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习题1.5

第 1 题

求下列变换所确定的向量值函数 \(\begin{pmatrix}u \\ v\end{pmatrix} = \begin{pmatrix}f_1(x, y) \\ f_2(x, y)\end{pmatrix}\) 的 Jacobi 矩阵 \(\dfrac{\partial(u, v)}{\partial(x, y)}\), 并指出在哪些区域 Jacobi 矩阵可逆.

(1) \(\begin{cases}u = \sqrt{x^2 + y^2}, \\ v = \arctan\dfrac{y}{x};\end{cases}\)

(3) \(\begin{cases}u = \dfrac{x}{x^2 + y^2}, \\ v = \dfrac{y}{x^2 + y^2};\end{cases}\)

解

(1)

\[ J= \begin{pmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{pmatrix} = \begin{pmatrix} \dfrac{2x}{2\sqrt{x^2 + y^2}} & \dfrac{2y}{2\sqrt{x^2 + y^2}} \\ \dfrac{1}{1 + \left(\dfrac{y}{x}\right)^2}\cdot\left(-\dfrac{y}{x^2}\right) & \dfrac{1}{1 + \left(\dfrac{y}{x}\right)^2}\cdot\dfrac{1}{x} \end{pmatrix} = \begin{pmatrix} \dfrac{x}{\sqrt{x^2 + y^2}} & \dfrac{y}{\sqrt{x^2 + y^2}} \\ \dfrac{-y}{x^2 + y^2} & \dfrac{x}{x^2 + y^2} \end{pmatrix} \]

且 \(\det{J} \neq 0 \iff x^2 + y^2 \neq 0\), 故在 \(\mathbb{R}^2 \backslash \{(0, 0)\}\) 处 Jacobi 矩阵可逆.

(3)

\[ J= \begin{pmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{pmatrix} = \begin{pmatrix} \dfrac{x^2 + y^2 - x\cdot 2x}{(x^2 + y^2)^2} & \dfrac{-x\cdot 2y}{(x^2 + y^2)^2} \\ \dfrac{-y\cdot 2x}{(x^2 + y^2)^2} & \dfrac{x^2 + y^2 - y\cdot 2y}{(x^2 + y^2)^2} \end{pmatrix} = \begin{pmatrix} \dfrac{y^2 - x^2}{(x^2 + y^2)^2} & \dfrac{-2xy}{(x^2 + y^2)^2} \\ \dfrac{-2xy}{(x^2 + y^2)^2} & \dfrac{x^2 - y^2}{(x^2 + y^2)^2} \end{pmatrix} \]

且 \(\det{J} \neq 0 \iff x^2 + y^2 \neq 0\), 故在 \(\mathbb{R}^2 \backslash \{(0, 0)\}\) 处 Jacobi 矩阵可逆.

第 2 题

求变换 \(\begin{cases}x = r\sin\theta\cos\phi, \\ y = r\cos\theta\cos\phi, \\ z=r\sin\phi,\end{cases}r>0, 0\le\theta \le 2\pi, 0 \le \phi \le pi\) 确定的向量值函数 \(\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}f_1(r, \theta, \phi) \\ f_2(r, \theta, \phi) \\ f_3(r, \theta, \phi)\end{pmatrix}\) 的 Jacobi 矩阵.

解

\[ J = \begin{pmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial \theta} & \dfrac{\partial x}{\partial \phi} \\ \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial \theta} & \dfrac{\partial y}{\partial \phi} \\ \dfrac{\partial z}{\partial r} & \dfrac{\partial z}{\partial \theta} & \dfrac{\partial z}{\partial \phi} \end{pmatrix} = \begin{pmatrix} \sin\theta & r\cos\theta\cos\phi & -r\sin\theta\sin\phi \\ \cos\theta\cos\phi & -r\sin\theta\cos\phi & -r\cos\theta\sin\phi \\ \sin\phi & 0 & r\cos\phi \end{pmatrix} \]

第 3 题

求下列复合函数的偏导数 \(\dfrac{\partial z}{\partial x}, \dfrac{\partial z}{\partial y}\) (已知 \(f\) 为可微函数).

