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Consider the oscillator equation $$\displaystyle\ddot{x}+F(x,\dot{x})\dot{x}+x=0$$ where $F(x,\dot{x})<0$ if $r\leq a$ and $F(x,\dot{x})>0$ if $r\geq b$ where $r^2=x^2+\dot{x}^2$. Show that there is at least one closed orbit in the region $a<r<b$.

Rescaling $\dot{x}=t$ the system of equations is as follows : $$\dot{x}=t=A(x,t) \\ \dot{t}=-F(x,t)t-x=B(x,t)$$ Now we use Bendixson-Dulac criteria as follows : $$\nabla(A,B)=-F(x,t)-t\frac{\partial F}{\partial t}(x,t)$$ Afterwards I'm confused as there is no way to show that $\nabla(A,B)$ is of one sign using the conditions given as above, even if it would only show that the equation has no closed orbits inside $r\leq a$ or outside of $r\geq b$. How to proceed for this one? Thanks in advance.

am_11235...
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1 Answers1

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Hint.

$$ \cases{\dot x_1 = x_2\\ \dot x_2 = - x_2 F(x_1,x_2)-x_1}\Rightarrow\cases{x_1\dot x_1 = x_1x_2\\ x_2\dot x_2 = - x_2^2 F(x_1,x_2)-x_1x_2} $$

adding the two equations we have

$$ \frac 12\left(x_1^2+x_2^2\right)'= -x_2^2F(x_1,x_2) $$

Cesareo
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