\(W\left ( t_{0},t_{1}\right ) ={\displaystyle \int _{t_{0}}^{t_{1}}} \Phi \left ( t_{0},\tau \right ) B\left ( \tau \right ) B^{T}\left ( \tau \right ) \Phi ^{T}\left ( t_{0},\tau \right ) d\tau \) is called the controllability Gramian. Is there a shorter method to check for controllability other having to build \(W\left ( t_{0},t_{1}\right ) \) and check it is not singular? The short cut will be sufficient for LTV. For LTI, this short cut with minor change. will become sufficient and necessary for controllability (it will lead to the rank condition on the pair \(A,B\)).
Lemma: Consider \(n\times m\) matrix, we called it \(F\left ( t\right ) \) of continuous and smooth functions on \(\left [ t_{0},t_{1}\right ] \). This means each column is a vector functions. Suppose further that the matrix \(\Im \left ( t\right ) =\begin{pmatrix} F\left ( t\right ) & F^{\prime }\left ( t\right ) & \cdots & F^{\left ( n-1\right ) }\left ( t\right ) \end{pmatrix} \). This is of size \(n\times nm\). If \(\Im \left ( t\right ) \) has rank \(n\) for some \(t^{\ast }\in \left [ t_{0},t_{1}\right ] \), then it follows that \(F^{\left ( i\right ) }\left ( t\right ) \) are all linearly independent time functions on \(\left [ t_{0},t_{1}\right ] \).
Proof by contradiction. Suppose we have such \(t^{\ast }\in \left [ t_{0},t_{1}\right ] \) where rank \(\Im \left ( t\right ) =n\,\), and assume \(F^{\left ( i\right ) }\left ( t\right ) \) are all linearly dependent time functions. Hence we can find \(\vec{\alpha }\neq 0\), s.t. \(\vec{\alpha }^{T}F\left ( t\right ) =0\). This mean \(F\left ( t\right ) =0,\) so that all columns of \(\Im \left ( t\right ) \) are zero. Hence rank of \(\Im \left ( t\right ) \) is not \(n\). Hence \(F^{\left ( i\right ) }\left ( t\right ) \) are all linearly independent time functions. QED.
Is this useful to study controllability? Use \[ F\left ( \tau \right ) =\Phi \left ( t_{0},\tau \right ) B\left ( \tau \right ) \] Hence \begin{align*} F^{\prime }\left ( \tau \right ) & =\frac{d\Phi \left ( t_{0},\tau \right ) }{d\tau }B\left ( \tau \right ) +\Phi \left ( t_{0},\tau \right ) \frac{dB\left ( \tau \right ) }{d\tau }\\ F^{\prime \prime }\left ( \tau \right ) & =\frac{d^{2}\Phi \left ( t_{0},\tau \right ) }{d\tau }B\left ( \tau \right ) +\frac{d\Phi \left ( t_{0},\tau \right ) }{d\tau }\frac{dB\left ( \tau \right ) }{d\tau }+\frac{d\Phi \left ( t_{0},\tau \right ) }{d\tau }\frac{dB\left ( \tau \right ) }{d\tau }+\Phi \left ( t_{0},\tau \right ) \frac{d^{2}B\left ( \tau \right ) }{d\tau ^{2}} \end{align*}
Reader: Generalize to many \(F^{\left ( n\right ) }\) using recursive formula.
Let \(M_{0}\left ( \tau \right ) =B\left ( \tau \right ) \), then \[ M_{k+1}\left ( \tau \right ) =-A\left ( \tau \right ) M_{k}\left ( \tau \right ) +\frac{dM_{k}\left ( \tau \right ) }{d\tau }\]
For \(k=0\cdots n-2.\)
Reader: Show the above is true.
Now we will use the above lemma. System is controllable at \(t_{0}\) if there exist \(t^{\ast }>t_{0}\) such that rank\(\left ( M\left ( t^{\ast }\right ) \right ) =n\)
Example: Let \(x^{\prime }=\begin{pmatrix} \cos t & t\\ e^{t} & 2 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} +\begin{pmatrix} 1\\ t \end{pmatrix} u\left ( t\right ) \). Is this controllable at \(t_{0}=0?\)\begin{align*} M_{0}\left ( t\right ) & =\begin{pmatrix} 1\\ t \end{pmatrix} \\ M_{1}\left ( t\right ) & =-\begin{pmatrix} \cos t & t\\ e^{t} & 2 \end{pmatrix}\begin{pmatrix} 1\\ t \end{pmatrix} +\begin{pmatrix} 0\\ 1 \end{pmatrix} \\ & =\begin{pmatrix} -\cos t+t^{2}\\ -\left ( e^{t}+2t\right ) +1 \end{pmatrix} \end{align*}
Hence \[ M\left ( t\right ) =\begin{pmatrix} 1 & -\cos t+t^{2}\\ t & -\left ( e^{t}+2t\right ) +1 \end{pmatrix} \] Check the determinant. If it is not zero, then rank is \(2\) and hence controllable. Need to use \(t^{\ast }>0\) to find numerical solution for determinant to check if zero or not. Basically we need to check if there exist \(t^{\ast }>t\) where the above matrix is not singular.
