Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 41C / b

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- Solution b

Solution

Differentiating the truncated Fourier expansion gives

u_{x} = \sum_{m = - d}^{d} iπm u_{m} e^{iπm x} .

The

n

th Fourier coefficient of the product

c u_{x}

is the discrete convolution

iπ \sum_{m = - d}^{d} m c_{n - m} u_{m} .

Projecting the advection equation onto modes

∣ n ∣ \leq d

therefore gives

\frac{d u _{n}}{d t} = - iπ \sum_{m = - d}^{d} m c_{n - m} u_{m} .

Thus

\frac{d u}{d t} = iπ B u, B_{nm} = - m c_{n - m} .

Equivalently, if

C_{nm} = c_{n - m}, D = diag (- d, - d + 1, \dots, d),

then

B = - C D .

This is the Fourier spectral method for variable-coefficient advection.

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