Lie Derivative interpretation for my system

Question

I have a system of the form:

$\dot{S}=f(S,u)=f_0(S)+\sum_{i=1}^Lf_i(S)u_i$

$y=h(S)$

In my case $L=2$ but I don't think it matters for the question I want to ask.

The definition of Lie Derivative of a field $\Lambda$ along the vector fields ${v_{i_1},...,v_{i_{k+1}}}$ is:

$L^0\Lambda = \Lambda,\;\;L_{v_{i_1},...,v_{i_{k+1}}}\Lambda=\nabla_S(L_{v_{i_1},...,v_{i_{k}}}\Lambda)\cdot v_{i_{k+1}}$

At some point in the paper which I am reading it says:

we remark that the Lie Derivatives $L_{f_{i_1},...,f_{i_k}}h$ , of the output function $h$ along vector fields $f_i$, quantify the impact of changes in the control input $u_i$ on the output function $h$.

I don't understand how you can say that and what it really means.

Thanks a lot for the help.

Answer 1

I write an answer becouse a comment would be inappropriate. Let $\mathcal S$ be the state space and let $h:\mathcal S\to\mathcal Y$, with $\mathcal Y\subset\mathcal S$, the output map. Let $g:\mathcal S\to\mathcal S$ be a vector field on $\mathcal S$. The Lie derivative of $h$ with respect to $g$, as you defined it, reads $$ L_g h(S) = \langle \dfrac{\partial h}{\partial S}(S)\,,\, g(S)\rangle $$ (with $\langle.,.\rangle$ I denoted an inner product on $\mathcal S$). Hence, as the definition says, it is nothing but the projection of the gradient of $h(S)$ on $g(S)$, or equivalently, it is the projection of $g(S)$ on $\nabla_Sh(S)$.

Now, take $y(t)=h(S(t))$. Differentiating $y$ yields (I will omit the time dependency) \begin{align*} \dot y &= \dfrac{\partial h}{\partial S}(S) \dot S = \dfrac{\partial h}{\partial S}(S)f(S,u) \\&= \dfrac{\partial h}{\partial S}(S)\left(f_0(S) + \sum_{i=1}^Lf_i(S)u_i\right) \end{align*} This tells you that, once fixed a trajectory and a control law $u(t)$, computing the derivative of the output at time $t$ is the same as evaluating $L_{f(S,u)} h(S)$ at the point $(S(t),u(t))$. Moreover this also says that the $i$-th input $u_i$ affects the first derivative of the output (or in rough terms, the i-th input affects the output with a dynamic relation). Assume now that you wish to quantify "how much", in a given instant $t$, the i-th input $u_i(t)$ affects $\dot y(t)$. you can look at the quantity $$ \dfrac{\partial \dot y}{\partial u_i}(t) = \langle \nabla_Sh(S(t))\,,\,f_i(S(t))\rangle $$ Thus the answer is equivalent to look at $$ L_{f_i(S)}h(S) $$ at the point $S(t)$. For instance, if $f_i(S(t))$ and $\nabla_Sh(S(t))$ are orthogonal at a certain $t$, then your i-th input is not affecting the output in that time instant.