Heat distribution in a metal plate, using Fourier's method
One notices that the Fourier series expansion of our function in example 1 looks much less simple than the formula ƒ(x) = x, and so it is not immediately apparent why one would need this Fourier series. While there are many applications, we cite Fourier's motivation of solving the heat equation. For example, consider a metal plate in the shape of a square whose side measures π meters, with coordinates (x, y) ∈ [0, π] × [0, π]. If there is no heat source within the plate, and if three of the four sides are held at 0 degrees Celsius, while the fourth side, given by y = π, is maintained at the temperature gradient T(x, π) = x degrees Celsius, for x in (0, π), then one can show that the stationary heat distribution (or the heat distribution after a long period of time has elapsed) is given by
Here, sinh is the hyperbolic sine function. This solution of the heat equation is obtained by multiplying each term of Eq.1 by sinh(ny)/sinh(nπ). While our example function f(x) seems to have a needlessly complicated Fourier series, the heat distribution T(x, y) is nontrivial. The function T cannot be written as a closed-form expression. This method of solving the heat problem was made possible by Fourier's work.
<== попередня сторінка
|
наступна сторінка ==>
|
Переглядів: 518
Не знайшли потрібну інформацію? Скористайтесь пошуком google: