高等微积分（2） 第7次作业

Chasse_neige

1

设 \(f(x, y) = (2 + \sin x) \cdot \sin y\)，令 \(D = \{(x, y)|0 < x, y < 2\pi\}\)。

(1) 求出 \(f\) 在 \(D\) 上的所有临界点。 $$ \frac{\partial}{\partial x} f = \cos x \sin y = 0 $$

\[ x = \frac{\pi}{2}, \frac{3 \pi}{2} \]

或 $$ y = \pi $$

\[ \frac{\partial}{\partial y} f = (2 + \sin x) \cos y = 0 \]

\[ y = \frac{\pi}{2} ,\frac{3 \pi}{2} \]

所以一共有四个临界点 $$ (x,y) = (\frac{\pi}{2}, \frac{\pi}{2}), (\frac{3 \pi}{2}, \frac{\pi}{2}), (\frac{\pi}{2}, \frac{3 \pi}{2}), (\frac{3 \pi}{2}, \frac{3 \pi}{2}) $$ (2) 判断上述每个临界点是否为 \(f\) 的极值点。如果是的话，请指出它是极大值点还是极小值点。 $$ \begin{aligned} \frac{\partial^{2}}{\partial x^{2}} f &= - \sin x \sin y \\ \frac{\partial^{2}}{\partial x \partial y} f &= \frac{\partial^{2}}{\partial y \partial x} f = \cos x \cos y \\ \frac{\partial^{2}}{\partial y^{2}} f &= - (2 + \sin x) \sin y \end{aligned} $$ 对于 \((x, y) = (\frac{\pi}{2}, \frac{\pi}{2})\) $$ \begin{aligned} \frac{\partial^{2}}{\partial x^{2}} f &= -1 < 0 \\ \frac{\partial^{2}}{\partial x \partial y} f &= 0 \\ \frac{\partial^{2}}{\partial y^{2}} f &= -3 < 0 \end{aligned} $$ 该点的 \(H_{f}\) 负定，所以该点为极小值点

对于 \((x, y) = (\frac{3 \pi}{2}, \frac{\pi}{2})\) $$ \begin{aligned} \frac{\partial^{2}}{\partial x^{2}} f &= 1 > 0 \\ \frac{\partial^{2}}{\partial x \partial y} f &= 0 \\ \frac{\partial^{2}}{\partial y^{2}} f &= -1 < 0 \end{aligned} $$ 该点的 \(H_{f}\) 不定，所以该点并非极值点

对于 \((x, y) = (\frac{\pi}{2}, \frac{3 \pi}{2})\) $$ \begin{aligned} \frac{\partial^{2}}{\partial x^{2}} f &= 1 > 0 \\ \frac{\partial^{2}}{\partial x \partial y} f &= 0 \\ \frac{\partial^{2}}{\partial y^{2}} f &= 3 > 0 \end{aligned} $$ 该点的 \(H_{f}\) 正定，所以该点为极大值点

对于 \((x, y) = (\frac{3 \pi}{2}, \frac{3 \pi}{2})\) $$ \begin{aligned} \frac{\partial^{2}}{\partial x^{2}} f &= -1 < 0 \\ \frac{\partial^{2}}{\partial x \partial y} f &= 0 \\ \frac{\partial^{2}}{\partial y^{2}} f &= 1 > 0 \end{aligned} $$ 该点的 \(H_{f}\) 不定，所以该点并非极值点

2

设 \(f(x, y, z) = x + y + z + xyz\)，令 \(B = \{(x, y, z)|x^2 + y^2 + z^2 \leq 1\}\)。

(1) 证明：\(f\) 在 \(B\) 上有最大值。

证明

显然 \(B\) 有界，并且 \(f\) 为 \(B\) 上的连续函数

再证明 \(B^{\complement}\) 为开集 $$ \forall \vec{x} \in B^{\complement}, \exists \delta = \frac{|\vec{x}| - 1}{2}, s.t. B_{\delta} (\vec{x}) \subseteq B^{\complement} $$ 所以 \(B^{\complement}\) 为开集，即 \(B\) 为闭集，所以 \(B\) 是紧致的。

