# Poincaré Map Nonlinear Pendulum with Torque

Suppose we have the following system $$\theta'=v$$ $$v'= -bv - \sin\theta + k.$$ I can find the equilibria and the bk-parameter plane of this system, but this question I don't know how to approach.

Suppose we have the line where $$\theta = 0$$, what can be said about the qualitative features of the Poincaré map along this line?

Consider the energy function $$E(\theta , v)=1/2 v^2 +1-cos(\theta)$$ .
And let $$p$$ is the Poincaré map. We have a periodic orbit if $$p (v)=v$$ for some v.
On the other hand $$dE/ d\theta = k-bv.$$ so a periodic orbit exist if the total energy is zero.
And hence $$\int_{0}^{2\pi} (dE/ d\theta ) d\theta =0$$ when $$\int_{0}^{2\pi} v d\theta = 2k\pi /b$$ . Notice that if $$v' then $$\int_{0}^{2\pi} v' d\theta$$ < $$\int_{0}^{2\pi} v d\theta$$ and same as if $$v'>v$$ , this means there is a unique v.
For some initial value $$(0,v_{0})$$ .
If $$v_{0}< v$$ ( the trajectory through the initial value is insitde the periodic orbit) then $$p$$ is increasing ( since E'>0). If $$v_{0}> v$$ (the trajectory is outside of the periodic orbit ) thenp is decreasing ( E'<0) In both two cases p tend to the fixed point v which yelds the periodic orbit.