高等微积分 （2） 第5次作业

Chasse_neige

1. 计算偏导数

(1) 设 \(f \in C^1(\mathbb{R}, \mathbb{R})\), \(z = xy + xf\left( \dfrac{y}{x} \right)\). 求 \(x \dfrac{\partial z}{\partial x} + y \dfrac{\partial z}{\partial y}\) $$ \begin{aligned} \frac{\partial z}{\partial x} &= y + f\left(\frac{y}{x}\right) + x f'\left(\frac{y}{x}\right) \cdot \left(-\frac{y}{x^2}\right) = y + f\left(\frac{y}{x}\right) - \frac{y}{x}f'\left(\frac{y}{x}\right) \\ \frac{\partial z}{\partial y} &= x + x f'\left(\frac{y}{x}\right) \cdot \frac{1}{x} = x + f'\left(\frac{y}{x}\right) \\ x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} &= x\left(y + f\left(\frac{y}{x}\right) - \frac{y}{x}f'\left(\frac{y}{x}\right)\right) + y\left(x + f'\left(\frac{y}{x}\right)\right) \\ &= xy + xf\left(\frac{y}{x}\right) - yf'\left(\frac{y}{x}\right) + xy + yf'\left(\frac{y}{x}\right) \\ &= 2xy + xf\left(\frac{y}{x}\right) = 2xy + z - xy = xy + z \end{aligned} $$

(2) 设 \(f \in C^2(\mathbb{R}^2, \mathbb{R})\), \(z = f(x, xy)\). 求 \(\dfrac{\partial^2 z}{\partial y^2}\)

\[ \begin{aligned} \frac{\partial z}{\partial y} &= f_{2}' (x, xy) \cdot x \\\\ \frac{\partial^2 z}{\partial y^2} &= \frac{\partial}{\partial y} \left(x f_{2}'(x, xy)\right) = x \cdot \left(f_{22}'' (x, xy) \cdot x\right) = x^2 f_{22}'' (x, xy) \end{aligned} \]

(3) 设 \(f, g \in C^2(\mathbb{R}, \mathbb{R})\), \(z = xf\left( \dfrac{y}{x} \right) + yg\left( \dfrac{x}{y} \right)\). 求 \(\dfrac{\partial^2 z}{\partial x \partial y}\) $$ \begin{aligned} \frac{\partial z}{\partial y} &= x f'\left(\frac{y}{x}\right) \cdot \frac{1}{x} + g\left(\frac{x}{y}\right) + y g'\left(\frac{x}{y}\right) \cdot \left(-\frac{x}{y^2}\right) = f'\left(\frac{y}{x}\right) + g\left(\frac{x}{y}\right) - \frac{x}{y}g'\left(\frac{x}{y}\right) \\ \frac{\partial^2 z}{\partial x \partial y} &= \frac{\partial}{\partial x} \left( f'\left(\frac{y}{x}\right) + g\left(\frac{x}{y}\right) - \frac{x}{y}g'\left(\frac{x}{y}\right) \right) \\ &= f''\left(\frac{y}{x}\right)\left(-\frac{y}{x^2}\right) + g'\left(\frac{x}{y}\right)\frac{1}{y} - \left[ \frac{1}{y}g'\left(\frac{x}{y}\right) + \frac{x}{y} g''\left(\frac{x}{y}\right) \frac{1}{y} \right] \\ &= -\frac{y}{x^2}f''\left(\frac{y}{x}\right) - \frac{x}{y^2}g''\left(\frac{x}{y}\right) \end{aligned} $$

(4) 设 \(f \in C^1(\mathbb{R}, \mathbb{R})\), \(z = \dfrac{y}{f(x^2 - y^2)}\). 求 \(\dfrac{1}{x} \dfrac{\partial z}{\partial x} + \dfrac{1}{y} \dfrac{\partial z}{\partial y}\) $$ \begin{aligned} \frac{\partial z}{\partial x} &= y \cdot \left(-\frac{1}{f^2(x^2-y^2)}\right) \cdot f'(x^2-y^2) \cdot (2x) = -\frac{2xy f'(x^2-y^2)}{f^2(x^2-y^2)} \\ \frac{\partial z}{\partial y} &= \frac{1 \cdot f(x^2-y^2) - y \cdot f'(x^2-y^2) \cdot (-2y)}{f^2(x^2-y^2)} = \frac{f(x^2-y^2) + 2y^2 f'(x^2-y^2)}{f^2(x^2-y^2)} \\ \frac{1}{x} \frac{\partial z}{\partial x} + \frac{1}{y} \frac{\partial z}{\partial y} &= \frac{1}{x}\left(-\frac{2xy f'}{f^2}\right) + \frac{1}{y}\left(\frac{f + 2y^2 f'}{f^2}\right) \\ &= -\frac{2y f'}{f^2} + \frac{1}{yf} + \frac{2y f'}{f^2} \\ &= \frac{1}{yf(x^2-y^2)} = \frac{z}{y^2} \end{aligned} $$

