Poincaré Map Nonlinear Pendulum with Torque

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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?

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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'<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.