for $p(x)$ which corresponds to the edge of the
bowl (\ref{eq:pd_sim}). (b) The corresponding cumulative expected
wolf density $(u(x)+v(x))$ and cumulative scent mark density
$(p(x)+q(x))$. The function $(1+\mu u)(1+\mu v)$ is constant across
the domain as predicted by (\ref{eq:gam1_cons}).
Note the the elevated scent mark density and buffer zone at $x=0.5$.
Numerical solution used the Method of Lines and Gear's Method.
\item[Figure 7:]
Analytical solution to the territoriality ODEs
(\ref{eq:su})--(\ref{eq:sq}) subject to the integral constraints
(\ref{eq:icon}) with $c(q)=c_\infty H(q-q_c)$ (Figure 4(c))
and $m(q)=\mu q$ (Figure 4(b)). The solution for $u(x)$
and $v(x)$ is given by (\ref{eq:usw})--(\ref{eq:vsw}) with
$u(0)$ and $x_c$ determined by (\ref{eq:icon1})--(\ref{eq:q_c}).
The corresponding solutions for $p(x)$ and $q(x)$ are given by
(\ref{eq:p_ss})--(\ref{eq:q_ss}). Note the width of the buffer
zone.
\item[Figure 8:]
Stable time-independent equilibria for (\ref{eq:sq})--(\ref{eq:sq})
with $m(q)=m_\infty H(q-q_m)$. Region A: $(p,q)=(u,v)$,
region B: $(p,q)=(u,(1+m_\infty)v)$,
region C: $(p,q)=((1+m_\infty)u,v)$ and
region D: $(p,q)=((1+m_\infty)u,(1+m_\infty)v)$. Note that
if $q_m/(1+m_\infty)__ q_m$
(\ref{eq:mswitch}) then the indeterminate region can bypassed in phase space.
A sample trajectory is shown with three legs of the trajectory
as given in (\ref{eq:mswitch}).
\item[Figure 9:]
An analytical solution to the territoriality ODEs
(\ref{eq:su})--(\ref{eq:sq}) subject to the integral constraints
(\ref{eq:icon}) with $m(q)=c_\infty H(q-q_m)$ (Figure 4(d))
and $c(q)=\gamma q$ (Figure 4(a)). The corresponding solution
trajectory is given in Figure 8. The solution method is the same
as described in Section 6.1 except with the relationship between
$u(x)$ and $v(x)$ in each subregion given by (\ref{eq:mswitch}).
Note the abrupt jumps in $p(x)$ and $q(x)$.
\item[Figure 10:]
Two dimensional territories. Each surface describes a group
of wolves and their scent marks. The height of the surface
describes the expected density of wolves and the color of the surface
describes the expected scent mark density. The shading follows
colors of the rainbow with red being low and violet being high.
Solutions were generated by solving (\ref{eq:w1a})--(\ref{eq:w2a}),
(\ref{eq:w1_zf})--(\ref{eq:w2_zf}), (\ref{eq:p1a})--(\ref{eq:q1a})
numerically using finite differences and an alternating direction
implicit method until the a stationary solution was effectively reached.
The domain size is $[0,1]\times[0,1]$ and the mesh size is 40$\times$40.
The movement function is given by
\begin{displaymath}
c(q,r)=\frac{c_\infty}{2}\left(1+\tanh(\gamma(q-q_c))\right)\tanh(\alpha r)
\end{displaymath}
and the scent marking function is given by
\begin{displaymath}
m(q)=\frac{m_\infty}{2}\left(1+\tanh(\mu(q-q_m))\right).
\end{displaymath}
In the case with three interacting packs, an additional
wolf movement and scent marking equation is added and the functions
$m$ and $c$ are modified to depend on the cumulative foreign scent
mark density.
(a) Two interacting packs with dens at opposite corners of the domain
($d=0.3$, $c=3$, $\alpha=100$, $\gamma=2$, $m_\infty=1.5$, $c_\infty=1$,
$\mu=10$, $q_c=1$, $q_m=1$ and the color scale ranges from 0 to 4).
(b) The cumulative expected wolf density and expected scent mark
density surface for (a) above. (c) Three interacting packs with
dens in two adjacent corners and at the midpoint of the opposite side
($d=0.3$, $c=3$, $\alpha=100$, $\gamma=2$, $m_\infty=1.5$, $c_\infty=1$,
$\mu=7$, $q_c=0.75$, $q_m=1$ and the color scale ranges from 0 to 5).
(d) The cumulative expected wolf density and expected scent mark
density surface for (c) above.
\end{description}
__