Determination of Maximum Eigenvalue

Calculaton of $v^*$ is on firm ground when the matrices involved are finite. For the fungal problem, however, the matrices concerned are infinite dimensional and calculation of the maximum eigenvalue of $\textbf{H} (s)$ is not straightforward. Recalling that we have taken $M(s)$ to be the (scalar) moment generating function for the dispersal kernel $K(x)$, and taking $LP=5$ to be definite, we can write

$$\textbf{H} (s) = \left( \begin{array}{ccccccccccc} 0 & 0 & ... ...ots & \vdots & \vdots & \vdots & \ddots\\ \end{array}\right),$$ (17)

where $R$ is defined by (5).

In order to analyze the spectrum of $\textbf{H} (s)$, we consider the linear operator $\textbf{H}:l^1\to l^1$ defined by

$$(x_1,x_2,x_3,\cdots)\to \left(\rho \sum_{k=6}^{\infty} (k-5... ...s \right), \qquad \rho \stackrel{\mbox{\tiny def}}{=}R\, M(s),$$ (18)

where $l^1$ is the Banach space of all real sequences $\textbf{x}\stackrel{\mbox{\tiny def}}{=}(x_1,x_2,x_3,\cdots)$ such that $\sum \vert x_k\vert<\infty$. The matrix $\textbf{H} (s)$ is the representation of $\textbf{H}$ in the standard basis $\textbf{e}_k\stackrel{\mbox{\tiny def}}{=}\delta_{ik}$, where $\delta_{ik}$ is the Kronecker symbol. Our choice of the space $l^1$ is natural, since the total number of lesions and spores is always finite. The domain of $\textbf{H}$ is

$$\textrm{Dom } \textbf{H} = \left\{ \textbf{x}\in l^1 : \left\vert\sum_{k=6}^{\infty} (k-5) x_k \right\vert <\infty \right\},$$

which also reflects the previous natural assumption, since the summation in the definition in the domain is proportional to the total number of spores produced (large, but finite). The operator $\textbf{H}$ is unbounded, invertible, and its inverse is the left-shift operator $\textbf{H}^{-1}$ defined by

$$(x_1, \ x_2, \ x_3, \ \cdots)\to (x_2, \ x_3, \ x_4,\cdots).$$

Clearly, $\textbf{H}^{-1}$ is bounded, so $\textbf{H}$ is a closed operator, and we can use classical analysis of closed operators (see [2, §2.6 and §2.7]). For each $\lambda$, the operator $\textbf{H}_\lambda \stackrel{\mbox{\tiny def}}{=}\textbf{H}-\lambda I$ is defined by

$$(x_1, \ x_2, \ x_3, \ \cdots)\to \left(-\lambda x_1+ \rho \s... ... x_2, \ x_2 -\lambda x_3, \ x_3-\lambda x_4, \ \cdots \right).$$ (19)

The point spectrum of $\textbf{H}$ is the set $P \sigma (\textbf{H})$ of all points $\lambda$ for which $\textbf{H}_\lambda$ has no inverse. Each element of $P \sigma (\textbf{H})$ is the eigenvalue of $\textbf{H}$. For each eigenvalue $\lambda$, each $\textbf{x}\in \textrm{Dom } \textbf{H}$ such that $\textbf{H}_\lambda \textbf{x}=0$, is the corresponding eigenvector. Thus, equating the right hand side of (19) to zero gives

$$x_k=\lambda x_{k+1}, \quad k=1,2,3,\cdots.$$ (20)

For the first component, by using (20) and induction, we have

$$-\lambda x_1+ \rho \sum_{k=6}^{\infty} (k-5) \frac{x_1}{\lambda^{k-1}}=0.$$ (21)

From (20) it follows that any non-trivial solution of $\textbf{H}_\lambda \textbf{x}=0$ must satisfy $x_1\neq 0$, so (21) implies

$$-\lambda + \rho \sum_{k=6}^{\infty} (k-5) \frac{1}{\lambda^{... ...{1}{\lambda^3} \sum_{k=1}^{\infty} \frac{k}{\lambda^{k+1}}= 0.$$

Since we are looking for the largest eigenvalue, we confine ourselves to the case $\vert\lambda\vert>1$. By using differentiation of geometric series, we have

$$-\lambda +\rho\, \frac{1}{\lambda^3} \frac{1}{(\lambda-1)^2}=0.$$

Therefore, the eigenvalues of $\textbf{H}$ are the zeros of the polynomial

$$\lambda^6-2\lambda^5+\lambda^4-\rho=0,$$ (22)

which also satisfy $\vert\lambda\vert>1$. From (20) we see that the corresponding eigenvectors are

$$\textbf{x}=\left(1,\frac{1}{\lambda},\frac{1}{\lambda^2},\frac{1}{\lambda^3},\cdots \right).$$

It is obvious that $\textbf{x}\in l^1$, but since $\vert\lambda\vert>1$ also implies $\sum (k-5)/\vert\lambda^k\vert < \infty$, we also have $\textbf{x}\in \textrm{Dom } \textbf{H}$. Interestingly, the roots of the polynomial (22) can be computed exactly in terms of radicals (e.g. by Mathematica), and they all lie outside the unit circle. Inspecting all six roots, we see that the largest magnitude root of the polynomial (22) is

$$\lambda_1(s)= \frac{1}{3}+\frac{2^{1/3}}{3\left(2+27\sqrt{\r... ...+ \sqrt{108\sqrt{\rho}+729\rho}\right)^{1/3}}{3\cdot 2^{1/3}}.$$ (23)

Since in deriving this expression we have assumed that $\vert\lambda\vert>1$, we conclude that (23) gives the largest eigenvalue of the operator $\textbf{H} (s)$ from (17). Since $\rho = R M(s)$, expression (23), together with (10), allows for prediction of rates of invasion as a function of parameters describing the fecundity, dispersal, and infectiousness of P. infestans. In addition, since the maximum eigenvalue $\lambda _1(s)$ behaves like $O(\rho^{1/6})$ from (23), we also conclude that the predicted upper bound $v^*$ from (10) is stable in the sense that small changes of the parameters from Table 1 or entries in the matrix cause only small changes in $v^*$.