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

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version