Assume $a>0$ and $b>0$ and $g(x)=0$ when $|x|>1$ , $g(x)=k$ when $|x|\le1$ .
Now show that in system of differential equation
$$x'=y $$ $$y'=-[2b-g(x)]ay-ay^2$$
if $k<2b$ then it hasn't periodic solution and if $k>2b $ then it has a periodic solution.
my stategy: I know that if an orbit is periodic one, then poincare index of corresponding equation on the orbit in clockwise direction is $+1$ , I want use the poincare index but i can't. can someone help or give a new way.
If the starting point $(x_0,y_0)$ is such that $y_0\gt0$ then $x$ increases until, if the case arises, $y=0$. Then $(x,y)$ stops since $x'=y'=0$ when $y=0$. Thus, no periodic solution may intersect the halfplane $y\gt0$. The same reasoning shows that no periodic solution may intersect the halfplane $y\lt0$. Finally, every point on $y=0$ is a fixed point hence, for every $(a,b,k)$ as in the exercise, this dynamical system has no periodic solution at all.
A related dynamical system which could have periodic solutions is $$x'=y,\qquad y'=k\mathbf 1_{|x|\leqslant1}-c-y,$$ for some $k\gt c\gt0$. An approach to study this system is to consider some starting points $(1,y_0)$ with $|y_0|$ small. The solution $(x,y)$ meets again the line $x=1$ after a path in $x\gt1$ (if $y_0\gt0$) or in $x\lt1$ (if $y_0\lt0$), at a point $(0,\varphi(y_0))$. Every $y_0$ such that $\varphi(\varphi(y_0))=y_0$ yields a periodic solution.
Another approach would be to consider the starting point $(x_0,y_0)=(1,-c)$. If the solution hits the line $y=k-c$ at some $(x_1,k-c)$ with $x_1\geqslant-1$, then the next part of the solution runs along the segment between $(-1,k-c)$ and $(1,k-c)$, until the point $(1,k-c)$. The solution starting from $(1,k-c)$ is such that $y$ is decreasing, and hits the line $y=-c$ at some point $(x_2,-c)$. If $x_2\geqslant1$, the solution starting from $(1,-c)$ (or from $(1,k-c)$) is periodic. Note that $x_1$ and $x_2$ are functions of $(k,c)$ only.