Few words about analytic functions: Of we have smooth function \(f\left ( t\right ) \) and \(f\left ( t\right ) =0\) on some region \(\left [ t_{0},t_{1}\right ] \) with \(t_{0}\neq t\), then this means \(f\left ( t\right ) =0\) on all time \(t\) and not just in this region. This is because an analytical function can not have any place with its derivative does not exist (no sharp corners). Also, for analytic functions, we can expand them locally around a point using Taylor series. Now we make a bridge to LTI. We need more result about linear independence.
Lemma: Suppose \(F\left ( t\right ) \) is analytic on \(\left [ t_{0},t_{1}\right ] \), (this is the new addition for LTI, which we did not use for LTV), define \[ \Im \left ( t\right ) =\begin{pmatrix} F\left ( t\right ) & F^{\prime }\left ( t\right ) & \cdots & F^{\left ( n-1\right ) }\left ( t\right ) & \cdots \end{pmatrix} \] Notice, there are infinite many columns now. Unlike with LTV. Now the lemma says: \(F_{i}\left ( t\right ) \) are L.I. on \(\left [ t_{0},t_{1}\right ] \) iff rank \(\Im \left ( t\right ) =n\) for some \(t^{\ast }\in \left [ t_{0},t_{1}\right ] .\)
Proof: sufficiency: \(\Longleftarrow \). Assume rank \(\Im \left ( t\right ) =n\) for some \(t^{\ast }\in \left [ t_{0},t_{1}\right ] \) we need to show that \(F_{i}\left ( t\right ) \) are L.I. on \(\left [ t_{0},t_{1}\right ] \). By contradiction: Assume \(\Im \left ( t\right ) =n\) for some \(t^{\ast }\in \left [ t_{0},t_{1}\right ] \) but \(F_{i}\left ( t\right ) \) are L.D. on \(\left [ t_{0},t_{1}\right ] \). Hence there exist vector \(\vec{\alpha }\neq 0\) such that \(\vec{\alpha }F_{i}\left ( t\right ) =0\). Hence \(F\left ( t\right ) =0\) and it follows that \(F^{\prime }\left ( t\right ) =0\) etc.. for all columns of \(\Im \left ( t\right ) \). Hence rank \(\Im \left ( t\right ) \neq n\). QED.
necessity: \(\Longrightarrow \). Assume \(F_{i}\left ( t\right ) \) are L.I. on \(\left [ t_{0},t_{1}\right ] \) we need to show that there exist \(t^{\ast }\in \left [ t_{0},t_{1}\right ] \) such that rank \(\Im \left ( t^{\ast }\right ) =n.\) Proof by contradiction. Assume \(F_{i}\left ( t\right ) \) are L.I. on \(\left [ t_{0},t_{1}\right ] \,\ \)but no such \(t^{\ast }\) exist, so rank \(\Im \left ( t^{\ast }\right ) <n\) for all \(\left [ t_{0},t_{1}\right ] \). Pick any \(t\) in the range and expand \(F\left ( t\right ) \) around \(t^{\ast }\) using Taylor (since analytic) \begin{equation} F\left ( t\right ) ={\displaystyle \sum \limits _{k=0}^{\infty }} \frac{F^{\left ( k\right ) }\left ( t-t^{\ast }\right ) ^{k}}{k!} \tag{1} \end{equation} Since rank rank \(\Im \left ( t^{\ast }\right ) <n\), then there exist vector \(\vec{\alpha }\neq 0\) such that \(\vec{\alpha }\Im \left ( t^{\ast }\right ) =0\). Now multiply (1) by \(\vec{\alpha }^{T}\) we get \(\vec{\alpha }^{T}F\left ( t\right ) =0\) (Reader) on \(\left [ t^{\ast }-\varepsilon ,t^{\ast }+\varepsilon \right ] \). But since we assumed \(F\left ( t\right ) \) analytic, then \(F\left ( t\right ) =0\) everywhere. Which contradicts that \(F_{i}\left ( t\right ) \) are L.I. on \(\left [ t_{0},t_{1}\right ] \). QED.