(1) \(z = \arctan \dfrac{u}{v}, u = x^2 + y^2, v = xy\);

(3) \(z = f(x^2 - y^2, e^{xy})\);

(5) \(z = xy + \dfrac{y}{x}f(xy)\);

解

(1) \(\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x} + \dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x} = \dfrac{\dfrac{1}{v}}{1 + \left(\dfrac{u}{v}\right)^2}\cdot 2x + \dfrac{-\dfrac{u}{v^2}}{1 + \left(\dfrac{u}{v}\right)^2}\cdot y = \dfrac{y(x^2 - y^2)}{x^4 + 3x^2y^2 + y^4}\).

\(\dfrac{\partial z}{\partial y} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y} + \dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y} = \dfrac{\dfrac{1}{v}}{1 + \left(\dfrac{u}{v}\right)^2}\cdot 2y + \dfrac{-\dfrac{u}{v^2}}{1 + \left(\dfrac{u}{v}\right)^2}\cdot x = \dfrac{x(y^2 - x^2)}{x^4 + 3x^2y^2 + y^4}\).

(3) \(\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x} + \dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x} = f_1'(x^2 - y^2, e^{xy})\cdot 2x + f_2'(x^2 - y^2, e^{xy})\cdot ye^{xy} = 2xf_1' + ye^{xy}f_2'\).

\(\dfrac{\partial z}{\partial y} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y} + \dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y} = f_1'(x^2 - y^2, e^{xy})\cdot (-2y) + f_2'(x^2 - y^2, e^{xy})\cdot xe^{xy} = -2yf_1' + xe^{xy}f_2'\).

(5) \(\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x} + \dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x} = y + \dfrac{y^2f'(xy)x-yf(xy)}{x^2} = y + \dfrac{y^2f'(xy)}{x} - \dfrac{yf(xy)}{x^2}\).

\(\dfrac{\partial z}{\partial y} = \dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y} + \dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y} = x + \dfrac{f(xy) + yxf'(xy)}{x} = x + yf'(xy) + \dfrac{f(xy)}{x}\).

第 4 题

已知函数 \(z = u\ln(u - v)\), 其中 \(u = e^{-x}, v = \ln x\), 求 \(\dfrac{\text{d}{z}}{\text{d}{x}}\).

解

\[ \begin{aligned} \dfrac{\text{d}{z}}{\text{d}{x}} & = \dfrac{\partial z}{\partial u}\dfrac{\text{d}{u}}{\text{d}{x}} + \dfrac{\partial z}{\partial v}\dfrac{\text{d}{v}}{\text{d}{x}} \\ & = \left(\ln(u - v) + \dfrac{u}{u - v}\right)\cdot (-e^{-x}) + \dfrac{u}{u - v}\cdot\left(-\dfrac{1}{x}\right) \\ & = -e^{-x}\ln(e^{-x}-\ln x) - \dfrac{e^{-2x}}{e^{-x} - \ln x} - \dfrac{e^{-x}}{x(e^{-x} - \ln x)} \end{aligned} \]

第 5 题

已知函数 \(u = f(x, y)\), 其中 \(x = r\cos \theta, y = r\sin\theta\), \(f\) 可微, 证明:

\[ \left(\dfrac{\partial u}{\partial r}\right)^2 + \left(\dfrac{1}{r}\dfrac{\partial u}{\partial \theta}\right)^2 = \left(\dfrac{\partial u}{\partial x}\right)^2 + \left(\dfrac{\partial u}{\partial y}\right)^2. \]

