6C (Mathematical Biology)

Solution

n_{t} = \frac{D}{r} \frac{\partial}{\partial r} (r n \frac{\partial n}{\partial r}) .

Therefore, assuming the population density and radial flux decay at infinity and are regular at the origin,

\frac{d}{d t} (2 π \int_{0}^{\infty} n r d r) = 2 π D [r n n_{r}]_{0}^{\infty} = 0.

Thus

N

is conserved; it is the total population obtained by integrating density over the plane. This is the global form of the equation's conservation law.

n (r, t) = (\frac{N}{D t})^{1/2} f (x), x = \frac{r}{( N D t ) ^{1/4}} .

Then

n_{t} = (\frac{N}{D})^{1/2} t^{- 3/2} (- \frac{1}{2} f - \frac{1}{4} x f^{'}),

while the radial diffusion term has the same prefactor and equals

D \nabla \cdot (n \nabla n) = (\frac{N}{D})^{1/2} t^{- 3/2} \frac{1}{x} \frac{d}{d x} (x f f^{'}) .

Equating them and multiplying by

x

gives

\frac{d}{d x} (x f f^{'} + \frac{1}{4} x^{2} f) = 0,

Regularity at

x = 0

makes the constant of integration zero. Wherever

f > 0

f^{'} = - \frac{x}{4}, f (x) = \frac{x _{0}^{2} - x ^{2}}{8}

for some

x_{0} > 0

. The nonnegative weak solution continues as

f = 0

once this parabola reaches zero:

f (x) = [\frac{x _{0}^{2} - x ^{2}}{8}]_{+} .

Indeed, conservation requires

1 = 2 π \int_{0}^{\infty} f (x) x d x = \frac{π x _{0}^{4}}{16},

x_{0} = (16/ π)^{1/4}

if desired. At time

t

the populated disk has radius

x_{0} (N D t)^{1/4}

and hence area

A (t) = π x_{0}^{2} (N D t)^{1/2} .

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