Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 5 / iii

Table of contents
- Solution iii

Solution

The convolution theorem turns the product

e^{- k^{2} t} θ (k, 0)

into a spatial convolution. Its kernel is

G (z, t) = \frac{1}{2 π} \int_{R} e^{- t k^{2} + ik z} d k, t > 0.

Complete the square:

- t k^{2} + ik z = - t (k - \frac{i z}{2 t})^{2} - \frac{z ^{2}}{4 t} .

After rescaling the variable, the supplied translated Gaussian integral gives

G (z, t) = \frac{1}{4 π t} e^{- z^{2} / (4 t)}, θ (x, t) = \frac{1}{4 π t} \int_{R} θ (y, 0) e^{- (x - y)^{2} / (4 t)} d y .

This is the heat-kernel solution for diffusivity one. Its kernel integrates to one, preserving the total temperature integral, and approaches a Dirac delta function as

t ↓ 0

. Its variance is

2 t

; the factor is appropriate to

θ_{t} = θ_{xx}

, rather than the alternative convention with diffusivity

1/2

.

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