Can you please help me to prove these properties of the given dynamical system? $$\begin{align} \frac{dN}{dt}=&aN\left(1-\frac{N}{k}\right)-bNP\\ \frac{dP}{dt}=&cNP-DP\\\text{so that:}\\ N(0)\geq& 0\\P(0)\geq& 0 \end{align}$$
Prove that $R^+$ is positive invariant set for the system.$\\$
Prove that the system is dissipative.
Thanks a lot!
Assuming that $a,k,b,c$ and $D$ are positive constants, the system can be written as \begin{align} \frac{dN}{dt} &= \big(a - \frac{a}{k} \, N - b\, P\big)\, N \\ \frac{dP}{dt} &= \big(c\, N - D \big)\, P \end{align} Observe that if you take a solution $N(t)$ of the equation $$\frac{dN}{dt} = a\, N - \frac{a}{k} \, N^2$$ then a direct check shows that $(N(t), 0)$ is a solution of the original system \begin{align} \frac{dN}{dt} &= \big(a - \frac{a}{k} \, N - b\, P\big)\, N \\ \frac{dP}{dt} &= \big(c\, N - D \big)\, P \end{align} which means that the line $\{(N,P) \, : \, P=0 \}$ is invariant under the flow, i.e. it is a trajectory of the system (the unstable manifold of the equilibrium $(0,0)$). Analogously, if one takes a solution $P(t)$ of the equation $$\frac{dP}{dt} = D\, P$$ then a direct check shows that $(0,P(t))$ is a solution to the original system. In other words, the line $\{(N,P) \, : \, N=0 \}$ is invariant under the flow, i.e. it is a trajectory of the system (the stable manifold of the equilibrium $(0,0)$). Thus, if a solution $(N(t), P(t))$ of the original system, starting from the positive quadrant, i.e. $(N(0), P(0)) \in \mathbb{R}_+^2$, traverses a trajectory for $t \in \mathbb{R}$ that always stays in $\mathbb{R}_+^2= \{(N,P) \, : \, N>0,\, P>0\}$ because in order to leave $\mathbb{R}_+^2$ that trajectory needs to intersect one of the trajectories lying on either $\{(N,P) \, : \, P=0 \}$ or $\{(N,P) \, : \, N=0 \}$ which is impossible because two different trajectories of an autonomous system like the given one do not intersect. Therefore, $\mathbb{R}_+^2 $ is an invariant domain of the system.
For the second question, I assume you have to manufacture a function $V(N,P)$ in $\mathbb{R}_+^2$ with the property that $\frac{d}{dt} V(N,P) \leq 0$. In order to do that, first calculate the equilibrium points of the original system \begin{align} \frac{dN}{dt} &= a\, N - \frac{a}{k} \, N^2 - b\, NP \\ \frac{dP}{dt} &= c\, NP - D\, P \end{align} i.e. solve the equations \begin{align} 0 &= a\, N - \frac{a}{k} \, N^2 - b\, NP = \big( a\, - \frac{a}{k} \, N - b\, P\big)\, N \\ 0 &= c\, NP - D\, P = \big(c\, N - D\big)\, P \end{align} If $N=0$ then $P=0$ too. Assume $N \neq 0$. Then either $P=0$ and in this case $N = k$ or $$N= \frac{D}{c} \,\,\text{ and } \,\, P = \frac{a}{b} - \frac{aD}{bck}$$ which is located in $\mathbb{R}^2_+$ if and only if $ck > D$. The only equilibrium in the interior of $\mathbb{R}^2_+$ is $\Big( \frac{D}{c}, \, \frac{a}{b} - \frac{aD}{bck}\Big)$. Now, rewrite your system in the form \begin{align} \frac{dN}{dt} &= a\, N - \frac{aD}{ck} \, N - b\, NP + \frac{aD}{ck} \, N - \frac{a}{k} \, N^2\\ \frac{dP}{dt} &= c\, NP - D\, P \end{align} I claim that the closely related system (in fact a truncated system, obtained from the latter one by removing the last two terms from the first equation) \begin{align} \frac{dN}{dt} &= a\, N - \frac{aD}{ck} \, N - b\, NP\\ \frac{dP}{dt} &= c\, NP - D\, P \end{align} is integrable, i.e. it is conservative because it has a conserved a quantity, i.e. a first integral of motion. Rewrite the system as \begin{align} \frac{dN}{\left(\, a\, N - \frac{aD}{ck} \, N - b\, NP \, \right)} &= dt\\ \frac{dP}{\left( c\, NP - D\, P \right)} &= dt \end{align} which after eliminating the $dt$ from both equations reduces to \begin{align} \frac{dN}{\left(\, a\, N - \frac{aD}{ck} \, N - b\, NP \, \right)} = \frac{dP}{\left( c\, NP - D\, P \right)} \end{align} which can be represented also as \begin{align} \frac{dN}{\left(\, a - \frac{aD}{ck} - b\, P \, \right) N} = \frac{dP}{\left( c\, N - D\, \right) P} \end{align} After reorganizing the terms of this latter differential form gets \begin{align} \frac{ \left( c\, N - D\, \right) \, dN}{N} = \frac{ \left(\, a - \frac{aD}{ck} - b\, P \, \right) \, dP}{ P} \end{align} which is exact, i.e. is the differential of a function i.e. there exists $V(N,P)$ such that \begin{align} dV &= \frac{ \left( c\, N - D\, \right) \, dN}{N} - \frac{ \left(\, a - \frac{aD}{ck} - b\, P \, \right) \, dP}{ P} = \left( c - \frac{D}{N}\right) \, dN - { \left(\, \Big(a - \frac{aD}{ck}\Big) \frac{1}{P} - b \, \right) \, dP} \end{align} which after integration turns into $$V(N,P) = c\, N - D \, \log(N) + b\, P - \Big(a - \frac{aD}{ck}\Big) \, \log(P) + V_0 $$ where the constant $V_0$ is chosen so that the value of the function $V(N,P)$ at the equilibrium point $\Big( \frac{D}{c}, \, \frac{a}{b} - \frac{aD}{bck}\Big)$ is zero. Also the value of the differential $dV$ et the equilibrium point is $\Big( \frac{D}{c}, \, \frac{a}{b} - \frac{aD}{bck}\Big)$ by construction, so the function $V$ has a critical point at this equilibrium. If you calculate the matrix of the second derivatives of $V$ at that equilibrium, you see that the matrix is diagonal with positive entries, so positive definite, which means the equilibrium is a minimum of the function $V$, i.e. the minimum energy of the modified conservative system occurs at the equilibrium.
Now, take a solution $(N(t), P(t))$ in $\mathbb{R}^2_+$ of the original system and calculate \begin{align} \frac{d}{dt} \, V\big(N(t), P(t)\big) &= \left( c - \frac{D}{N}\right) \, \frac{dN}{dt} - { \left(\, \Big(a - \frac{aD}{ck}\Big) \frac{1}{P} - b \, \right) \, \frac{dP}{dt}}\\ & = \left( c - \frac{D}{N}\right) \,\left( a\, N - \frac{a}{k} \, N^2 - b\, NP\right) - { \left(\, \Big(a - \frac{aD}{ck}\Big) \frac{1}{P} - b \, \right) \, \left(c\, NP - D\, P\right) }\\ &= - \frac{a}{ck}\big(c\, N - D\big)^2 \leq 0 \end{align} which means that $V$ is a global Lyapunov (kind of like an energy) function for your system in the domain $\mathbb{R}^2$, the equilibrium point there is asymptotically stable, and the system itselfis dissipative in the first quadrant.
