No lecture next Tuesday. Second midterm next Thursday at 6 pm.
Keywords for test 2: Not cumulative. Covers material from first exam. Test 2, starts with vector spaces, definition of vector space, norms, sequences, convergence. We used space of bounded function \(B\left ( \left [ t_{0},t_{1}\right ] ,\Re ^{n}\right ) \). This is a function space. Can be made of vector functions. The default convergence in this space is uniform convergence. But there is a weaker convergence called pointwise. This is important for integrals (we can move the limit inside if the function converges uniformly).
Picard iterations. M-test to test for uniform convergence. Solution of state equation using Picard iterations. We proofed many things about Picard, such as convergence and uniqueness. We used Granwall’s inequality.
We looked at negative aspects of Picard iterations. We want closed form. Using fundamental matrix \(\Psi \left ( t\right ) \). Once we have \(\Psi \left ( t\right ) \) we have solution for any input. \(\Psi \left ( t\right ) \) is not unique, but \(\Phi \left ( t,\tau \right ) \) is. \(\Phi \left ( t,\tau \right ) =\Psi \left ( t\right ) \Psi ^{-1}\left ( \tau \right ) \). We also talked at LTI. It simplifies. We used \(\Psi \left ( t\right ) =e^{At}\). We looked at three methods to find \(e^{At}.\)
Back to lecture. We are talking about controllability of \(x^{\prime }\left ( t\right ) =A\left ( t\right ) x\left ( t\right ) +B\left ( t\right ) u\left ( t\right ) \). We started talking about physical controllability. This is the ability to take the system from \(x\left ( t_{0}\right ) \) to \(x\left ( t_{1}\right ) \) for \(t_{1}>t_{0}\) by using some \(u\left ( t\right ) \).
How to test controllability of LTV system? Define \(W={\displaystyle \int } FF^{T}dt\) and check of \(W\) is not singular. In our case \(F=\Phi \left ( t_{0},\tau \right ) B\left ( \tau \right ) \).
Theorem: LTV system is controllable at \(t_{0}\) iff the rows of \(F=\Phi \left ( t_{0},\tau \right ) B\left ( \tau \right ) \) are linearly independent time functions on \(\left [ t_{0},t_{1}\right ] \) for some \(t_{1}\geq t_{0}\). From last lecture, we defined \[ W\left ( t_{0},t_{1}\right ) ={\displaystyle \int _{t_{0}}^{t_{1}}} \Phi \left ( t_{0},\tau \right ) B\left ( \tau \right ) B^{T}\left ( \tau \right ) \Phi ^{T}\left ( t_{0},\tau \right ) d\tau \] \(W\left ( t_{0},t_{1}\right ) \) must be not singular for the system to be controllable at \(t_{0}\).
Proof: Sufficiency \(\Longleftarrow \). We need to proof this: If \(W\) not singular, then LTV is controllable at \(t_{0}\). For necessity \(\Longrightarrow \) we need to proof this: If LTV is controllable at \(t_{0}\) then \(W\) is not singular.
We start with Sufficiency. Let \(W\) be not singular. Let \(x\left ( t_{0}\right ) ,x\left ( t_{1}\right ) \) be arbitrarily given states, where \(x\left ( t_{1}\right ) \) is the target state at time \(t_{1}\). We must be able to construct \(u\left ( t\right ) \) that steers the system from \(x\left ( t_{0}\right ) \) to \(x\left ( t_{1}\right ) \). i.e. we want \(x\left ( t_{1}\right ) =\Phi \left ( t_{1},t_{0}\right ) x\left ( t_{0}\right ) +{\displaystyle \int \limits _{t_{0}}^{t}} \Phi \left ( t,\tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau \). Pre-multiplying both sides by \(\Phi \left ( t_{0},t_{1}\right ) \) gives
\begin{align*} \Phi (t_0,t_1) x(t_1) &= \overbrace{\Phi (t_0,t_1) \Phi (t_1,t_0)}^{I} x(t_0) +{\displaystyle \int \limits _{t_{0}}^{t}} \overbrace{\Phi (t_0,t_1) \Phi (t,\tau )}^{\Phi (t_0,\tau )} B(\tau ) u(\tau ) d\tau \\ \Phi (t_0,t_1) x(t_1) & =x\left (t_{0}\right ) +{\displaystyle \int \limits _{t_{0}}^{t}} \Phi (t_0,\tau ) B(\tau ) u(\tau ) d\tau \\ \Phi (t_0,t_1) x(t_1) -x(t_0) & ={\displaystyle \int \limits _{t_{0}}^{t}} \Phi (t_0,\tau ) B(\tau ) u(\tau ) d\tau \end{align*}
Let \[ u(\tau ) = - B^T(\tau ) \Phi ^T(t_1,\tau ) W^{-1}(t_0,t_1) \left [ \Phi (t_1,t_0) x(t_0) -x(t_1) \right ] \] Reader Show that this \(u(t)\) leads to \(x(t_0) \longrightarrow x(t_1)\)
