A question on the qualitative analysis of solution of a system of ODEs [closed]

Question

Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a non-zero smooth vector field satisfying $\text{div} f \ne 0.$ Which of the following are necessarily true for the ODE: $\dot{\mathbf{x}}=f(\mathbf{x})$?

(a) There are no equilibrium points
(b) There are no periodic solutions
(c) All the solutions are bounded
(d) All the solutions are unbounded

My attempt: This is actually a system of 2 ODEs.

Notation: $\mathbf{x}=(x,y).$

For option (a), set $f(x,y)=(x,y).$ See that $\text{div} f \ne 0.$ Now, we can clearly see that $(0,0)$ is an equilibrium point for the system of ODE $\dot{\mathbf{x}}=f(\mathbf{x}).$

So, option (a) is NOT true in general meaning that it is NOT necessary.

I am confused as to how to proceed for the remaining options.

The KEY ANSWERS says that (a), (b) and (d) are the correct answers for this question.

Please help on how to proceed.

Answer 1

Indeed, the answer key is correct. The options (a), (b) and (d) are true.

(a) Notice that, in the question, it is said that $f$ is non-zero, thus there are no equilibrium points.

(b) Suppose there was a periodic solution of period $T$, $\gamma(t)=(x(t),y(t))$, which consists of a closed curve in $\mathbb{R}^2$, say $C$. Let $R$ be the region enclosed by this curve. Then, if $f=(f_1,f_2)$, we can use Green's theorem to obtain (omitting $t$): $$\iint_{R}\text{div} f\,dA =\oint_{C}(-f_2,f_1)\,dS=\int_{t_0}^{t_0+T}(-f_2(\gamma),f_1(\gamma)) \cdot(\dot{x},\dot{y})\, dt\\ =\int_{t_0}^{t_0+T}-f_2(\gamma)f_1(\gamma)+f_1(\gamma)f_2(\gamma)\,dt=0$$ However, since div$f \neq 0$ and since $\mathbb{R}^2$ is connected, div$f$ has constant sign, and thus the left-most integral cannot be $0$, and we have arrived at a contradiction.

(d) If there was a bounded solution, then it would be contained in some closed ball $B$, which is compact. By Poincaré-Bendixson's theorem, there would need to exist either equilibrium points or periodic solutions in $B$. However, we have seen that this is not the case.