{"topic":{"id":14878,"title":"变分法","slug":"topic","userId":4668,"categoryId":4,"createdAt":"2025-07-04 16:58:49.702961","bumpedAt":"2025-07-04 18:32:13.220606","views":227,"postsCount":2,"highestPostNumber":2,"likeCount":2,"closed":false,"pinned":false,"pinnedGlobally":false,"pinnedAt":null,"pinnedUntil":null,"isBanner":false,"excerpt":"tag: 速成发一些适合速成的教学视频 https://www.bilibili.com/video/BV1yt4y1W7kt/","deletedAt":null,"isAnonymous":false,"isPrivate":false,"isMembersOnly":false,"category":{"id":4,"name":"谈笑风生","slug":"general","color":"25AAE2"},"author":{"id":4668,"username":"IO","name":null,"avatarUrl":null},"realAuthor":{"id":4668,"username":"IO","name":null,"avatarUrl":null},"tags":[]},"posts":[{"id":59709,"postNumber":1,"raw":"tag: 速成\n\n发一些适合速成的教学视频\n\nhttps://www.bilibili.com/video/BV1yt4y1W7kt/\n\n![ef84be2f763f70fa16ea3d33a4eed65b|690x322](/uploads/image/38/a1/38a1fef17c38f4460e279f9c811de28cf7f4c7111cdae139c15d23d9fd0d5aae.avif)","cooked":"

