## 3 Preliminaries
### Graph Theory  
We use an undirected graph \(\mathcal{G}=(\mathcal{V},\mathcal{E})\) to describe the interactions among a group of n fixed-wing UAVs, where \(\mathcal{V}=\{1,\cdots,n\}\) is a node set, \(\mathcal{E}\subseteq\mathcal{V}\times\mathcal{V}\) is a set of ordered pairs of nodes, called edges. The adjacency matrix \(\mathbf{A}=[a_{ij}]_{n\times n}\) is defined as \(a_{ii}=0\) and \(a_{ij}=a_{ji}=1\) if there is an edge between node i and node j. Diagonal matrix \(\mathbf{D}=diag\{d_{11},d_{22},\cdots,d_{m}\}\) is the degree matrix of \(\mathcal{G}\), where the element of \(d_{ii}=\sum_{j=1}^{n}a_{ij}\). The Laplacian matrix of the graph \(\mathcal{G}\) is defined as \(\mathbf{L}=\mathbf{D}-\mathbf{A}\). Without loss of generality, the UAVs are denoted by nodes \(1,2,\cdots,n\), and the target vehicle is denoted by node \(0\). We use another diagonal matrix \(\mathbf{B}=diag\{b_{11},b_{22},\cdots,b_{nm}\}\) to describe the connection between the \(i\)-th UAV and the target vehicle, where \(b_{ii}>0\) if the state of the target is available to the \(i\)-th UAV and \(b_{ii}=0\) otherwise. The following assumption is made on digraph \(\mathcal{G}\). **Assumption 1**: _The undirected graph \(\mathcal{G}\) is connected and there exists at least one UAV that can obtains the state of the target._  
Based on the Assumption 1, there are two important lemmas as follows  
**Lemma 1**: _Let \(\mathbf{L}=(l_{ij})\in\mathbb{R}^{n\times n}\) be a Laplacian matrix of a connected undirected graph. Then the matrix_  
\[\tilde{\mathbf{L}}=\mathbf{L}+\mathbf{B}= \begin{bmatrix}l_{11}+b_{11}&\cdots&l_{1n}\\ \vdots&\ddots&\vdots\\ l_{n1}&\cdots&l_{nn}+b_{nn}\end{bmatrix} , \tag{1}\]  
_is positive definite, if there exists an \(i\) such that \(b_{ii}>0\)._  
Proof of Lemma 1.: For \(\forall\mathbf{z}\in\mathbb{R}^{n}\backslash\{\mathbf{0}\}\), one can obtain that \(\mathbf{z}^{T}\tilde{\mathbf{L}}\mathbf{z}=\mathbf{z}^{T}\mathbf{L}\mathbf{z}+\mathbf{z}^{T}\mathbf{B}\mathbf{z}\geq 0\), since \(\mathbf{L}\) and \(\mathbf{B}\) are both positive semi-definite. When \(\mathbf{z}^{T}\mathbf{L}\mathbf{z}=0\), one can obtain that \(\mathbf{z}\in span\{\mathbf{1}\}\backslash\{\mathbf{0}\}\), and \(\mathbf{z}^{T}\mathbf{B}\mathbf{z}>0\). Thus, we always have \(\mathbf{z}^{T}\tilde{\mathbf{L}}\mathbf{z}>0\) for \(\mathbf{z}\
eq 0\), which implies the conclusion. **Lemma 2**: _Let \(\tilde{\mathbf{L}}\in\mathbb{R}^{n\times n}\) be a positive definite and symmetric matrix. Then the matrix_  
\[\mathbf{M}= \begin{bmatrix}\mathbf{0}&I_{n}\\ -\beta_{1}\tilde{\mathbf{L}}&-\beta_{2}\tilde{\mathbf{L}}\end{bmatrix} \in\mathbb{R}^{2 n\times 2n}, \tag{2}\]  
_is negative definite, where \(\beta_{1},\beta_{2}>0\)._Proof of Lemma 2.: To prove that \(\mathbf{M}\prec\mathbf{0}\), we can solve the equation \(\det(\lambda\mathbf{I}_{2n}-\mathbf{M})=0\), where \(\det(\lambda\mathbf{I}_{2n}-\mathbf{M})\) is the characteristic polynomial of matrix \(\mathbf{M}\). Please note that  
\[ \begin{array}{ll}\det(\lambda\mathbf{I}_{2n}-\mathbf{M})&=\det\biggl{(}\left[\begin{array} []{cc}\lambda\mathbf{I}_{n}&-\mathbf{I}_{n}\\ \beta_{1}\tilde{\mathbf{L}}&\lambda\mathbf{I}_{n}+\beta_{2}\tilde{\mathbf{L}}\end{array} \right]\biggr{)}\\ &=\det\biggl{(}\left[\begin{array}{cc}0&-\mathbf{I}_{n}\\ \lambda^{2}\mathbf{I}_{n}+(\beta_{1}+\beta_{2}\lambda)\mathbf{L}&\lambda\mathbf{I}_{n}+ \beta_{2}\mathbf{L}\end{array}\right]\biggr{)}\\ &=\det\bigl{(}\lambda^{2}\mathbf{I}_{n}+(\beta_{1}+\beta_{2}\lambda)\tilde{\mathbf{L }}\bigr{)}\end{array} , \tag{3}\]  
Additionally note that the eigenvalues of \(\tilde{\mathbf{L}}\) satisfy the following property  
\[\det\bigl{(}\lambda\mathbf{I}_{2n}+\tilde{\mathbf{L}}\bigr{)}=\prod_{i=1}^{n}(\lambda +\rho_{i}) \tag{4}\]  
where \(\rho_{i}\) is the \(i\)-th eigenvalues of \(\tilde{\mathbf{L}}\).