Show positivity and boundedness of a non-linear system

[dgf]

I want to show the boundedness of the nonlinear system below. Assume, $N, P, K, t > 0$ and all other parameters to be positive.

$$\frac{{d}N}{{d}t} ~=~ rN\left(1 - \frac{N}{K}\right) - \frac{a(1 - m)NP}{1 + \gamma(1 - m)N}$$

$$\frac{{d}P}{{d}t} ~=~ \frac{b(1 - m)NP}{1 + \gamma(1 - m)N} - cP$$

$$\frac{{d}K}{{d}t} ~=~ \frac{(K - \alpha)(p - K)}{q + K}$$

My attempt is this:

We show that the system is bounded by showing each equation is bounded.

For, $\frac{{d}K}{{d}t} = \frac{(K - \alpha)(p - K)}{q + K}$, the

denominator $q+K>0$ ensures it is well-defined. Now if $\alpha<p$ then $K$

is bounded between $\alpha$ and $p$ and if $\alpha>p$, then $K$ is bounded between

$p$ and $\alpha$. Therefore, $\min(\alpha,p)\le K \le \max(\alpha,p)$.

Now for, $\frac{{d}N}{{d}t} = rN\left(1 - \frac{N}{K}\right) - \frac{a(1 - m)NP}{1 + \gamma(1 - m)N} \implies \frac{{d}N}{{d}t} \le rN\left(1 - \frac{N}{K}\right)$. Now since $K$ is bounded, we choose the upper and lower bounds to be $K_{max}$ and $K_{min}$ respectively. Then

$K\le K_{max} \implies - \frac{N}{K} \le \frac{N}{K_{max}}$. Therefore, for any

$\epsilon >0$, we choose $N_{upper} = K_{max} + \epsilon$. So we have,

$\frac{{d}N}{{d}t} \le rN\left(1 - \frac{N}{K_{max}}\right) < -rN\frac{\epsilon}{K_{max}} < 0$. This implies it is always bounded above.

Now for, $\frac{{d}P}{{d}t} ~=~ \frac{b(1 - m)NP}{1 + \gamma(1 - m)N} - cP$,

we consider the Lyapunov function $V(t)$ defined as:

$$V(t) = N(t) + \frac{a}{b}P(t)$$

Since $N(t) \ge 0$ and $P(t) \ge 0$, it follows that $V(t) \ge 0$.

Let's compute the derivative of $$V(t)$$ with respect to time:

$$ \frac{dV}{dt} = \frac{dN}{dt} + \frac{a}{b}\frac{dP}{dt} $$

Substitute the expressions for $$\frac{dN}{dt}$$ and $$\frac{dP}{dt}$$ from equations (1) and (2):

$$\begin{align*}

\frac{dV}{dt} &= \left[ rN\left(1 - \frac{N}{K}\right) - \frac{a(1 - m)NP}{1 + \gamma(1 - m)N} \right] + \frac{a}{b}\left[ \frac{b(1 - m)NP}{1 + \gamma(1 - m)N} - cP \right] \\

\frac{dV}{dt} &= rN\left(1 - \frac{N}{K}\right) - \frac{a(1 - m)NP}{1 + \gamma(1 - m)N} + \frac{ab(1 - m)NP}{b(1 + \gamma(1 - m)N)} - \frac{ac}{b}P \\

\frac{dV}{dt} &= rN\left(1 - \frac{N}{K}\right) - \frac{a(1 - m)NP}{1 + \gamma(1 - m)N} + \frac{a(1 - m)NP}{1 + \gamma(1 - m)N} - \frac{ac}{b}P

\end{align*}$$

The terms cancel out:

$$ \frac{dV}{dt} = rN\left(1 - \frac{N}{K}\right) - \frac{ac}{b}P $$

We know that $K(t) \ge K_{min}$ and $N(t) \ge 0$. The function $f(N) = rN\left(1 - \frac{N}{K}\right)$ is a parabola opening downwards with roots at $N=0$ and $N=K$. Its maximum value is achieved at $N = K/2$, with the value $r(K/2)(1 - 1/2) = rK/4$.

Since $K(t) \le K_{max}$, the maximum value of $rN\left(1 - \frac{N}{K}\right)$ for all possible $K(t)$ is bounded by $rK_{max}/4$.

Therefore, we can write:

$$ \frac{dV}{dt} \le \frac{rK_{max}}{4} - \frac{ac}{b}P $$

Let $$M_V = \frac{rK_{max}}{4}$$, which is a positive constant. Then:

$$ \frac{dV}{dt} \le M_V - \frac{ac}{b}P $$

Now, consider the case where $P(t)$ becomes sufficiently large. If $P(t) > \frac{M_V b}{ac}$, then:

$$ M_V - \frac{ac}{b}P < M_V - \frac{ac}{b}\left(\frac{M_V b}{ac}\right) = M_V - M_V = 0 $$

So, if $P(t) > \frac{M_V b}{ac}$, then $\frac{dV}{dt} < 0$. This implies that if $P(t)$ exceeds this threshold, the value of $V(t)$ will start to decrease. Since $V(t) = N(t) + \frac{a}{b}P(t)$ and $N(t) \ge 0$, a decrease in $V(t)$ when $P(t)$ is large means that $P(t)$ must also decrease. Thus, $P(t)$ cannot grow indefinitely.

Therefore, $P(t)$ is ultimately bounded above. Since $P(t) \ge 0$, it is also bounded below. So, there exists a constant $P_{upper} > 0$ such that $0 \le P(t) \le P_{upper}$ for all sufficiently large $t$.

 

[anuttarasammyak]

The first equation is, when N>>1
\frac{dN}{dt} \approx -\frac{r}{K}N^2 &lt; 0
So N seems to be upper bounded.

The third equation is, when K>>1
\frac{dK}{dt} \approx -K &lt; 0
So K seems to be upper bounded.

The second equation is
\frac{dP}{dt} =-D P
where
D=c-\frac{b(1-m)N}{1+\gamma(1-m)N}
Can we expect D>0 in your system so that P is upper bounded ?

 

[pasmith]

Can you find a Liapunov function for the system?