Disease invasion on community networks with environmental pathogen movement

Abstract

The ability of disease to invade a community network that is connected by environmental pathogen movement is examined. Each community is modeled by a susceptible–infectious–recovered (SIR) framework that includes an environmental pathogen reservoir, and the communities are connected by pathogen movement on a strongly connected, weighted, directed graph. Disease invasibility is determined by the basic reproduction number $\mathcal{{R}}_0$ for the domain. The domain $\mathcal{{R}}_0$ is computed through a Laurent series expansion, with perturbation parameter corresponding to the ratio of the pathogen decay rate to the rate of water movement. When movement is fast relative to decay, $\mathcal{{R}}_0$ is determined by the product of two weighted averages of the community characteristics. The weights in these averages correspond to the network structure through the rooted spanning trees of the weighted, directed graph. Clustering of disease “hot spots” influences disease invasibility. In particular, clustering hot spots together according to a generalization of the group inverse of the Laplacian matrix facilitates disease invasion.

Appendices

Appendix A: Graph-theoretic background

For the convenience of the reader, we present some definitions and standard results from graph theory used throughout the paper. Further information can be found in graph theory textbooks such as West (2001). We include also a statement of the matrix tree theorem for weighted, directed graphs. A proof can be found in Moon (1970).

A directed graph (digraph) $\mathcal {G}=(V,E)$ consists of a set $V=\{1, 2, \ldots , n\}$ of vertices and a set $E=E(\mathcal {G})$ of directed arcs $(i,j)$ from vertex $i$ to vertex $j$. A directed graph $\mathcal {G}$ is weighted if each arc $(j,i)$ is assigned a positive weight $m_{ij}$; the weighted, directed graph is denoted as $(\mathcal {G}, M)$, with nonnegative weight matrix $M=[m_{ij}]$ and $m_{ij}>0$ if and only if $(j,i)\in E(\mathcal {G})$. For example, the weight matrix $M$ may correspond to the movement matrix in the community network, and thus the weighted directed graph $(\mathcal {G}, M)$ corresponds to the community network. A directed graph is strongly connected if for any ordered pair of vertices, there exists a directed path from one to the other. A weighted directed graph $(\mathcal {G}, M)$ is strongly connected if and only if the weight matrix $M$ is irreducible (Berman and Plemmons 1979).

A rooted spanning tree (in-tree) $\mathcal {T}$ is a subgraph of $\mathcal {G}$ on $n$ vertices such that $\mathcal {T}$ is connected with no cycles, and has a root vertex such that every directed path between a non-root vertex and the root is oriented towards the root. The weight of a rooted spanning tree is the product of all arcs in the rooted spanning tree. To illustrate, consider the star graph shown in Fig. 5, with arc weights $a$ from periphery to center, and $b$ from center to periphery. Figure 5b shows the single spanning in-tree rooted at the center. This tree possesses $n$ arcs with weight $a$, giving a tree weight of $a^{n}$. Each peripheral vertex roots a single spanning in-tree, shown in Fig. 5c. Trees rooted at the periphery have $n-1$ arcs with weight $a$ and a single arc with weight $b$, giving a tree weight of $a^{n-1}b$.

Let $(\mathcal {G}, M)$ be a weighted, directed graph. The Laplacian matrix $L$ of the graph is then defined as:

$$\begin{aligned} L = \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} \Sigma _{j \ne 1} m_{j1} &{} -m_{12} &{} \ldots &{} -m_{1n} \\ -m_{21} &{} \Sigma _{j \ne 2} m_{j2} &{} \ldots &{} -m_{2n} \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ -m_{n1} &{} -m_{n2} &{} \ldots &{} \Sigma _{j \ne n} m_{jn} \\ \end{array} \right) . \end{aligned}$$

(37)

Note that each of the columns of $L$ sum to zero, implying both that $L$ is singular, and that the cofactors of $L$ are the same within each column. The matrix tree theorem states that any cofactor corresponding to column $k$ of $L$ is equal to the sum of the weights of the spanning in-trees rooted at vertex $k$. For a proof see Moon (Section 5.5 and Theorem 5.5, Moon 1970).

Theorem 2

(Matrix Tree Theorem)

Let $(\mathcal {G}, M)$ be a weighted, directed graph, and let $L$ be the Laplacian matrix of $(\mathcal {G}, M)$. Let $c_{kk}$ denote the $(k,k)$ cofactor of $L$. Then the cofactors of $L$ are related to the rooted spanning trees of $\mathcal{G}$ by the following:

$$\begin{aligned} c_{kk} = \sum _{\mathcal{T} \in \mathbb {T}_k} \prod _{(j,i) \in E(\mathcal{T})} m_{ij}, \end{aligned}$$

(38)

where $\mathbb {T}_k$ is the set of all spanning in-trees rooted at vertex $k, E(\mathcal {T})$ is the arc set of rooted spanning in-tree $\mathcal {T}$, and $m_{ij}$ the weight of the arc from $j$ to $i$.

A direct consequence of the matrix tree theorem is a relationship between $\ker L$ and the rooted spanning trees of $\mathcal{G}$. This is pointed out, for example, in Lemma 2.1 of Guo et al. (2006). Consider the vector of cofactors $c = (c_{11}, \dots , c_{nn})^T$. Then $(Lc)_i = \det L = 0$ for $i=1,\dots ,n$, so $c$ belongs to $\ker L$. The dimension of the nullspace of $L$ is equal to the number of connected components of $\mathcal{G}$, so for strongly connected digraphs, $L$ has a one dimensional nullspace spanned by $c$. For strongly connected digraphs, there is at least one in-tree rooted at each vertex, and thus the entries of $c$ are all positive for digraphs with non-negative arc weights. Let $u_i = c_{ii} / \sum _{j=1}^n c_{jj}$, and let $u = (u_1, \dots , u_n)^T$. Then $u$ provides a basis vector for $\ker L$, with all positive entries and normalized so that $\sum _{i=1}^n u_i = 1$.

