This question may seem pretty dumb. But I really want to know. I have the following linear dynamical system for Dubin's path,
\begin{align*} \phi &= \begin{bmatrix} x \\ y \\ \theta \end{bmatrix}\\ \frac{d\phi}{dt} &= f(\phi, u) = \begin{bmatrix} v_s \cos{\theta} \\ v_s \sin{\theta} \\ u \end{bmatrix} \end{align*}
where $v_s$ is the average velocity, $\frac{dx}{dt} = v_s \cos{\theta}$, $\frac{dy}{dt} = v_s \sin{\theta}$ and $\frac{d\theta}{dt} = u$. Now clearly we can say for two states $(x_1,y_1,\theta_1)$ and $(x_2,y_2,\theta_2)$, if $\theta_2 = \theta_1$ then the input $u = 0$. But I just want to know what will be the input condition for stopping the robot at $(x_n,y_n,\theta_n)$? Should it be $0$ if $\theta_n = \theta_{n-1}$? We only need to make $v_s=0$ for stopping regardless of the input right? But $v_s$ is constant for the other states. So how I should handle this? This may be due to my lacking of understanding and a dumb question but it will be really helpful if anybody can explain this to me.
I don't think your question is dumb, but the design of the ODE system is quite dumb. The speed for moving along $x,y$ plane is a constant, $v_s$, while $u$ only controls rotation. In this design, there is no way to make this robot stop dancing around the best it could do is to rotating around a fix point at its speed $v_s$...
Unless the question adds another input to accelerate or decelerate the robot (which changes $v_s$).