又因为 \(f\) 为 \(B\) 上的连续函数，所以\(f\) 在 \(B\) 上有最大值。

(2) 利用拉格朗日（Lagrange）乘子法求出 \(f\) 在 \(B\) 上的最大值。

证明 \(f\) 在 \(B\) 上的最大值在 \(\partial B\) 上

在 \(\overset{\circ}{B}\) 上，\(f\) 的临界点满足 $$ \begin{aligned} 1 + yz &= 0 \\ 1 + xz &= 0 \\ 1 + xy &= 0 \end{aligned} $$ 所以 \(xy = xz = yz = -1\) ，即 \((xyz)^{2} = -1\)，在实数范围内无解，所以 \(f\) 在 \(\overset{\circ}{B}\) 上没有极值点，所以最大值在 \(\partial B\) 上

利用拉格朗日乘子法 $$ L = x + y + z + xyz - \lambda (x^{2} + y^{2} + z^{2} - 1) $$ 所以极值点满足 $$ \begin{aligned} 1 + yz - 2 \lambda x &= 0 \\ 1 + xz - 2 \lambda y &= 0 \\ 1 + xy - 2 \lambda z &= 0 \\ x^{2} + y^{2} + z^{2} &= 1 \end{aligned} $$ 所以 $$ \begin{aligned} 3 + xy + xz + yz &= 2 \lambda (x + y + z) \\ \frac{5}{2} + \frac{1}{2} (x + y + z)^{2} - 2 \lambda (x + y + z) &= 0 \\ (x + y + z)^{2} - 4 \lambda (x + y + z) + 5 &= 0 \end{aligned} $$ 解得 $$ x + y + z = 2 \lambda \pm \sqrt{4 \lambda^{2} - 5} $$ 由上方程组的对称性，解应当满足 \(x = y = z\)，所以极值点有 $$ (x, y, z) = (\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}) \text{or} (-\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}) $$ 带入原函数 $$ \begin{aligned} f (\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}) &= \sqrt{3} + \frac{\sqrt{3}}{9} = \frac{10}{9} \sqrt{3} \\ f (-\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}) &= - \frac{10}{9} \sqrt{3} < \frac{10}{9} \sqrt{3} \end{aligned} $$ 所以 \(f\) 在 \(B\) 上的最大值为 \(\frac{10}{9} \sqrt{3}\)

3

设 \(x, y, z\) 满足两个约束条件 \(x + y + z = 1, x^2 + y^2 + z^2 = 1\)。求函数 \(f(x, y, z) = xyz\) 的最小值。 $$ L = xyz - \lambda_{1} (x + y + z - 1) - \lambda_{2} (x^{2} + y^{2} + z^{2} - 1) $$ 所以极值点满足 $$ \begin{aligned} yz - \lambda_{1} - 2 \lambda_{2} x &= 0 \\ xz - \lambda_{1} - 2 \lambda_{2} y &= 0 \\ xy - \lambda_{1} - 2 \lambda_{2} z &= 0 \\ x + y + z &= 1 \\ x^{2} + y^{2} + z^{2} &= 1 \end{aligned} $$ 所以 $$ \begin{aligned} xy + xz + yz &= 3 \lambda_{1} + 2 \lambda_{2} \\ (x + y + z)^{2} &= 1 + 6 \lambda_{1} + 4 \lambda_{2} = 1 \end{aligned} $$ 所以 $$ 3 \lambda_{1} + 2 \lambda_{2} = 0 $$ 然后进行暴力计算 $$ \begin{aligned} (xyz)^{2} &= (\lambda_{1} + 2 \lambda_{2} x) (\lambda_{1} + 2 \lambda_{2} y) (\lambda_{1} + 2 \lambda_{2} z) \\ &= \lambda_{1}^{3} + 8 \lambda_{2}^{3} xyz + 2 \lambda_{1}^{2} \lambda_{2} (x + y + z) + 4 \lambda_{1} \lambda_{2}^{2} (xy + xz + yz) \\ &= \lambda_{1}^{3} + 8 \lambda_{2}^{3} xyz + 2 \lambda_{1}^{2} \lambda_{2} = \lambda_{1}^{3} (1 - 27 xyz - 3) \end{aligned} $$

\[ \begin{aligned} xyz (x + y + z) &= (\lambda_{1} + 2 \lambda_{2} x) (\lambda_{1} + 2 \lambda_{2} y) + (\lambda_{1} + 2 \lambda_{2} x) (\lambda_{1} + 2 \lambda_{2} z) + (\lambda_{1} + 2 \lambda_{2} y) (\lambda_{1} + 2 \lambda_{2} z) \\\\ xyz &= 3 \lambda_{1}^{2} + 4 \lambda_{1} \lambda_{2} = -3 \lambda_{1}^{2} \end{aligned} \]

