Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 11E (Mathematical Biology)

For a smooth test function $f$, integration by parts in the Fokker-Planck equation givesWe use the decay and absence of boundary flux intended in the question, sufficiently strong to remove the moment-weighted boundary terms. For $f=x_k$, only ${\partial }_{i}f={\delta }_{ik}$ survives, giving $d⟨x_k⟩/dt=⟨A_k⟩$. For $f=x_k{x}_l$, the first derivative is ${\delta }_{ik}{x}_l+{\delta }_{il}{x}_k$ and the second is ${\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk}$. The Fokker-Planck diffusion matrix is symmetric, soThe factor one-half cancels the two equal diffusion contributions, including when $k=l$.

Let $\xi =x-⟨x⟩$ and $C_{ij}=⟨{\xi }_i{\xi }_j⟩$. In the linear noise approximation, $A_i=a_{ik}{\xi }_k$ and $B_{ij}=b_{ij}$. At the stationary mean, the preceding moment identity gives ${\dot{C}}_{ij}=a_{ik}{C}_{kj}+a_{jk}{C}_{ik}+b_{ij}$. Hence the stationary covariance matrix solves the Continuous Lyapunov equationwhich is precisely the requested component formula because $C$ is symmetric.

The three jump vectors and their rates are $(1,0)$ at rate $\lambda$, $(-1,1)$ at rate $\gamma x_1$, and $(0,-1)$ at rate $\beta x_2$. Summing rate times jump gives the Fokker-Planck drift vector, while summing rate times the outer product of the jump gives the Fokker-Planck diffusion matrix:The maturation event simultaneously decreases larvae and increases adults; this accounts for the negative off-diagonal entries. Since $A$ is affine, its expectation is its value at the mean. Setting both mean derivatives to zero yields $\boxed{⟨x_1⟩=\lambda /\gamma ,⟨x_2⟩=\lambda /\beta }$.

At the stationary mean,The $(1,1)$ entry of the Continuous Lyapunov equation gives $-2\gamma C_{11}+2\lambda =0$, so $C_{11}=\lambda /\gamma$. Its $(1,2)$ entry gives $\gamma C_{11}-(\gamma +\beta )C_{12}-\lambda =0$, hence $C_{12}=0$. Finally $(2,2)$ gives $2\gamma C_{12}-2\beta C_{22}+2\lambda =0$. ThusBoth variances equal their means; the absence of stationary covariance does not mean that the maturation jumps are independent events in the two populations.