1 Like
3 Dislike
kai龍(苗撚)
2021-11-16 09:23:12
kai龍(苗撚)
2021-11-16 09:31:49
kai龍(苗撚)
2021-11-16 09:36:04
kai龍(苗撚)
2021-11-16 09:47:52
kai龍(苗撚)
2021-11-16 09:55:14
kai龍(苗撚)
2021-11-16 10:55:21
kai龍(苗撚)
2021-11-16 10:56:42
手一黏便緊(UTC+9
2021-11-16 12:16:19
kai龍(苗撚)
2021-11-16 12:34:44
孟加拉虎王
2021-11-16 13:36:49
kai龍(苗撚)
2021-11-16 13:38:34
method of characteristics喎
你目標係solve左條integral curve先
kai龍(苗撚)
2021-11-16 14:03:53
條 integral curve (x(t), y(t)) 會satisfy
x'(t) = x(t) - y(t)
y'(t) = 0
而且 u(x(t), y(t)) 會satisfies
(d/dt) u(x(t), y(t)) = 3u(x(t), y(t)) + 2[x(t)]^2
解哂三條ODE就ok
手一黏便緊(UTC+9
2021-11-16 14:23:19
唔太關事既野:
1. 用鉛筆
2. 你點由y’=-y/x落下一步的
kai龍(苗撚)
2021-11-16 14:26:09
btw in case你唔知堆野點黎...
Given
(x - y)u_x - 3u - 2x^2 = 0
我地可以變做
∇u.(x - y, 0) - 3u - 2x^2 = 0
假設有條curve a(t) = (x(t), y(t))
咁 (d/dt) (u(a(t)) = ∇u.a'(t)
係咪好似第一個term?
所以我地可以set
a'(t) = (x - y, 0)
於是就有我上講果堆野