所以 $$ \begin{aligned} 9 \lambda_{1}^{4} + \lambda_{1}^{3} (2 - 81 \lambda_{1}^{2}) &= 0 \\ \lambda_{1} &= \frac{1 \pm 3}{18} \end{aligned} $$ 所以函数 \(f(x, y, z) = xyz\) 的最小值为 $$ xyz = -3 (\frac{1 + 3}{18})^{2} = - \frac{4}{27} $$

4

给定整数 \(n \geq 2\)，定义 \((n - 1)\) 维球面为 $$ S = {(x_1, ..., x_n) \in \mathbb{R}^n | \sum_{i=1}^{n} x_i^2 = 1} $$ 设 \(f : \mathbb{R}^n \to \mathbb{R}\) 是 \(C^1\) 光滑映射，\((\overline{x}_1, ..., \overline{x}_n) \in S\) 是 \(f\) 在 \(S\) 上的最大值点，即对任何 \((x_1, ..., x_n) \in S\)，有

\[ f(x_1, ..., x_n) \leq f(\overline{x}\_1, ..., \overline{x}\_n) \]

证明：\(f\) 在 \((\overline{x}_1, ..., \overline{x}_n)\) 处的梯度方向平行于向量 \((\overline{x}_1, ..., \overline{x}_n)\)，即存在实数 \(\lambda\)，使得

\[ \left( \frac{\partial f}{\partial x_1}, ..., \frac{\partial f}{\partial x_n} \right) |\_{(\overline{x}_1, ..., \overline{x}_n)} = \lambda (\overline{x}_1, ..., \overline{x}_n) \]

证明\(S\) 是紧致的，显然 \(S\) 有界，再说明 \(S\) 为闭集，即 \(S^{\complement}\)是开集，即证明 \(S^{\complement}\)中的每一个点均为其内点即可。 $$ S^{\complement} = {(x_1, ..., x_n) \in \mathbb{R}^n | \sum_{i=1}^{n} x_i^2 \neq 1} $$ 对于 \(\sum_{ i = 1}^{n} x_{i}^{2} < 1\) 的点 \(x_{0}\)，取 \(r_{0} = \frac{1 - d(0, x_{0})}{2}\) ，则 $$ \forall a \in B_{r_{0}} (x_{0}), d(0, a) \leq d(0, x_{0}) + r_{0} = 1 - r_{0} < 1 $$ 所以 \(x_{0}\) 为 \(S^{\complement}\) 内点

对于 \(\sum_{ i = 1}^{n} x_{i}^{2} > 1\) 的点 \(x_{0}\)，取 \(r_{0} = - \frac{1 - d(0, x_{0})}{2}\) ，则 $$ \forall a \in B_{r_{0}} (x_{0}), d(0, a) \geq d(0, x_{0}) - r_{0} = 1 - r_{0} > 1 $$ 所以 \(x_{0}\) 为 \(S^{\complement}\) 内点

所以 \(S\) 紧致，显然 \(g\) 对于坐标的偏导不全为0，所以对于题给条件，拉格朗日定理成立，即极值点\((\overline{x}_1, ..., \overline{x}_n) \in S\) 满足 $$ \begin{aligned} L (x_{1}, x_{2}, \cdots, x_{n}, \lambda) &= f (x_{1}, x_{2}, \cdots, x_{n}) - \lambda (\sum_{i = 1}^{n} x_{i}^{2} - 1) \\ \nabla L &= \vec{0} \end{aligned} $$ 所以 $$ \nabla f = 2 \lambda (\overline{x}_1, ..., \overline{x}_n)^{T} $$ 即 $$ \left( \frac{\partial f}{\partial x_1}, ..., \frac{\partial f}{\partial x_n} \right) |_{(\overline{x}_1, ..., \overline{x}_n)} = \lambda (\overline{x}_1, ..., \overline{x}_n) $$