证明

由于 \(\dfrac{\partial u}{\partial r} = \dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial r} + \dfrac{\partial u}{\partial y}\dfrac{\partial y}{\partial r} = \cos\theta \dfrac{\partial u}{\partial x} + \sin\theta \dfrac{\partial u}{\partial y}, \dfrac{\partial u}{\partial \theta} = \dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial \theta} + \dfrac{\partial u}{\partial y}\dfrac{\partial y}{\partial \theta} = -r\sin\theta \dfrac{\partial u}{\partial \theta} + r\cos\theta \dfrac{\partial u}{\partial \theta}\), 所以代入得

\[ \begin{aligned} \left(\dfrac{\partial u}{\partial r}\right)^2 + \left(\dfrac{1}{r}\dfrac{\partial u}{\partial \theta}\right)^2 & = \left(\cos\theta \dfrac{\partial u}{\partial x} + \sin\theta \dfrac{\partial u}{\partial y}\right)^2 + \left(-\sin\theta \dfrac{\partial u}{\partial \theta} + \cos\theta \dfrac{\partial u}{\partial \theta}\right)^2 \\ & = \left(\dfrac{\partial u}{\partial x}\right)^2 + \left(\dfrac{\partial u}{\partial y}\right)^2. \end{aligned} \]

命题得证.

第 6 题

设 \(f\) 可微, \(u = xy + xf\left(\dfrac{y}{x}\right)\), 证明: \(x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = u + xy\).

证明

由于 \(\dfrac{\partial u}{\partial x} = y + f\left(\dfrac{y}{x}\right) - \dfrac{y}{x}f'\left(\dfrac{y}{x}\right), \dfrac{\partial u}{\partial y} = x + f'\left(\dfrac{y}{x}\right)\), 所以代入得

\[ \begin{aligned} x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} & = xy + xf\left(\dfrac{y}{x}\right) - yf'\left(\dfrac{y}{x}\right) + xy + yf'\left(\dfrac{y}{x}\right) \\ & = 2xy + xf\left(\dfrac{y}{x}\right) \\ & = u + xy \end{aligned} \]

命题得证.

第 7 题

设 \(f \in C^2(\mathbb{R}^2)\) 满足 Laplace 方程 \(\dfrac{\partial ^2f}{\partial x^2} + \dfrac{\partial ^2f}{\partial y^2} = 0\), 证明: \(u(x, y) = f\left(\dfrac{x}{x^2 + y^2}, \dfrac{y}{x^2 + y^2}\right)\) 也满足 Laplace 方程.

证明

\(\dfrac{\partial u}{\partial x} = u_1'\dfrac{y^2 - x^2}{(x^2 + y^2)^2} + u_2'\dfrac{-2xy}{(x^2 + y^2)^2}\).

\[ \begin{aligned} \dfrac{\partial^2 u}{\partial x^2} & = \dfrac{\partial}{\partial x}\left(u_1'\dfrac{y^2 - x^2}{(x^2 + y^2)^2}\right) + \dfrac{\partial}{\partial x}\left(u_2'\dfrac{-2xy}{(x^2 + y^2)^2}\right) \\ & = u_{11}'\left(\dfrac{y^2 - x^2}{(x^2 + y^2)^2}\right)^2 - (u_{12}' + u_{21}')\dfrac{2xy(y^2 - x^2)}{(x^2 + y^2)^4} + u_{22}'\left(\dfrac{-2xy}{(x^2 + y^2)^2}\right)^2 \\ & + u_1'\dfrac{-2x(3y^2 - x^2)}{(x^2 + y^2)^3} + u_2'\dfrac{2y(3x^2 - y^2)}{(x^2 + y^2)^3} \end{aligned} \]

同理可得

\[ \begin{aligned} \dfrac{\partial^2 u}{\partial y^2} & = u_{11}'\left(\dfrac{-2xy}{(x^2 + y^2)^2}\right)^2 + (u_{12}' + u_{21}')\dfrac{2xy(y^2 - x^2)}{(x^2 + y^2)^4} + u_{22}'\left(\dfrac{y^2 - x^2}{(x^2 + y^2)^2}\right)^2 \\ & + u_2'\dfrac{-2y(3x^2 - y^2)}{(x^2 + y^2)^3} + u_1'\dfrac{2x(3y^2 - x^2)}{(x^2 + y^2)^3} \end{aligned} \]

