13C (Mathematical Biology)

Table of contents
- a 13C
  - Solution a
- b 13C
  - Solution b
- c 13C
  - Solution c
- d 13C
  - Solution d
- e 13C
  - Solution e
- f 13C
  - Solution f

J_{i} = A_{i} P - \frac{1}{2} \frac{\partial}{\partial x _{j}} (B_{ij} P) .

(1)

Then

- \frac{\partial J _{i}}{\partial x _{i}} = - \frac{\partial}{\partial x _{i}} (A_{i} P) + \frac{1}{2} \frac{\partial ^{2}}{\partial x _{i} \partial x _{j}} (B_{ij} P),

(2)

so the Fokker-Planck equation is exactly the continuity equation

\partial_{t} P + \nabla \cdot J = 0

b

 0  0

Solution

 0  0

Integrating the continuity equation over state space and applying the divergence theorem gives

\frac{d}{d t} \int P d V = - \int \nabla \cdot J d V = - \int_{\partial V} J \cdot d S .

(1)

Assume the probability density function and its derivatives decay sufficiently rapidly at infinity, or impose a zero-flux boundary condition. The boundary integral then vanishes, so

\int P d V

is constant. Since it is initially one, the distribution remains normalized for all time.

c

 0  0

Solution

 0  0

Assume again that all boundary terms vanish. Multiplying the Fokker-Planck equation by

x_{k}

and integrating by parts gives

\frac{d}{d t} ⟨ x_{k} ⟩ = - \int x_{k} \partial_{i} (A_{i} P) d V + \frac{1}{2} \int x_{k} \partial_{i} \partial_{j} (B_{ij} P) d V = \int δ_{ik} A_{i} P d V = ⟨ A_{k} ⟩ .

(1)

The diffusion term vanishes after the second integration by parts.

Similarly,

\partial_{i} (x_{k} x_{l}) = δ_{ik} x_{l} + δ_{i l} x_{k}

and

\partial_{i} \partial_{j} (x_{k} x_{l}) = δ_{ik} δ_{j l} + δ_{i l} δ_{jk} .

(2)

Because

B

is symmetric,

\frac{d}{d t} ⟨ x_{k} x_{l} ⟩ = ⟨ x_{l} A_{k} + x_{k} A_{l} + B_{k l} ⟩ .

(3)

These are the First two moments of a multivariate Fokker-Planck equation.

d

 0  0

Solution

 0  0

With

A_{i} = a_{ij} x_{j}

and constant

B_{ij} = b_{ij}

, the Fokker-Planck probability current becomes

J_{i} = a_{ij} x_{j} P - \frac{1}{2} b_{ij} \frac{\partial P}{\partial x _{j}} .

(1)

In the stationary state we impose the stronger zero-current condition

J_{i} = 0

, corresponding to detailed balance.

e

 0  0

Solution

 0  0

For

P (x) = Z^{- 1} exp (- \frac{1}{2} D_{ij} x_{i} x_{j})

(1)

with symmetric

D

\frac{\partial P}{\partial x _{j}} = - D_{jk} x_{k} P .

(2)

Substitution into the current from part (d) yields

J_{i} = (a_{ik} + \frac{1}{2} b_{ij} D_{jk}) x_{k} P .

(3)

It vanishes for all

x

exactly when

a + \frac{1}{2} b D = 0, D = - 2 b^{- 1} a .

(4)

Thus the stated multivariate normal distribution is a stationary solution whenever the zero-current assumption makes

D

symmetric positive definite. This is the Stationary Gaussian density for a linear Fokker-Planck equation.

f

 0  0

Solution

 0  0

Comparison with the multivariate normal density shows that

D

is its precision matrix. Since the mean is zero,

C_{ij} = ⟨ x_{i} x_{j} ⟩ = (D^{- 1})_{ij} .

(1)

Multiplying

a + \frac{1}{2} b D = 0

on the right by

C = D^{- 1}

gives

a C = - b /2

. Taking the transpose and using the symmetry of

b

and

C

gives

C a^{T} = - b /2

. Hence the stationary covariance obeys the Continuous Lyapunov equation

a C + C a^{T} + b = 0.

(2)

The same equation follows directly by setting the time derivative of the second-moment equation from part (c) to zero.

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Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 13C (Mathematical Biology)

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