# 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An alternative way of describing the system, pursued in an unpublished paper by Richard Law and R. Daniel Morton, is in terms of the convex hull $$C$$ and the set $$D$$ where the densities of all species are non-increasing.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). An example 2-species system showing the interior and boundary equilibria, the convex hull C, and set D where all densities are non-increasing, and the distance d which is the shortest distance between C and D.

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Consider the four-species system shown in Figure 2 with the parameter values below, adapted from an unpublished paper by Richard Law and R. Daniel Morton, with: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
$f(\hat{x}^{}) = r + A\hat{x}^{b}$
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.
$\mathbf{n} = pA \\ = \begin{bmatrix} -4.5568\times10^{-4} & -1.5835\times10^{-4} & -4.4597\times10^{-2} & -7.3481\times10^{-2} \end{bmatrix} \\ = \begin{bmatrix} 0.0062014 & 0.0021550 & 0.6069248 & 1 \end{bmatrix}.$

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”

1. Jailbreak says:

Hey, I just stopped in to visit your website and thought I’d say I enjoyed myself.

1. nadiah says:

Thank you Jailbreak, I rather enjoy measuring the distance from the convex hull to the interior equilibrium myself.