We now do proof of necessity \(\Longrightarrow \) If LTV is controllable at \(t_{0}\) then \(W\) is not singular. Equivalently, show that if LTV is controllable at \(t_{0}\) then rows of \(\Phi \left ( t_{0},\tau \right ) B\left ( \tau \right ) \) are linearly independent. Proof by contradiction: Assume \(W\) is singular but LTV is controllable at \(t_{0}\) and show a contradiction. Since \(W\) is singular, then there exist a vector \(\vec{\alpha }\neq 0\) s.t. \(\vec{\alpha }^{T}\Phi \left ( t_{0},\tau \right ) B\left ( \tau \right ) =0\) for all \(\tau \in \left [ t_{0},t_{1}\right ] \). Now construct \(x\left ( t_{0}\right ) ,x\left ( t_{1}\right ) \). Let \(x\left ( t_{0}\right ) =\vec{\alpha }\) and let \(x\left ( t_{1}\right ) =0\) (i.e. the origin vector in the state space). Since LTV is controllable, then there exist \(u\left ( t\right ) \) such that
\begin{align*} x\left ( t_{1}\right ) & =\Phi \left ( t_{1},t_{0}\right ) x\left ( t_{0}\right ) +{\displaystyle \int \limits _{t_{0}}^{t_{1}}} \Phi \left ( t_{1},\tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau \\ & =\Phi \left ( t_{1},t_{0}\right ) \vec{\alpha }+{\displaystyle \int \limits _{t_{0}}^{t_{1}}} \Phi \left ( t_{1},\tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau \end{align*}
Pre-multiply by \(\Phi \left ( t_{0},t_{1}\right ) \) both sides gives\[ \Phi \left ( t_{0},t_{1}\right ) x\left ( t_{1}\right ) =\overbrace{\Phi \left ( t_{0},t_{1}\right ) \Phi \left ( t_{1},t_{0}\right ) }^I\vec{\alpha }+{\displaystyle \int \limits _{t_{0}}^{t_{1}}} \overbrace{\Phi \left ( t_{0},t_{1}\right ) \Phi \left ( t_{1},\tau \right ) }^{\Phi \left ( t_{0},\tau \right ) }B\left ( \tau \right ) u\left ( \tau \right ) d\tau \]
So the above becomes
\[ \Phi \left ( t_{0},t_{1}\right ) x\left ( t_{1}\right ) =\vec{\alpha }+{\displaystyle \int \limits _{t_{0}}^{t}} \Phi \left ( t_{0},\tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau \] Now pre-multiplying both sides by \(\vec{\alpha }^{T}\)\[ \vec{\alpha }^{T}\Phi \left ( t_{0},t_{1}\right ) x\left ( t_{1}\right ) =\vec{\alpha }^{T}\vec{\alpha }+{\displaystyle \int \limits _{t_{0}}^{t}} \vec{\alpha }^{T}\Phi \left ( t_{0},\tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau \]
But \(\vec{\alpha }^{T}\vec{\alpha }=\left \Vert \alpha \right \Vert ^{2}\) and \(\vec{\alpha }^{T}\Phi \left ( t_{0},t_{1}\right ) x\left ( t_{1}\right ) =0\) since \(x\left ( t_{1}\right ) =0\), hence the above becomes\[ 0=\left \Vert \alpha \right \Vert ^{2}+{\displaystyle \int \limits _{t_{0}}^{t}} \vec{\alpha }^{T}\Phi \left ( t_{0},\tau \right ) B\left ( \tau \right ) u\left ( \tau \right ) d\tau \]
But we assumed \(\vec{\alpha }^{T}\Phi \left ( t_{0},\tau \right ) B\left ( \tau \right ) =0\) above, hence the above reduces to\[ 0=\left \Vert \alpha \right \Vert ^{2}\]
Which means \(\vec{\alpha }=0\). But this contradicts our assumption that \(\vec{\alpha }\neq 0\). Hence our assumption that \(W\) is singular but LTV is controllable at \(t_{0}\) has been found to produce a contradiction. Hence it must be that out assumption of \(W\) being singular is not valid. QED.
Advanced reader: The proof of necessity above contains an error. Try to find it.
Remark: There are infinite many controls \(u\left ( t\right ) \) that can take \(x\left ( t_{0}\right ) \) to \(x\left ( t_{1}\right ) .\) We considered one of them in the above proof \(u\left ( t\right ) =B^{T}\left ( \tau \right ) \Phi ^{T}\left ( t_{0},\tau \right ) W^{-1}\left ( t_{0},t_{1}\right ) \left [ \Phi \left ( t_{0},t_{1}\right ) x\left ( t_{1}\right ) -x\left ( t_{0}\right ) \right ] \). In this \(u\) special in some sense?
Minimum energy theorem: Suppose \({\displaystyle \sum } \) is controllable at \(t_{0},\) i.e, controllability \(W\left ( t_{0},t_{1}\right ) \) is not singular. So we can steer \(x\left ( t_{0}\right ) \) to \(x\left ( t_{1}\right ) \). So given pair \(x\left ( t_{0}\right ) ,x\left ( t_{1}\right ) \) and as associated control law \(u\left ( t\right ) \), define the energy \(E\left ( u\right ) ={\displaystyle \int \limits _{t_{0}}^{t_{1}}} u^{T}udt={\displaystyle \int \limits _{t_{0}}^{t_{1}}} \left \Vert u\right \Vert ^{2}dt\). This gives energy needed. \(u\left ( t\right ) \) used above in the proof minimizes \(E\left ( u\right ) .\) Next lecture we will show that \(u\left ( t\right ) =B^{T}\left ( \tau \right ) \Phi ^{T}\left ( t_{0},\tau \right ) W^{-1}\left ( t_{0},t_{1}\right ) \left [ \Phi \left ( t_{0},t_{1}\right ) x\left ( t_{1}\right ) -x\left ( t_{0}\right ) \right ] \) is energy minimizer.