How to know the shape of trajectories of Prey Predator Model and existence of orbit Consider the Prey Predator Model,
$$
\begin{align}
\frac{du}{dt}&=(a_1-b_1 v)u,\\
\frac{dv}{dt}&=(-a_2+b_2 u)v\\
u(0)&=u_{10},v(0)=v_{20}
\end{align}\tag 1
$$
Although the model $(1)$ looks simple, a complete analytic solution of these equations are
not possible but for every $a_1, a_2, b_1, b_2, u_{10}, v_{20}$. By variable separation,
$$
\begin{align}
\frac{du}{dv}&=\frac{(a_1-b_1 v)u}{(-a_2+b_2 u)v}\\
\left(\frac{u}{u_{10}}\right)^{-a_2}e^{b_2(u-u_{10})}&=\left(\frac{v}{v_{20}}\right)^{a_1}e^{-b_1(v-v_{20})} \quad\textrm{ Integrating from 0 to t}\\
e^{b_2u+b_1v}&=\lambda v^{a_1}u^{a_2} \quad\text{ Where }\lambda=\frac{v_{20}^{-a_1}e^{b_1v_{20}}}{v_{10}^{a_2}e^{-b_2u_{10}}}\tag 2
\end{align}
$$

Both the $u$ and $v$ axes are orbits of $(1)$ and the orbits are the family of curves defined by $(2)$ and these curves are closed. This implies that each solution $u(t)$ and $v(t)$ of $(1)$ which starts in the first quadrant $u>0,v>0$ at time $t=t_0$ will remain there for all future time $t\geq t_0$.

I didn't understand this paragraph. How they said about orbits and closed without any justification. Especially, "each trajectory is bounded closed curve". This system has two equilibrium points. The zero equilibrium point $(0,0)$ and the positive equilibrium point $(a_2/b_2,a_1/b_1)$. In fact, I couldn't find any way to conclude this.

*

*How could I get some idea of the shapes of the trajectories?

*How could I know there exist a orbit?

We have recently introduced with population Dynamics and have only lecture notes to follow. That's why so many unclear paragraph need to resolve before exam :)
Any help will be appreciated. TIA

The terms $b_1v-a_2\log v$ and $b_2u-a_2\log u$ in the "logarithmic" first integral are both convex functions with a minimum $m_v$ and $m_u$ and going to $+\infty$ moving towards $0$ and $\infty$. If the sum of both is equal to some constant $C$, then we know that the first term is bounded above by $C-m_u$ and the second by $C-m_v$. Both bounds give a finite interval where they are valid. This means that the level sets have to consist of closed curves.
A change of direction requires a stationary point, which do not exist outside the one equilibrium point at the center/minimum. So the level curves are followed in only one orientation.
As the first integral is also jointly convex as a function of both variables, all level sets are convex, so the level curves all have only one component.