5

设 \(n\) 元函数 \(f(x_1, ..., x_n), g(x_1, ..., x_n)\) 与一元函数 \(x_1(t), ..., x_n(t)\) 都是 \(C^2\) 光滑的定义函数 $$ h(t) = f(x_1(t), ..., x_n(t)) $$ (1) 求 \(h''(t)\)，请用 \(f(x_1, ..., x_n)\) 与 \(x_1(t), ..., x_n(t)\) 的高阶（偏）导函数表示。
$$ h' (t) = \sum_{i = 1}^{n} x_{i}' (t) \frac{\partial}{\partial x_{i}} f $$

\[ h(t)'' = \sum_{i = 1}^{n} x_{i}'' (t) \frac{\partial}{\partial x_{i}} f + \sum_{i = 1}^{n} \sum_{j = 1}^{n} x_{i}' (t) x_{j}' (t) \frac{\partial^{2}}{\partial x_{j} \partial x_{i}} f \]

(2) 令 \(p = (x_1(0), ..., x_n(0))\)。假设 \(p\) 是函数 \(f(x_1, ..., x_n)\) 在约束条件 \(g(x_1, ..., x_n) = 0\) 下的条件极值点。请叙述此情形下的拉格朗日乘子法。
$$ L = f(x_1, ..., x_n) - \lambda g(x_1, ..., x_n) $$ 条件极值点满足 $$ \begin{aligned} \left. \nabla f \right|_{x_{i} = x_{i} (0), 1 \leq i \leq n } &= \lambda \left. \nabla g \right|_{x_{i} = x_{i} (0), 1 \leq i \leq n } \\ g (x_{1} (0), \cdots, x_{n} (0)) &= 0 \end{aligned} $$ (3) 设 \(\lambda \in \mathbb{R}\) 满足 (2) 中所述拉格朗日乘子法的结论，定义 \(n\) 元函数 \(F\) 为 $$ F(x_1, ..., x_n) = f(x_1, ..., x_n) - \lambda g(x_1, ..., x_n) $$ 证明：如果对任何 \(t\)，都有 \(g(x_1(t), ..., x_n(t)) = 0\)，则

\[ h(0)'' = \left. \sum_{i=1}^n \sum_{j=1}^n \frac{\partial^2 F}{\partial x_i \partial x_j} \right|\_\mathbf{p} x_i(0)' x_j(0)' \]

证明：已知 $$ h'' (0) = \left. \sum_{i = 1}^{n} x_{i}'' (0) \frac{\partial}{\partial x_{i}} f \right|_{\mathbf{p}} + \left. \sum_{i = 1}^{n} \sum_{j = 1}^{n} x_{i}' (0) x_{j}' (0) \frac{\partial^{2}}{\partial x_{j} \partial x_{i}} f \right|_{\mathbf{p}} $$ 由于 $$ \left. \frac{\partial^2 F}{\partial x_i \partial x_j} \right|_\mathbf{p} = \left. \frac{\partial^2 f}{\partial x_i \partial x_j} \right|_\mathbf{p} - \lambda \left. \frac{\partial^2 g}{\partial x_i \partial x_j} \right|_\mathbf{p} $$ 所以 $$ h'' (0) = \left. \sum_{i = 1}^{n} x_{i}'' (0) \frac{\partial}{\partial x_{i}} f \right|_{\mathbf{p}} - \lambda \left. \sum_{i = 1}^{n} \sum_{j = 1}^{n} x_{i}' (0) x_{j}' (0) \frac{\partial^{2}}{\partial x_{j} \partial x_{i}} g \right|_{\mathbf{p}} + \left. \sum_{i = 1}^{n} \sum_{j = 1}^{n} x_{i}' (0) x_{j}' (0) \frac{\partial^{2}}{\partial x_{j} \partial x_{i}} F \right|_{\mathbf{p}} $$ 因为 \(n\) 元函数 \(f(x_1, ..., x_n), g(x_1, ..., x_n)\) 与一元函数 \(x_1(t), ..., x_n(t)\) 都是 \(C^2\) 光滑的定义函数 ，所以导数顺序可以交换

因为对任何 \(t\)，都有 \(g(x_1(t), ..., x_n(t)) = 0\)，所以 $$ g'' (t) = 0 $$ 所以 $$ g'' (t) = \sum_{i = 1}^{n} x_{i}'' (t) \frac{\partial}{\partial x_{i}} g + \sum_{i = 1}^{n} \sum_{j = 1}^{n} x_{i}' (t) x_{j}' (t) \frac{\partial^{2}}{\partial x_{j} \partial x_{i}} g = 0 $$