(5) 设 \(f \in C^1(\mathbb{R}^3, \mathbb{R})\), \(u = f(x, xy, xyz)\). 求 \(\dfrac{\partial u}{\partial x}, \dfrac{\partial u}{\partial y}, \dfrac{\partial u}{\partial z}\) $$ \begin{aligned} \frac{\partial u}{\partial x} &= f_{1}'(x, xy, xyz) \cdot 1 + f_{2}'(x, xy, xyz) \cdot y + f_{3}'(x, xy, xyz) \cdot yz \\ \frac{\partial u}{\partial y} &= f_{1}'(x, xy, xyz) \cdot 0 + f_{2}'(x, xy, xyz) \cdot x + f_{3}'(x, xy, xyz) \cdot xz = x f_{2}' + xz f_{3}' \\ \frac{\partial u}{\partial z} &= f_{1}'(x, xy, xyz) \cdot 0 + f_{2}'(x, xy, xyz) \cdot 0 + f_{3}'(x, xy, xyz) \cdot xy = xy f_{3}' \end{aligned} $$ 这里, 我们称函数 \(f : \mathbb{R}^n \rightarrow \mathbb{R}\) 是 \(C^k\) 光滑的, 记作 \(f \in C^k(\mathbb{R}^n, \mathbb{R})\), 如果 \(f\) 的各个 \(k\) 阶 (偏) 导函数都存在且连续。

2.

给定 \(C^1\) 光滑的函数 \(F : \mathbb{R}^3 \rightarrow \mathbb{R}\). 求函数 $$ F(u^2 - x^2, u^2 - y^2, u^2 - z^2) $$ 对 \(x, y, z, u\) 的偏导数 $$ \begin{aligned} \frac{\partial F}{\partial x} &= F_{1}'(u^2-x^2, \dots) \cdot (-2x) = -2x F_{1}' \\ \frac{\partial F}{\partial y} &= F_{2}'(u^2-x^2, \dots) \cdot (-2y) = -2y F_{2}' \\ \frac{\partial F}{\partial z} &= F_{3}'(u^2-x^2, \dots) \cdot (-2z) = -2z F_{3}' \\ \frac{\partial F}{\partial u} &= F_{1}' \cdot (2u) + F_{2}' \cdot (2u) + F_{3}' \cdot (2u) = 2u (F_{1}' + F_{2}' + F_{3}') \end{aligned} $$

3.

给定 \(n \times n\) 的对称实矩阵 \((A_{ij})_{1 \leq i,j \leq n}\)(即对任何 \(i, j\), 有 \(A_{ij} = A_{ji}\)). 定义二次函数 $$ Q(x_1, \ldots, x_n) = \sum_{i=1}^n \sum_{j=1}^n A_{ij} x_i x_j, \quad \forall (x_1, \ldots, x_n) \in \mathbb{R}^n $$

(1) 求 \(Q\) 的微分 $$ \begin{aligned} \frac{\partial Q}{\partial x_k} &= \sum_{i=1}^n \sum_{j=1}^n A_{ij} (\delta_{ik} x_j + x_i \delta_{jk}) = \sum_{j=1}^n A_{kj} x_j + \sum_{i=1}^n A_{ik} x_i = 2 \sum_{j=1}^n A_{kj} x_j \\ dQ &= \sum_{k=1}^n \frac{\partial Q}{\partial x_k} dx_k = 2 \sum_{k=1}^n \sum_{j=1}^n A_{kj} x_j dx_k \end{aligned} $$

(2) 设 \(f : \mathbb{R}^n \rightarrow \mathbb{R}\) 是 \(C^1\) 光滑的函数. 定义函数 $$ g(x_1, \ldots, x_n) = f(x_1, \ldots, x_n) e^{- \frac{1}{2} Q(x_1, \ldots, x_n)}. $$ 计算 \(g\) 的各个偏导数 \(\dfrac{\partial g}{\partial x_1}, \ldots, \dfrac{\partial g}{\partial x_n}\). $$ \begin{aligned} \frac{\partial g}{\partial x_{k}} &= \frac{\partial f}{\partial x_k} e^{- \frac{1}{2} Q} + f \cdot e^{- \frac{1}{2} Q} \cdot \left(-\frac{1}{2} \frac{\partial Q}{\partial x_k}\right) \\ &= f_{k}' e^{- \frac{1}{2} Q} - \frac{1}{2} f e^{- \frac{1}{2} Q} \left(2 \sum_{j=1}^n A_{kj} x_j\right) \\ &= \left( f_{k}' - f \sum_{j=1}^n A_{kj} x_j \right) e^{- \frac{1}{2} Q} \end{aligned} $$

4.

