TITLE: Abel differential equation with periodic coefficient
QUESTION [5 upvotes]: Consider differential equation $$y'=a_3(x)y^3+a_2(x)y^2+a_1(x)y+a_0(x)$$ where $a_i(x)$ is continuous and periodic with period $2\pi$, i=0, 1, 2, 3.
Assume that $a_3(x)\ge0$ and $a_3(x)$ is not equal to 0 for all x.
Prove that the equation has at most three different periodic solutions with period $2\pi$.

REPLY [2 votes]: Assume $y_1(x)>y_2(x)>y_3(x)>y_4(x)$ are four $2\pi-$ periodic solutions.
Then we get $$ \frac{y_1'-y_2'}{y_1-y_2}-\frac{y_2'-y_3'}{y_2-y_3}=a_3(y_1-y_3)(y_1+y_2+y_3)+a_2(y_1-y_3)$$
Integrate both side from $0$ to $2\pi$, we get $$ 0=\int_0^{2\pi}\left[ a_3(y_1-y_3)(y_1+y_2+y_3)+a_2(y_1-y_3)\right] \mathrm{d}x$$
Similarly we can get $$ 0=\int_0^{2\pi}\left[ a_3(y_1-y_3)(y_1+y_4+y_3)+a_2(y_1-y_3)\right] \mathrm{d}x$$
Subtract two equations, we get $$ 0=\int_0^{2\pi} a_3(y_1-y_3)(y_2-y_4) \mathrm{d}x$$
It's clear that RHS is larger than $0$. Contradiction!