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Lotka-Volterra Dynamics

The most obvious thing about equation (1) is its fixed point

\begin{displaymath}
\hat{\mbox{\boldmath {$n$}}} = -\mbox{\boldmath {$\beta$}}^{-1}\mbox{${\bf r}$},
\end{displaymath} (2)

where $\dot{\mbox{\boldmath {$n$}}}=0$. For this point to be biologically meaningful, all components of $\hat{\mbox{\boldmath {$n$}}}$ must be positive, giving rise to the following inequalities:
\begin{displaymath}
\hat n_i = \left(\mbox{\boldmath {$\beta$}}^{-1}\mbox{${\bf r}$}\right)_i>0, \forall i
\end{displaymath} (3)

The stability of this point is related to the negative definiteness of derivative of $\dot{\mbox{\boldmath {$n$}}}$ at $\hat{\mbox{\boldmath {$n$}}}$. The components of the derivative are given by
\begin{displaymath}
\frac{\partial\dot{n}_i}{\partial n_j} =
\delta_{ij}\left(r_i+\sum_k\beta_{ik}n_k\right) + \beta_{ij}n_i
\end{displaymath} (4)

Substituting eq (2) gives
\begin{displaymath}
\left.\frac{\partial\dot{n}_i}{\partial n_j}\right\vert _{\h...
...\left(\mbox{\boldmath {$\beta$}}^{-1}\mbox{${\bf r}$}\right)_i
\end{displaymath} (5)

Stability of the fixed point requires that this matrix should be negative definite. Since the $\left(\mbox{\boldmath {$\beta$}}^{-1}\mbox{${\bf r}$}\right)_i$ are all negative by virtue of (3), each minor determinant of this matrix is equal to a minor determinant of $\beta$ multiplied by a positive number, stability of the equilibrium is equivalent to $\beta$ being negative definite.

A weaker condition is to require that the system remain bounded with time:

\begin{displaymath}
\sum_i\dot{n_i}=\mbox{${\bf r}$}\cdot\mbox{\boldmath {$n$}}+...
...< 0, \;\forall \mbox{\boldmath {$n$}}:
\sum_in_i>N \;\exists N
\end{displaymath} (6)

As $n$ becomes large in any direction, this functional is dominated by the quadratic term, so this implies that $\mbox{\boldmath {$n$}}\cdot\mbox{\boldmath {$\beta$}}\mbox{\boldmath {$n$}}\leq0
\; \forall\mbox{\boldmath {$n$}}: n_i>0$. Negative definiteness of $\beta$ is sufficient, but not necessary for this condition. For example, the predator-prey relations (heavily normalised) have the following matrix as $\beta$: $\mbox{\boldmath {$\beta$}}=\left(\begin{array}{cc}
-1 & 2\\
-2 & 0\\
\end{array}\right)$ which has eigenvalues $3/2, -5/2$. If we let $\mbox{\boldmath {$n$}}=(x,y), x,y\geq0$, then $\mbox{\boldmath {$n$}}\cdot\mbox{\boldmath {$\beta$}}\mbox{\boldmath {$n$}}=-2x^2$, which is clearly non-positive for all $x$.

Consider adding a new row and column to $\beta$. What is condition is the new row and column required to satisfy such that equation (6) is satisfied. Break up $\beta$ in the following way:

\begin{displaymath}
\left(
\mbox{
\begin{tabular}{c\vert c}
$\begin{array}{cc...
... \vdots\end{array}$\\
\hline
$n_2$
\end{tabular} }
\right)
\end{displaymath}

Condition (6) becomes:

\begin{displaymath}
{\bf n_1}\cdot{\bf A}{\bf n_1} + {\bf n_1}\cdot({\bf B}+{\bf C})n_2 +
Dn_2^2 \leq 0
\end{displaymath} (7)

Let

\begin{displaymath}
a=\max_{n=1} \mbox{\boldmath {$n$}}\cdot A\mbox{\boldmath {$n$}},\mbox{ and } b=\max_{i}B_i+C_i.
\end{displaymath}

Then a sufficient but not necessary condition for condition (7) is

\begin{displaymath}
an_1^2+bn_1n_2+Dn_2^2\leq0
\end{displaymath}

The maximum value with respect to $n_2$ is $an_1^2-(bn_1)^2/4D$, so this requires that

\begin{displaymath}
b \geq 2\sqrt{aD}
\end{displaymath} (8)


next up previous contents index
Next: Mutation Up: The Eco Lab Model Previous: The Eco Lab Model   Contents   Index
Russell Standish 2016-09-02