A weighted directed graph $(\mathcal {G}, M)$ is balanced if the net inflow is equal to the net outflow at each vertex, i.e. for each $i, \sum _{j\not =i} m_{ij} = \sum _{j\not =i}m_{ji}$. For balanced community networks, both the row sums and column sums of $L$ are equal to zero, giving that all cofactors of $L$ are equal. This in turn implies that all entries of $u$ are equal to $1/n$ for balanced networks.

Appendix B: $X_0$ and the group inverse

In the equal pathogen decay case, $G = (L+\varepsilon Id)$. The resolvent expansion for matrices of this form is considered by Rothblum (1981), who shows that the zeroth order term in the series is given by the Drazin inverse of $L$. For matrices of index 1, the Drazin inverse is equal to the group inverse (e.g. Ben-Israel and Greville 2003, Chapter 4). As pointed out by Schweitzer and Stewart (1993), $X_0$ in the general case is a generalization of the Drazin (here, group) inverse. It is interesting to consider the relationship between $X_0$ in Proposition 2 and the group inverse in more detail.

The group inverse of a square matrix $M$ of index 1 is the unique matrix $M^\#$ that satisfies the following three conditions (Ben-Israel and Greville 2003, Section 4.4):

$$\begin{aligned} MM^\#M&= M, \end{aligned}$$

(39)

$$\begin{aligned} M^\#MM^\#&= M^\#, \end{aligned}$$

(40)

$$\begin{aligned} MM^\#&= M^\#M. \end{aligned}$$

(41)

The matrix $X_0$ in Proposition 2 satisfies conditions (39) and (40) with $M=L$, but in general not (41). In the language of generalized inverses, $X_0$ is said to be a “$\lbrace 1,2 \rbrace$-inverse” of $L$ (Ben-Israel and Greville 2003).

We first show that $X_0$ acts as a right inverse of $L$ on $\mathop {\mathrm {range}}L$.

Lemma 5

Let $X_0$ be given as in Proposition 2. Then

$$\begin{aligned} LX_0x = \left\{ \begin{array}{l@{\quad }l} x, &{}\quad x \in \mathop {\mathrm {range}}L \\ 0, &{}\quad x \in R_{1,0}. \end{array} \right. \end{aligned}$$

Proof

Let $x \in \mathop {\mathrm {range}}L$. As $L: N_{1,0} \rightarrow \mathop {\mathrm {range}}L$ a bijection (Theorem 1(c)), there exists $y \in N_{1,0}$ such that $Ly = x$. Then

$$\begin{aligned} L X_0 x&= L X_0 Ly \nonumber \\&= Ly \nonumber \\&= x. \end{aligned}$$

(42)

Finally, note that $X_0$ sends $R_{1,0}$ to 0 (Theorem 1(e)), and thus $LX_0x = 0$ for all $x \in R_{1,0}$. $\square$

We can now show that $X_0$ is a $\lbrace 1,2 \rbrace$-inverse of $L$, and determine where $X_0$ and the group inverse coincide.

Proposition 3

Let $X_0$ be given as in Proposition 2. Then $X_0$ is a $\lbrace 1,2 \rbrace$-inverse of $L$ (i.e. satisfies conditions (39) and (40)), and commutes with $L$ on $N_{1,0} \cap \mathop {\mathrm {range}}L$.

Proof

For any $x \in \mathbb {R}^n$, take $x = y+z$, where $y \in N_{1,0}$ and $z \in \ker L$. From Proposition 2,

$$\begin{aligned} X_0 L x&= X_0 L (y+z) \nonumber \\&= X_0 L y \nonumber \\&= y. \end{aligned}$$

(43)

Multiplying on the left by $L$ then gives

$$\begin{aligned} L X_0 L x&= Ly \nonumber \\&= L(y+z) \nonumber \\&= Lx, \end{aligned}$$

(44)

and thus $X_0$ satisfies (39).

Similarly, for any $x \in \mathbb {R}^n$ take $x = \tilde{y} + \tilde{z}$, where $\tilde{y} \in \mathop {\mathrm {range}}L$ and $\tilde{z} \in R_{1,0}$. Using Lemma 5,

$$\begin{aligned} X_0 L X_0 x&= X_0 L X_0 (\tilde{y}+\tilde{z}) \nonumber \\&= X_0 \tilde{y} \nonumber \\&= X_0 (\tilde{y} + \tilde{z}) \nonumber \\&= X_0 x, \end{aligned}$$

(45)

so $X_0$ satisfies (40).

Finally, consider $x \in N_{1,0} \cap \mathop {\mathrm {range}}L$. As $x \in \mathop {\mathrm {range}}L$, Lemma 5 gives $x = L X_0 x$. But $x \in N_{1,0}$ as well, so property (e) in Theorem 1 gives $x = X_0 L x$, and thus $X_0$ and $L$ commute on $N_{1,0} \cap \mathop {\mathrm {range}}L$. $\square$

Proposition 3 shows that $X_0$ and the group inverse coincide on $N_{1,0} \cap \mathop {\mathrm {range}}L$. Of course, this intersection may be trivial. In the case where $D=Id$ (i.e. equal pathogen decay rates), $N_{1,0} = \mathop {\mathrm {range}}L$, and $X_0 = (Id-X_{-1}D)(L-X_{-1}D)^{-1}$ is equal to the group inverse of $L$.