# Permanence and the distance from the convex hull to the interior equilibrium

Background

To use permanence, Lotka-Volterra dynamics have to be assumed, because it is only in this case that a sufficient condition for permanence is known:

$
dot{x}_{i} = x_i cdot f_i(x) = x_i cdot (r_i + (A cdot x)_i) quad forall i = 1, ldots, n.
$

Such a dynamical system is permanent if two conditions hold (Hofbauer and Sigmund 1988, The theory of evolution and dynamical systems p. 98). (1) It is dissipative; this is true for Lotka-Volterra systems where all the basal species are self-limiting and heterotrophs cannot survive without their prey. (2) That:
$
P(x) = prod_i x_{i}^{p_i}
$

for some $p_i>0$ is an average Lyapunov function.

In dissipative Lotka-Volterra systems, the test for permanence reduces to a linear programming problem (Jansen 1987, J. Math. Biol.; Law & Morton 1996, Ecology).

An alternative way of describing the system, pursued in an unpublished paper by Richard Law and R. Daniel Morton, is in terms of the convex hull $C$ and the set $D$ where the densities of all species are non-increasing. The convex hull of the boundary equilibria is the smallest convex set of which every boundary equilibrium is a member. If this set is disjoint from $D$, then the system is permanent. So to test for permanence, one can find a hyperplane passing through the interior equilibrium that has a greater value than that of the hyperplane passing through every boundary equilibrium. One can also measure the ‘strength’ of permanence by $d$, which is the shortest distance from $C$ to $D$, which is always the shortest distance from $C$ to the interior equilibrium $hat{x}$ (Figure 1).

Example

You can download the complete example for Octave from my website: mortongen.m.

Consider the four-species system shown in Figure 2 with the parameter values below, adapted from an unpublished paper by Richard Law and R. Daniel Morton, with:
$
A =
begin{bmatrix}
-0.000610 & -0.000207 & -0.052365 & -0.001789 \
-0.000400 & -0.000139 & -0.001260 & -0.064501 \
0.005740 & 0.000033 & 0 & -0.096816 \
0.000010 & 0.000151 & 0.000399 & 0 \
end{bmatrix}
$

and
$
r^{T} = begin{bmatrix} 0.220737 & 0.164163 &
-0.087756 & -0.096189 end{bmatrix}.
$

It has an interior equilibrium point:
$
hat{x} =
begin{bmatrix}
28.230 & 631.560 & 1.356 & 0.983
end{bmatrix}
$

The first step is to find all subsystems $b = (0,0,0),(1,0,0), ldots, (1,2,0),(1,3,0),(1,4,0),(2,3,0), ldots, (2,3,4)$, and then identify which of these have a positive boundary equilibrium $hat{x}^{b}$.

If we had all subsystems in subStore, then equilibria would be evaluated by:

for subCnt = 2:noSubs;
% Acquire list of species absent and present
present = subStore(subCnt,:);
present(find(present == 0)) = [];
absent = 1:nospp; absent(present) = [];

% Evaluate boundary equilibrium
ANow = A;
ANow(absent,:) = []; ANow(:,absent) = [];
rNow = r; rNow(absent) = [];
if rank(ANow) == size(ANow,1);
xSub = ANow-rNow;
% etc.


For our example system, the positive boundary equilibria are:
$
hat{x}^{(1)} =
begin{bmatrix}
362 & 0 & 0 & 0
end{bmatrix} \
hat{x}^{(2)} =
begin{bmatrix}
0 & 1181 & 0 & 0
end{bmatrix}\
hat{x}^{(1,3)} =
begin{bmatrix}
15.3 & 0 & 4 & 0
end{bmatrix}\
hat{x}^{(2,4)} =
begin{bmatrix}
0 & 637 & 0 & 1.2
end{bmatrix}\
hat{x}^{(1,2,4)} =
begin{bmatrix}
148 & 627 & 0 & 0.3
end{bmatrix}
$

