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\begin{description}
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\item[Leader-follower network] \marginnote{Leader-follower network}
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Consider agents partitioned into $N_f$ followers and $N-N_f$ leaders.
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Consider $N$ agents partitioned into $N_f$ followers and $N-N_f$ leaders.
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The state vector can be partitioned as:
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\[ \x = \begin{bmatrix} \x_f \\ \x_l \end{bmatrix} \]
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\]
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where $\lap_f$ is the followers' Laplacian, $\lap_l$ the leaders', and $\lap_{fl}$ is the part in common.
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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:
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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:
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\[
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\begin{bmatrix} \dot{\x}_f(t) \\ \dot{\x}_l(t) \end{bmatrix} =
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- \begin{bmatrix} \lap_f & \lap_{fl} \\ \lap_{fl}^T & \lap_l \end{bmatrix}
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@ -30,7 +30,7 @@
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\begin{bmatrix}
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\dot{x}_1(t) \\ \dot{x}_2(t) \\ \dot{x}_3(t) \\ \dot{x}_4(t)
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\end{bmatrix} =
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\begin{bmatrix}
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- \begin{bmatrix}
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\begin{tabular}{ccc|c}
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1 & -1 & 0 & 0 \\
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-1 & 2 & -1 & 0 \\
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\x_f^T \lap_f \x_f &\geq 0 & & \forall \x_f
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\end{aligned}
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\]
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\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 $\forall \x_f: \x_f^T \lap_f \x_f \neq 0$.
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\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 $\forall \x_f \neq 0: \x_f^T \lap_f \x_f \neq 0$.
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\end{enumerate}
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Therefore, $\lap_f$ is positive definite as $\forall \x_f \neq 0: \x_f^T \lap_f \x_f > 0$.
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\end{proof}
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Therefore, we have that:
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\[
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\begin{aligned}
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\left( \sum_{j=1}^N a_{ij} \right) x_{E,i} &= \sum_{j=1}^N a_{ij} x_{E,j} & & \forall i \in \{ 1, \dots, N_f \} \\
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\left( \sum_{k=1}^N a_{ik} \right) x_{E,i} &= \sum_{j=1}^N a_{ij} x_{E,j} & & \forall i \in \{ 1, \dots, N_f \} \\
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x_{E,i} &= \sum_{j=1}^N \frac{a_{ij}}{\sum_{k=1}^N a_{ik}} x_{E,j} & & \forall i \in \{ 1, \dots, N_f \} \\
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\end{aligned}
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\]
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\end{description}
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\begin{theorem}[Containment with non-static leaders non-equilibrium]
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Naive containment with non-static leaders do not have an equilibrium.
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Naive containment with non-static leaders does not have an equilibrium.
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\begin{proof}
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Ideally, the equilibria for followers' and leader's dynamics are:
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@ -235,7 +235,7 @@
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\end{split}
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\]
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By inspecting the value of the containment error $\vec{e}(t)$ when it reaches equilibrium we have that:
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By inspecting the value of the containment error $\vec{e}(t)$ when it reaches equilibrium, we have that:
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\[
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\begin{split}
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0 &= \dot{\vec{e}}(t) \\
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\begin{description}
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\item[Containment with discrete-time] \marginnote{Containment with discrete-time}
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Containment can be discretized using the forward-Eurler discretization. Its dynamics is defined as:
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Containment can be discretized using the forward-Euler discretization. Its dynamics is defined as:
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\[
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\begin{aligned}
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\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\} \\
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\[
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\dot{\x}(t) = - \lap \otimes \matr{I}_d \x(t)
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\]
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where $\otimes$ is the Kronecker product.
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where $\otimes$ is the Kronecker product (i.e., apply the same matrix across each dimension).
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\end{description}
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