Phase portrait for time-optimal control problem (controllable region??)

Question

I need to find the time-optimal control to the origin for the system: $$\dot x_1=3x_1+x_2$$ $$\dot x_2=4x_1+3x_2+u$$ where $|u|\le1$. I have used Pontryagin's Minimum Principle and linear algebra to find the family of curves solution to be: $$ \begin{pmatrix} x_1 \\ x_2 \\ \end{pmatrix} =A\begin{pmatrix} -1 \\ 2 \\ \end{pmatrix}e^t+B\begin{pmatrix} 1 \\ 2 \\ \end{pmatrix}e^{5t}\pm\begin{pmatrix} \frac{1}{5} \\ \frac{-3}{5} \\ \end{pmatrix}$$

Since the two distinct real eigenvalues are both positive, I know $\pm\begin{pmatrix} \frac{1}{5} \\ \frac{-3}{5} \\ \end{pmatrix}$ are unstable nodes. Also, the parabolas in the phase portrait will hug the $\begin{pmatrix} 1 \\ 2 \\ \end{pmatrix}$ vector, and we can switch from a $u=1$ to $u=-1$ solution at most once. I have drawn a phase portrait here (not 100% sure it is correct), with the unique switching curve intersecting the origin in red (POQ). I'm not sure how to describe the controllable region, and it seems if you're to the right of the $u=1$ fixed point or to the left of the $u=-1$ fixed point, you'd be in an uncontrollable region that can never reach the origin. So I suspect the only controllable region is the rectangular one created by the eigenvectors (dotted lines in my diagram). Can anyone further elaborate? Thanks

Answer 1

Given the general system

$$ \dot{x} = A\,x + B\,u, \tag{1} $$

which you want to drive to $x(T)=0$ while minimizing $T$ and satisfying $|u| \leq u_\max = 1$. Minimizing time for controllable LTI systems I believe always yields a bang-bang controller, with for a two-dimensional systems at most one switch (though I do not have a source for this). All initial conditions for which no switching is required can be obtained by solving the dynamics backwards in time

$$ \begin{bmatrix} x(0) \\ \pm u_\max \end{bmatrix} = e^{-\begin{bmatrix}A & B \\ 0 & 0\end{bmatrix} T} \begin{bmatrix}x(T) \\ \pm u_\max\end{bmatrix},\ \forall\ T\in(0,\infty]. \tag{2} $$

If $A$ is non-singular $(2)$ can also be written as

$$ x(0) = e^{-A\,T} x(T) \pm A^{-1} (e^{-A\,T} - I) B\,u_\max. \tag{3} $$

It can be noted that if all eigenvalues of $A$ have a positive real part then $e^{-A\,T}$ vanishes as $T\to\infty$, which implies that

$$ \lim_{T\to\infty} x(0) = \mp A^{-1} B\,u_\max. \tag{4} $$

All initial conditions for which one switch in control input is required can be obtained by again solving the dynamics backwards in time and ending at $(2)$

$$ \begin{bmatrix} x(0) \\ \mp u_\max \end{bmatrix} = e^{-\begin{bmatrix}A & B \\ 0 & 0\end{bmatrix} T_1} \underbrace{\begin{bmatrix}I & 0 \\ 0 & -1\end{bmatrix}}_\text{invert $u$} \underbrace{e^{-\begin{bmatrix}A & B \\ 0 & 0\end{bmatrix} T_2} \begin{bmatrix}x(T_1+T_2) \\ \pm u_\max\end{bmatrix}}_{\begin{bmatrix}x(T_1) \\ \pm u_\max\end{bmatrix}},\ \forall\ T_1,T_2\in(0,\infty], \tag{5} $$

with $T = T_1 + T_2$.

In your case all eigenvalues of $A$ have a positive real part, which means that $x(0)$ from $(5)$ remains bounded for all $T_1,T_2\in(0,\infty]$. The border of this controllable region can be obtained by using $\lim_{T_2\to\infty}$, which by using $(3)$ and $(4)$ can be simplified to

$$ x(0) = \mp A^{-1} (2\,e^{-A\,T_1} - I) B\,u_\max,\ \forall\ T_1\in(0,\infty]. \tag{6} $$

Plotting this together with the no switching curves yields for your system the following plot. So your guess of a rectangular region was close, but also includes a couple of initial conditions which can't be driven to the origin.

It can be noted that for $A$ with all eigenvalues having a positive real part and some non-zero imaginary parts the minimum time trajectory is not directly obvious, since in that case $(2)$ and $(6)$ can cross each other infinitely many times.