DAS small changes

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2025-04-26 18:08:17 +02:00
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\begin{description}
\item[Leader-follower network] \marginnote{Leader-follower network}
Consider agents partitioned into $N_f$ followers and $N-N_f$ leaders.
Consider $N$ agents partitioned into $N_f$ followers and $N-N_f$ leaders.
The state vector can be partitioned as:
\[ \x = \begin{bmatrix} \x_f \\ \x_l \end{bmatrix} \]
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\]
where $\lap_f$ is the followers' Laplacian, $\lap_l$ the leaders', and $\lap_{fl}$ is the part in common.
Assume that leaders and followers run the same Laplacian-based distributed control law (i.e., an normal averaging system), the system can be formulated as:
Assume that leaders and followers run the same Laplacian-based distributed control law (i.e., a normal averaging system), the system can be formulated as:
\[
\begin{bmatrix} \dot{\x}_f(t) \\ \dot{\x}_l(t) \end{bmatrix} =
- \begin{bmatrix} \lap_f & \lap_{fl} \\ \lap_{fl}^T & \lap_l \end{bmatrix}
@ -30,7 +30,7 @@
\begin{bmatrix}
\dot{x}_1(t) \\ \dot{x}_2(t) \\ \dot{x}_3(t) \\ \dot{x}_4(t)
\end{bmatrix} =
\begin{bmatrix}
- \begin{bmatrix}
\begin{tabular}{ccc|c}
1 & -1 & 0 & 0 \\
-1 & 2 & -1 & 0 \\
@ -122,7 +122,7 @@
\x_f^T \lap_f \x_f &\geq 0 & & \forall \x_f
\end{aligned}
\]
\item The only case when $\x^T \lap \x = 0$ for $\x \neq 0$ is with $\x = \alpha\vec{1}$ for $\alpha \neq 0$. As $\forall \x_f: \bar{\x} \neq \alpha\vec{1}$, it holds that $\forall \x_f: \x_f^T \lap_f \x_f \neq 0$.
\item The only case when $\x^T \lap \x = 0$ for $\x \neq 0$ is with $\x = \alpha\vec{1}$ for $\alpha \neq 0$. As $\forall \x_f: \bar{\x} \neq \alpha\vec{1}$, it holds that $\forall \x_f \neq 0: \x_f^T \lap_f \x_f \neq 0$.
\end{enumerate}
Therefore, $\lap_f$ is positive definite as $\forall \x_f \neq 0: \x_f^T \lap_f \x_f > 0$.
\end{proof}
@ -182,7 +182,7 @@
Therefore, we have that:
\[
\begin{aligned}
\left( \sum_{j=1}^N a_{ij} \right) x_{E,i} &= \sum_{j=1}^N a_{ij} x_{E,j} & & \forall i \in \{ 1, \dots, N_f \} \\
\left( \sum_{k=1}^N a_{ik} \right) x_{E,i} &= \sum_{j=1}^N a_{ij} x_{E,j} & & \forall i \in \{ 1, \dots, N_f \} \\
x_{E,i} &= \sum_{j=1}^N \frac{a_{ij}}{\sum_{k=1}^N a_{ik}} x_{E,j} & & \forall i \in \{ 1, \dots, N_f \} \\
\end{aligned}
\]
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\end{description}
\begin{theorem}[Containment with non-static leaders non-equilibrium]
Naive containment with non-static leaders do not have an equilibrium.
Naive containment with non-static leaders does not have an equilibrium.
\begin{proof}
Ideally, the equilibria for followers' and leader's dynamics are:
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\end{split}
\]
By inspecting the value of the containment error $\vec{e}(t)$ when it reaches equilibrium we have that:
By inspecting the value of the containment error $\vec{e}(t)$ when it reaches equilibrium, we have that:
\[
\begin{split}
0 &= \dot{\vec{e}}(t) \\
@ -331,7 +331,7 @@
\begin{description}
\item[Containment with discrete-time] \marginnote{Containment with discrete-time}
Containment can be discretized using the forward-Eurler discretization. Its dynamics is defined as:
Containment can be discretized using the forward-Euler discretization. Its dynamics is defined as:
\[
\begin{aligned}
\dot{\x}_i(t) &= - \sum_{j \in \mathcal{N}_i} a_{ij} (x_i(t) - x_j(t)) & & \forall i \in \{1, \dots, N_f\} \\
@ -373,5 +373,5 @@
\[
\dot{\x}(t) = - \lap \otimes \matr{I}_d \x(t)
\]
where $\otimes$ is the Kronecker product.
where $\otimes$ is the Kronecker product (i.e., apply the same matrix across each dimension).
\end{description}