傅里叶变换求解 KdV 方程

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最新推荐文章于 2024-03-10 14:30:00 发布 · 4.6k 阅读

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本文深入探讨了KdV方程的历史起源及其数学解析过程，详细解释了瞬态波形下KdV方程的积分变换，以及通过傅里叶变换求解偏微分方程的方法，并附带Matlab仿真代码。

KdV方程的出处： Dr. D. J. Korteweg & Dr. G. de Vries (1895) XLI. On the change of form of long waves advancing in a rectangular canal, and on a new type of long stationary waves, The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, 39:240, 422-443, DOI: 10.1080/14786449508620739

@article{doi:10.1080/14786449508620739,
	author = { Dr.   D. J.   Korteweg  and  Dr.   G.   de Vries },
	title = {XLI. On the change of form of long waves advancing in a rectangular canal, and on a new type of long stationary waves},
	journal = {The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science},
	volume = {39},
	number = {240},
	pages = {422-443},
	year  = {1895},
	publisher = {Taylor & Francis},
	doi = {10.1080/14786449508620739},
	URL = {https://doi.org/10.1080/14786449508620739},
	eprint = {https://doi.org/10.1080/14786449508620739},
}

$\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + \frac{\partial^3 u}{\partial x^3} = 0,$

瞬态波形

当固定 $t$ 时, $\frac{\partial u}{\partial t} = 0$ . 得
$u\frac{\partial u}{\partial x} + \frac{\partial^3 u}{\partial x^3} = 0, \tag{1}$
对(1)关于 $x$ 积分，或者
$\frac{\partial }{\partial x}\left( \frac{1}{2}u^2+ \frac{\partial^2 u}{\partial x^2} \right)= 0,$