Stability Analysis
Here, we describe several ways of computing threshold conditions for this basic malaria model:
\[ \begin{array}{rl} \frac{dX}{dt} = F_X(Y) &= bfq\frac{Y}{H}(H-X) - r X \\ \frac{dY}{dt} = F_Y(X) &= cfq \frac{X}{H} (M-Y) - g Y \end{array} \]
Steady States
We note that the system has two steady states. The disease-free steady state is at \(X=Y=0.\)
To compute the endemic equilibrium, let \(x = X/H\) and \(y=Y/M.\) At the steady state,
\[y = \frac{cfqx}{g + cfqx}\]
and substituting this back, we get
\[\frac{dx}{dt} = bcf^2 q^2 \frac{x}{g+cfqx}(1-x) - rx = 0\]
We let
\[R_0 = bc \frac{M}{H} \frac{f^2 q^2}{gr}\]
A recurring term is the expected number of times a mosquito would get infected in its lifetime, assuming humans are perfectly infectious: \[s=cfq/g\]
Since we’re not at \(x=0\), we can divide by \(x,\) rewrite and simplify
\[R_0 (1-x) = 1+s\] Collecting and solving for \(x\), we get
\[ x = \frac{R_0-1}{R_0 + s} \] substituting this back, we get
\[ y = \frac{R_0-1}{R_0} \frac{s}{1+s} \]
Stability Analysis
To evaluate the stability of these equilbria, we compute the matrix of partial derivatives:
\[ R = \left[ \begin{array}{rl} \frac{\partial F_X}{dX} & \frac{\partial F_X}{dY} \\ \frac{\partial F_Y}{dX} & \frac{\partial F_Y}{dY} \end{array} \right] = \left[ \begin{array}{rl} - bfq\frac{Y}{H} - r& bfq \frac{H-X}{H} \\ cfq\frac{M-Y}{H} & - cfq \frac{X}{H} - g \end{array} \right] \]
Effectively, we are looking at the stability of the linearized system at these steady states.
If we evaluate \(R\) at the disease-free equilibrium, \(X_0=Y_0=0.\)
\[ \left. R \right|_{X=Y=0} = \left[ \begin{array}{rl} - r& bfq \\ cfq\frac{M}{H} & - g \end{array} \right] \] The eigenvalues of a matrix are negative if and only if the trace is negative and the determinant is positive. Since the trace of \(R\) is negative (using mathematical notation, \(\mbox{Tr}\left(R\right) = - r - g\)), the determinant must be positive. The disease free state is stable (meaning the parasite will not increase) only if: \[ \mbox{det}\left(R\right) = rg - bc f^2 q^2 \frac{M}{H} > 0 \] It follows that the disease free state is unstable (and the parasite will increase) only if:
\[ R_0 = bc \frac{f^2 q^2}{rg} \frac{M}{H} > 1. \]
Eigenvalues & Eigenvectors
We can find the eigenvalues (vectors \(E\) and scalars \(\ell\) that satisfy \(R \cdot x = \ell x\)), by computing
\[ \mbox{det}\left[ \begin{array}{rl} - r-\ell& bfq \\ cfq\frac{M}{H} & - g -\ell \end{array} \right] = \ell^2 + \ell \; \mbox{Tr}\left(R\right) + \mbox{det}\left(R\right) = 0 \] or
\[ \ell = \frac{- \mbox{Tr}\left(R\right) \pm \sqrt{\mbox{Tr}\left(R\right)^2 - 4 \; \mbox{det}\left(R\right)} }{2} \]
If we let \(E_1\) be the eigenvector associated with the larger eignvalue (\(\ell_1\)) and \(E_2\) the other one (associated with eigenvalue \(\ell_2\)), then we can write any initial vector \(X_0, Y_0\) as a linear combination:
\[\left[\begin{array}{r} X_0 \\ Y_0 \end{array} \right] = \xi_1 E_1 + \xi_2 E_2 \]
Then
\[e^{Rt} = \xi_1 \ell_1^t E_1 + \xi_2 \ell_2^t E_2\]