tag: 速成

发一些适合速成的教学视频

\n\n

$\"ef84be2f763f70fa16ea3d33a4eed65b\"$

\n","createdAt":"2025-07-04 16:58:49.796648","updatedAt":"2025-07-04 16:58:49.796648","replyToPostNumber":null,"replyCount":0,"likeCount":0,"wordCount":17,"liked":false,"bookmarked":false,"polls":[],"reactions":[],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"foldedAt":null,"isAnonymous":false,"isOwner":false,"user":{"id":4668,"username":"IO","name":null,"avatarUrl":null,"admin":true,"moderator":false,"signatureRaw":"","signatureCooked":null}},{"id":59712,"postNumber":2,"raw":"布置的练习题\n\n![image|690x348](/uploads/image/df/55/df5558a08ac5a93aad3b8aba087fb566d8803bfb4fd414cd32dc2b5ae5ae8cb2.avif)\n\nGemini:\n\n好的，我们来依次求解这四个变分问题。\n\n### 问题 (1)\n\n**泛函：**\nV(y) = ∫₀¹ [(y')² + 12xy] dx\n给定 y(0) = 0, y(1) = 1\n\n**1. 求驻值方程 (欧拉 - 拉格朗日方程):**\n令 F(x, y, y') = (y')² + 12xy。\n欧拉 - 拉格朗日方程为:\n∂F/∂y - d/dx(∂F/∂y') = 0\n\n计算偏导数:\n∂F/∂y = 12x\n∂F/∂y' = 2y'\n\n代入欧拉 - 拉格朗日方程:\n12x - d/dx(2y') = 0\n12x - 2y'' = 0\n**y'' = 6x** (这就是驻值方程)\n\n**2. 解微分方程:**\n对 y'' = 6x 积分一次:\ny' = ∫6x dx = 3x² + C₁\n\n再积分一次:\ny = ∫(3x² + C₁) dx = x³ + C₁x + C₂\n\n使用边界条件确定常数 C₁ 和 C₂:\n* y(0) = 0 => 0³ + C₁(0) + C₂ = 0 => C₂ = 0\n* y(1) = 1 => 1³ + C₁(1) + 0 = 1 => 1 + C₁ = 1 => C₁ = 0\n\n**解为：y(x) = x³**\n\n---\n\n### 问题 (2)\n\n**泛函：**\nV(y) = ∫₀^(π/2) [(y')² - y²] dx\n给定 y(0) = 0, y(π/2) = 1\n\n**1. 求驻值方程:**\n令 F(x, y, y') = (y')² - y²。\n∂F/∂y = -2y\n∂F/∂y' = 2y'\n\n代入欧拉 - 拉格朗日方程:\n-2y - d/dx(2y') = 0\n-2y - 2y'' = 0\n**y'' + y = 0** (这就是驻值方程)\n\n**2. 解微分方程:**\n这是一个二阶常系数齐次线性微分方程。其特征方程为 r² + 1 = 0，解得 r = ±i。\n通解为:\ny(x) = C₁cos(x) + C₂sin(x)\n\n使用边界条件:\n* y(0) = 0 => C₁cos(0) + C₂sin(0) = 0 => C₁ = 0\n* y(π/2) = 1 => 0 ⋅ cos(π/2) + C₂sin(π/2) = 1 => C₂ ⋅ 1 = 1 => C₂ = 1\n\n**解为：y(x) = sin(x)**\n\n---\n\n### 问题 (3)\n\n**泛函：**\nV(y) = ∫₀¹ √[1 + (y')²] dx\n给定 y(0) = 0, y(1) = 1\n(这个泛函代表从点 (0,0) 到点 (1,1) 的曲线长度)\n\n**1. 求驻值方程:**\n令 F(x, y, y') = √[1 + (y')²]。\n∂F/∂y = 0\n∂F/∂y' = y' / √[1 + (y')²]\n\n代入欧拉 - 拉格朗日方程:\n0 - d/dx (y' / √[1 + (y')²]) = 0\n这意味着:\ny' / √[1 + (y')²] = C₁ (常数)\n\n**2. 解微分方程:**\n从上式可以解出 y' 也是一个常数。\n(y')² = C₁²(1 + (y')²)\n(y')²(1 - C₁²) = C₁²\ny' = C₂ (另一个常数)\n\n对 y' = C₂ 积分:\ny(x) = C₂x + C₃\n\n使用边界条件:\n* y(0) = 0 => C₂(0) + C₃ = 0 => C₃ = 0\n* y(1) = 1 => C₂(1) + 0 = 1 => C₂ = 1\n\n**解为：y(x) = x**\n(正如预期，两点之间的最短距离是直线。)\n\n---\n\n### 问题 (4)\n\n**泛函：**\nV(y) = ∫₁² (xy'² - y) dx\n给定 y(1) = 0, y(2) = 1\n\n**1. 求驻值方程:**\n令 F(x, y, y') = xy'² - y。\n∂F/∂y = -1\n∂F/∂y' = 2xy'\n\n代入欧拉 - 拉格朗日方程:\n-1 - d/dx(2xy') = 0\n**d/dx(2xy') = -1** (这就是驻值方程)\n\n**2. 解微分方程:**\n对 d/dx(2xy') = -1 积分一次:\n2xy' = -x + C₁\ny' = -1/2 + C₁/(2x)\n\n再积分一次:\ny(x) = ∫[-1/2 + C₁/(2x)] dx = -x/2 + (C₁/2)ln(x) + C₂\n\n为了方便，令 A = C₁/2, B = C₂:\ny(x) = -x/2 + A ln(x) + B\n\n使用边界条件:\n* y(1) = 0 => -1/2 + A ln(1) + B = 0 => -1/2 + 0 + B = 0 => B = 1/2\n* y(2) = 1 => -2/2 + A ln(2) + 1/2 = 1 => -1 + A ln(2) + 1/2 = 1 => A ln(2) = 3/2 => A = 3 / (2 ln(2))\n\n**解为：y(x) = -x/2 + [3 / (2 ln 2)] ln(x) + 1/2**","cooked":"

布置的练习题

$\"image\"$

Gemini:

好的，我们来依次求解这四个变分问题。

问题 (1)

泛函：
\nV(y) = ∫₀¹ [(y')² + 12xy] dx
\n给定 y(0) = 0, y(1) = 1

1. 求驻值方程 (欧拉 - 拉格朗日方程):
\n令 F(x, y, y') = (y')² + 12xy。
\n欧拉 - 拉格朗日方程为:
\n∂F/∂y - d/dx(∂F/∂y') = 0

计算偏导数:
\n∂F/∂y = 12x
\n∂F/∂y' = 2y'

代入欧拉 - 拉格朗日方程:
\n12x - d/dx(2y') = 0
\n12x - 2y'' = 0
\ny'' = 6x (这就是驻值方程)

