Can you please help me to prove these properties of the given dynamical system? $$\begin{align} \frac{dN}{dt}=&aN\left(1-\frac{N}{k}\right)-bNP\\ \frac{dP}{dt}=&cNP-DP\\\text{so that:}\\ N(0)\geq& 0\\P(0)\geq& 0 \end{align}$$

  1. Prove that $R^+$ is positive invariant set for the system.$\\$

  2. Prove that the system is dissipative.

Thanks a lot!


closed as off-topic by Did, Shailesh, Leucippus, C. Falcon, Daniel W. Farlow Apr 23 '17 at 2:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Shailesh, Leucippus, C. Falcon, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Are $a, k, b, c$ and $D$ constants? If yes, according to your definition of dissipative system, do you need to find a function $V(N,P)$ such that $\frac{d}{dt} \, V(N,P) < 0$? $\endgroup$ – Futurologist Apr 17 '17 at 17:13
  • $\begingroup$ And you have really no idea about how to attack even one part of this? How comes? $\endgroup$ – Did Apr 22 '17 at 22:06

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.

  • $\begingroup$ thanks a lot for the solution! $\endgroup$ – Ксения Цочева Apr 19 '17 at 4:06
  • 1
    $\begingroup$ @КсенияЦочева Няма защо :)... Все пак имай впредвид, че решението зависи и от определенията за дисипативни системи които са ви дали на лекции, така, че не съм сигурен дали се търси решение точно в този вид. Но това е един възможен подход. По принцип, на тази система лесно се изследва фазовия портрет, особено ако се разгледат правата по които векторното поле е хоризонтално насочено и правата по която полето е вертикално насочено. $\endgroup$ – Futurologist Apr 19 '17 at 5:48

Not the answer you're looking for? Browse other questions tagged or ask your own question.