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 $2\pi$.
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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!
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$\begingroup$ You have shown when $y_1(x)>y_2(x)>y_3(x)>y_4(x),\, \forall x$, they cannot all be solutions. But does that condition necessarily hold for any set of $4$ distinct solutions? $\endgroup$ – Hans Sep 24 '18 at 5:48
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$\begingroup$ @Hans Yes, it's true because of the existence and uniqueness theorem for ODE. $\endgroup$ – user263834 Sep 24 '18 at 7:46
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$\begingroup$ You are right. Two distinct solutions cannot cross each other, otherwise they should coalesce into one. Very nice. +1. What inspired you to think of this manipulation? Do you have other similar examples of this method? $\endgroup$ – Hans Sep 24 '18 at 8:54
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$\begingroup$ @Hans I think this is a classic result in ODE, but I can't find the source of this method. Maybe it's just a very special one. $\endgroup$ – user263834 Sep 25 '18 at 11:35
