13C (Mathematical Biology)

Setting

ϵ = 0

makes the inhibitor equation quasistatic:

0 = v_{xx} + u - v .

0 = - k^{2} v + u - v .

v (k, t) = \frac{u ( k , t )}{1 + k ^{2}} .

b

Since

f (0) = 0

and

f^{'} (0) = - r

f (u) = - r u + O (u^{2}) .

For a spatial Fourier mode of the perturbation, part a gives

\frac{\partial u}{\partial t} = [- D k^{2} - r + ρ (1 - \frac{1}{1 + k ^{2}})] u .

Thus

u (k, t) \propto e^{σ (k) t}

σ (k) = - D k^{2} - r + ρ \frac{k ^{2}}{1 + k ^{2}} .

Put

y = k^{2} \geq 0

. At the threshold of a Turing instability, the maximum of

σ (y) = - Dy - r + ρ \frac{y}{1 + y}

touches zero. Hence

σ (y_{c}) = 0, \frac{d σ}{d y} (y_{c}) = - D + \frac{ρ}{( 1 + y _{c} ) ^{2}} = 0.

ρ = D (1 + y_{c})^{2}

. Substitution in the first gives

0 = - D y_{c} - r + D y_{c} (1 + y_{c}) = D y_{c}^{2} - r .

Therefore

y_{c} = \frac{r}{D}, k_{c} = (\frac{r}{D})^{1/4},

and

ρ_{c} (r) = D (1 + \frac{r}{D})^{2} = (r + D)^{2} .

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