设 \(f : \mathbb{R}^3 \rightarrow \mathbb{R}\) 是 \(C^1\) 光滑的函数, 即 \(f\) 的各个偏导数都存在且连续.

(1) 对于给定的点 \((x, y, z) \in \mathbb{R}^3\), 考虑关于 \(t\) 的一元函数 $$ g(t) = f(tx, ty, tz). $$ 求 \(g'(t)\)

\[ g'(t) = f_{x}' (tx, ty, tz) \cdot x + f_{y}' (tx, ty, tz) \cdot y + f_{z}' (tx, ty, tz) \cdot z \]

(2) 证明: 对任何 \((x, y, z) \in \mathbb{R}^3\), 有 $$ f(x, y, z) = f(0, 0, 0) + x \int_0^1 f_x(tx, ty, tz) \, dt + y \int_0^1 f_y(tx, ty, tz) \, dt + z \int_0^1 f_z(tx, ty, tz) \, dt $$ 证明： 根据微积分基本定理，我们有 \(g(1) = g(0) + \int_0^1 g'(t) dt\)。 代入 \(g(t)\) 的定义： $$ \begin{aligned} f(x, y, z) &= f(0, 0, 0) + \int_0^1 \left( x f_{x}'(tx, ty, tz) + y f_{y}'(tx, ty, tz) + z f_{z}' (tx, ty, tz) \right) dt \\ &= f(0, 0, 0) + x \int_0^1 f_x(tx, ty, tz) \, dt + y \int_0^1 f_y(tx, ty, tz) \, dt + z \int_0^1 f_z(tx, ty, tz) \, dt \end{aligned} $$

在本题 (3), (4) 小问中假设 \(f\) 满足: 对任何 \((x, y, z) \in \mathbb{R}^3\) 都有 $$ x f_x(x, y, z) + y f_y(x, y, z) + z f_z(x, y, z) = n f(x, y, z), $$ 其中 \(n\) 是某个给定的正整数.

(3) 对于给定的点 \((x, y, z) \in \mathbb{R}^3\), 考虑关于 \(t\) 的一元函数 $$ h(t) = \dfrac{f(tx, ty, tz)}{t^n}. $$ 求 \(h'(t)\) $$ \begin{aligned} h'(t) &= \frac{g'(t) \cdot t^n - g(t) \cdot n t^{n-1}}{(t^n)^2} \\ &= \frac{t \cdot g'(t) - n \cdot g(t)}{t^{n+1}} \\ &= \frac{t(x f'_x + y f'_y + z f'_z) - n f(tx, ty, tz)}{t^{n+1}} \end{aligned} $$

(4) 证明: 对任何 \((x, y, z) \in \mathbb{R}^3\) 与 \(t > 0\), 都有 $$ f(tx, ty, tz) = t^n f(x, y, z) $$ 证明： 根据齐次函数的条件，令 \(x' = tx, y' = ty, z' = tz\)，我们有 $$ x' f_x(x', y', z') + y' f_y(x', y', z') + z' f_z(x', y', z') = n f(x', y', z') $$ 代入 \(g(t)\) 和 \(g'(t)\) 的表达式，得到： $$ t \cdot g'(t) = (tx) f_{x}' (tx,...) + (ty) f_{y}' (tx,...) + (tz) f_{z}' (tx,...) = n f(tx, ty, tz) = n g(t) $$ 这就给出了一个关于 \(g(t)\) 的微分方程：\(t g'(t) = n g(t)\)。 分离变量得： $$ \frac{dg}{g} = n \frac{dt}{t} $$ 对两边从 \(1\) 到 \(t\) 进行积分： $$ \begin{aligned} \int_{g(1)}^{g(t)} \frac{d\gamma}{\gamma} &= n \int_{1}^{t} \frac{d\tau}{\tau} \\ \ln|g(t)| - \ln|g(1)| &= n (\ln|t| - \ln|1|) \\ \ln\left(\frac{g(t)}{g(1)}\right) &= n \ln(t) = \ln(t^n) \\ g(t) &= g(1) \cdot t^n \end{aligned} $$ 代回 \(f\) 的定义，即得： $$ f(tx, ty, tz) = t^n f(x, y, z) $$