The transversal eigenvalues at the boundary equilibria
$
f(hat{x}^{}) = r + Ahat{x}^{b}
$

are found with code like:

rowofG = A*(xStore(subCnt,:)')+r;


and are:
$
f(hat{x}^{()}) =
begin{bmatrix}
0.22074 & 0.16416 & -0.08776 & -0.09619\
end{bmatrix}\
f(hat{x}^{(1)}) =
begin{bmatrix}
0.00000 & 0.01942 & 1.98934 & -0.09257\
end{bmatrix}\
f(hat{x}^{(2)}) =
begin{bmatrix}
-0.02374 & 0.00000 & -0.04878 & 0.08215\
end{bmatrix}\
f(hat{x}^{(1,3)}) =
begin{bmatrix}
0.00000 & 0.15296 & 0.00000 & -0.09443\
end{bmatrix}\
f(hat{x}^{(2,4)}) =
begin{bmatrix}
0.08678 & 0.00000 & -0.18024 & 0.00000\
end{bmatrix}\
f(hat{x}^{(1,2,4)}) =
begin{bmatrix}
0.00000 & 0.00000 & 0.75719 & 0.00000\
end{bmatrix}\
$

The linear programming problem is then to minimise $z$ subject to:
$
sum p_i cdot f(hat{x}^{b_1}) + z geq 0 \
sum p_i cdot f(hat{x}^{b_2}) + z geq 0 \
vdots \
p_i geq 0 \
$

where each positive equilibrium point on the boundary $hat{x}^{b_1},hat{x}^{b_2}, ldots$ gives rise to a constraint, and $p_i$ is part of the average Lyapunov function. So if one can find a solution with $z < 0$, then one can be certain that $P(x)$ is an average Lyapunov function.

An introduction on how to code linear programming problems in Octave can be found in a previous post. The vector $p$ for the Lyapunov function, as found using Jansen’s linear program method, is
$
p = begin{bmatrix} 0.835219 & 1 & 0.077317 & 0.999918 end{bmatrix}.
$

Hofbauer & Sigmund (1988, p. 176-177) states that the convex hull $C$ can be separated from $D$ by a hyperplane with a “separating functional” $pA$, where $p$ is the same $p$ used in the Lyapunov function. This separating functional is the vector normal to the hyperplane.
$
mathbf{n} = pA \
= begin{bmatrix} -4.5568times10^{-4} & -1.5835times10^{-4}
& -4.4597times10^{-2} & -7.3481times10^{-2} end{bmatrix} \
= begin{bmatrix} 0.0062014 & 0.0021550 & 0.6069248 & 1
end{bmatrix}.
$

The equation for the hyperplane is
$
0 = n_1x_1 + n_2x_2 + n_3x_3 + x_4 + b.
$

If one wants to go straight to finding $d$, then substitute in the boundary steady state for each positive subsystem, reducing the value of $b$ until the hyperplane is above or equal to every boundary steady state.

So for subsystem $(1)$,
$
hat{x}^{(1)} = begin{bmatrix} 361.86393 & 0 & 0 & 0 end{bmatrix},
$

which gives $b = -2.2441$

Then subsystem $(2)$ is next, with
$
hat{x}^{(2)} = begin{bmatrix} 0 & 1181 & 0 & 0 end{bmatrix}.
$

Substituting into the hyperplane equation gives
$
0.0021550 times 1181 -2.2441 = 0.30105
$

which implies that this boundary steady state is above the hyperplane, therefore it is used to find a revised value of $b$, $b =-2.5451$

This process is continued, checking that all other boundary steady
states are below the plane, and revising $b$ if they are not, which gives the final value of $b =-2.5451$.

Now the distance from the interior equilibrium to the plane can be
found. The interior equilibrium is
$
hat{x} = begin{bmatrix} 28.23008 & 631.55967 & 1.35636 & 0.98255 end{bmatrix}
$

and so the perpendicular distance from the hyperplane to $hat{x}$ is
$
d = (mathbf{n}' hat{x} + b)/|mathbf{n}| \
d = frac{(0.0062014 times 28.23008 + 0.0021550 times 631.55967 +
0.6069248 times 1.35636 + 0.98255 -2.5451)}{1.1698} \
d = 0.68108,
$

Alternatively $d$ can be calculated for each boundary steady state, and the smallest taken as the final value. If xStore stores the values of all the boundary equilibria, then:

d = (n'*xs-xStore(posSubs,:)*n)/norm(n);
disp('d = ')
disp(d)
mind = min(d);
disp('smallest distance = ')
disp(mind)


You can download the complete example for Octave from my website: mortongen.m.

## 3 thoughts on “Permanence and the distance from the convex hull to the interior equilibrium”

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