所以

\[ \lambda \left. \sum_{i = 1}^{n} \sum_{j = 1}^{n} x_{i}' (0) x_{j}' (0) \frac{\partial^{2}}{\partial x_{j} \partial x_{i}} g \right|\_{\mathbf{p}} = - \lambda \left. \sum_{i = 1}^{n} x_{i}'' (t) \frac{\partial}{\partial x_{i}} g \right|\_{\mathbf{p}} \]

因为

\[ \left. \nabla f \right|_{x_{i} = x_{i} (0), 1 \leq i \leq n } = \lambda \left. \nabla g \right|_{x_{i} = x_{i} (0), 1 \leq i \leq n } \]

所以

\[ - \lambda \left. \sum_{i = 1}^{n} x''_{i} (t) \frac{\partial}{\partial x_{i}} g \right|_{\mathbf{p}} = - \left. \sum_{i = 1}^{n} x''_{i} (t) \frac{\partial}{\partial x_{i}} f \right|_{\mathbf{p}} \]

所以

\[ \begin{aligned} h'' (0) &= \left. \sum_{i = 1}^{n} x''_{i} (0) \frac{\partial}{\partial x_{i}} f \right|_{\mathbf{p}} - \lambda \left. \sum_{i = 1}^{n} \sum_{j = 1}^{n} x'_{i} (0) x'_{j} (0) \frac{\partial^{2}}{\partial x_{j} \partial x_{i}} g \right|_{\mathbf{p}} + \left. \sum_{i = 1}^{n} \sum_{j = 1}^{n} x'_{i} (0) x'_{j} (0) \frac{\partial^{2}}{\partial x_{j} \partial x_{i}} F \right|_{\mathbf{p}} \\\\ &= \left. \sum_{i=1}^n \sum_{j=1}^n \frac{\partial^2 F}{\partial x_i \partial x_j} \right|_\mathbf{p} x_i'(0) x_j'(0) \end{aligned} \]

6

设 \(f, g \in C^1(\mathbb{R}^2, \mathbb{R})\)，令 \(D = \{(x, y)|g(x, y) \geq 0\}\)。设 \(g(x_0, y_0) = 0\) 且 \(g_x(x_0, y_0), g_y(x_0, y_0)\) 不全为零，且对任何 \((x, y) \in D\) 有 \(f(x_0, y_0) \leq f(x, y)\)。

证明：存在非负实数 \(\lambda\)，使得 $$ \begin{aligned} \begin{cases} f_x(x_0, y_0) - \lambda g_x(x_0, y_0) &= 0, \\ f_y(x_0, y_0) - \lambda g_y(x_0, y_0) &= 0. \end{cases} \end{aligned} $$ 证明 由于 \(g(x_0, y_0) = 0\) 且 \(\nabla g(x_0, y_0) \neq 0\) ，根据隐函数定理，存在邻域 \(U\) 和唯一 \(C^1\) 函数 \(y = \varphi(x)\)，使得 \(g(x, \varphi(x)) = 0\) 对所有 \(x \in U\) 成立。

\[ h'(x_0) = f_x(x_0, y_0) + f_y(x_0, y_0) \varphi'(x_0) = 0 \]

根据隐函数定理，\(\varphi'(x_0) = -\frac{g_x(x_0, y_0)}{g_y(x_0, y_0)}\) $$ f_x - f_y \cdot \frac{g_x}{g_y} = 0 $$

令 \(\lambda = \frac{f_{y} (x_{0}, y_{0})}{g_{y} (x_{0}, y_{0})}\)，则 $$ \begin{aligned} \begin{cases} f_x(x_0, y_0) - \lambda g_x(x_0, y_0) &= 0, \\ f_y(x_0, y_0) - \lambda g_y(x_0, y_0) &= 0. \end{cases} \end{aligned} $$ 由于 \((x_0, y_0)\) 是 \(f\) 在 \(D\) 上的极小值点，沿梯度 \(\nabla g\) 方向 $$ \nabla f \cdot \frac{\nabla g}{|\nabla g|} \geq 0. $$ 结合 \(\nabla f = \lambda \nabla g\) $$ \lambda |\nabla g|^2 \geq 0 \quad \implies \quad \lambda \geq 0. $$