2. 解微分方程:
\n对 y'' = 6x 积分一次:
\ny' = ∫6x dx = 3x² + C₁

再积分一次:
\ny = ∫(3x² + C₁) dx = x³ + C₁x + C₂

使用边界条件确定常数 C₁ 和 C₂:

y(0) = 0 => 0³ + C₁(0) + C₂ = 0 => C₂ = 0
y(1) = 1 => 1³ + C₁(1) + 0 = 1 => 1 + C₁ = 1 => C₁ = 0

解为：y(x) = x³

问题 (2)

泛函：
\nV(y) = ∫₀^(π/2) [(y')² - y²] dx
\n给定 y(0) = 0, y(π/2) = 1

1. 求驻值方程:
\n令 F(x, y, y') = (y')² - y²。
\n∂F/∂y = -2y
\n∂F/∂y' = 2y'

代入欧拉 - 拉格朗日方程:
\n-2y - d/dx(2y') = 0
\n-2y - 2y'' = 0
\ny'' + y = 0 (这就是驻值方程)

2. 解微分方程:
\n这是一个二阶常系数齐次线性微分方程。其特征方程为 r² + 1 = 0，解得 r = ±i。
\n通解为:
\ny(x) = C₁cos(x) + C₂sin(x)

使用边界条件:

y(0) = 0 => C₁cos(0) + C₂sin(0) = 0 => C₁ = 0
y(π/2) = 1 => 0 ⋅ cos(π/2) + C₂sin(π/2) = 1 => C₂ ⋅ 1 = 1 => C₂ = 1

解为：y(x) = sin(x)

问题 (3)

泛函：
\nV(y) = ∫₀¹ √[1 + (y')²] dx
\n给定 y(0) = 0, y(1) = 1
\n(这个泛函代表从点 (0,0) 到点 (1,1) 的曲线长度)

1. 求驻值方程:
\n令 F(x, y, y') = √[1 + (y')²]。
\n∂F/∂y = 0
\n∂F/∂y' = y' / √[1 + (y')²]

代入欧拉 - 拉格朗日方程:
\n0 - d/dx (y' / √[1 + (y')²]) = 0
\n这意味着:
\ny' / √[1 + (y')²] = C₁ (常数)

2. 解微分方程:
\n从上式可以解出 y' 也是一个常数。
\n(y')² = C₁²(1 + (y')²)
\n(y')²(1 - C₁²) = C₁²
\ny' = C₂ (另一个常数)

对 y' = C₂ 积分:
\ny(x) = C₂x + C₃

使用边界条件:

y(0) = 0 => C₂(0) + C₃ = 0 => C₃ = 0
y(1) = 1 => C₂(1) + 0 = 1 => C₂ = 1

解为：y(x) = x
\n(正如预期，两点之间的最短距离是直线。)

问题 (4)

泛函：
\nV(y) = ∫₁² (xy'² - y) dx
\n给定 y(1) = 0, y(2) = 1

1. 求驻值方程:
\n令 F(x, y, y') = xy'² - y。
\n∂F/∂y = -1
\n∂F/∂y' = 2xy'

代入欧拉 - 拉格朗日方程:
\n-1 - d/dx(2xy') = 0
\nd/dx(2xy') = -1 (这就是驻值方程)

2. 解微分方程:
\n对 d/dx(2xy') = -1 积分一次:
\n2xy' = -x + C₁
\ny' = -1/2 + C₁/(2x)

再积分一次:
\ny(x) = ∫[-1/2 + C₁/(2x)] dx = -x/2 + (C₁/2)ln(x) + C₂

为了方便，令 A = C₁/2, B = C₂:
\ny(x) = -x/2 + A ln(x) + B

使用边界条件:

y(1) = 0 => -1/2 + A ln(1) + B = 0 => -1/2 + 0 + B = 0 => B = 1/2
y(2) = 1 => -2/2 + A ln(2) + 1/2 = 1 => -1 + A ln(2) + 1/2 = 1 => A ln(2) = 3/2 => A = 3 / (2 ln(2))

解为：y(x) = -x/2 + [3 / (2 ln 2)] ln(x) + 1/2

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