由题知 \(f\) 满足 Laplace 方程, 则 \(u_{11}' + u_{22}' = 0\). 则

\[ \begin{aligned} \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} & = u_{11}'\left(\dfrac{y^2 - x^2}{(x^2 + y^2)^2}\right)^2 - (u_{12}' + u_{21}')\dfrac{2xy(y^2 - x^2)}{(x^2 + y^2)^4} + u_{22}'\left(\dfrac{-2xy}{(x^2 + y^2)^2}\right)^2 \\ & + u_1'\dfrac{-2x(3y^2 - x^2)}{(x^2 + y^2)^3} + u_2'\dfrac{2y(3x^2 - y^2)}{(x^2 + y^2)^3} \\ & + u_{11}'\left(\dfrac{-2xy}{(x^2 + y^2)^2}\right)^2 + (u_{12}' + u_{21}')\dfrac{2xy(y^2 - x^2)}{(x^2 + y^2)^4} + u_{22}'\left(\dfrac{y^2 - x^2}{(x^2 + y^2)^2}\right)^2 \\ & + u_2'\dfrac{-2y(3x^2 - y^2)}{(x^2 + y^2)^3} + u_1'\dfrac{2x(3y^2 - x^2)}{(x^2 + y^2)^3} \\ & = (u_{11}' + u_{22}')\left[\left(\dfrac{-2xy}{(x^2 + y^2)^2}\right)^2 +\left(\dfrac{y^2 - x^2}{(x^2 + y^2)^2}\right)^2\right] \\ & = 0 \end{aligned} \]

命题得证.

第 8 题

已知变换 \(\begin{cases} w=x+y+z, \\u=x, \\ v=x+y,\end{cases}\) 化简方程 \(\dfrac{\partial ^2z}{\partial x^2} - 2\dfrac{\partial ^2z}{\partial x \partial y} + \dfrac{\partial ^2z}{\partial y^2} + \dfrac{\partial z}{\partial x} - \dfrac{\partial z}{\partial y} = 0\), 以 \(w\) 为因变量, \(u, v\) 为自变量.

解

化简得 \(z = -x - y + w\). 因此

\[ \begin{aligned} \dfrac{\partial z}{\partial x} & = \dfrac{\partial w}{\partial x} - 1 = \dfrac{\partial w}{\partial u}\dfrac{\partial u}{\partial x} + \dfrac{\partial w}{\partial v}\dfrac{\partial v}{\partial x} - 1 = \dfrac{\partial w}{\partial u} + \dfrac{\partial w}{\partial v} - 1, \\ \dfrac{\partial z}{\partial y} & = \dfrac{\partial w}{\partial y} - 1 = \dfrac{\partial w}{\partial u}\dfrac{\partial u}{\partial y} + \dfrac{\partial w}{\partial v}\dfrac{\partial v}{\partial y} - 1 = \dfrac{\partial w}{\partial v} - 1. \end{aligned} \]

故 \(\dfrac{\partial^2 z}{\partial x^2} = \dfrac{\partial^2 w}{\partial u^2} + 2\dfrac{\partial^2 w}{\partial u\partial v} + \dfrac{\partial^2 w}{\partial v^2}, \dfrac{\partial^2 z}{\partial x\partial y} = \dfrac{\partial^2 w}{\partial u\partial v} + \dfrac{\partial^2 w}{\partial v^2}, \dfrac{\partial^2 z}{\partial y^2} = \dfrac{\partial^2 w}{\partial v^2}\).

则

\[ \begin{aligned} &\; \; \; \; \; \; \; \; \dfrac{\partial ^2z}{\partial x^2} - 2\dfrac{\partial ^2z}{\partial x \partial y} + \dfrac{\partial ^2z}{\partial y^2} + \dfrac{\partial z}{\partial x} - \dfrac{\partial z}{\partial y} \\ & = \left(\dfrac{\partial^2 w}{\partial u^2} + 2\dfrac{\partial^2 w}{\partial u\partial v} + \dfrac{\partial^2 w}{\partial v^2}\right) - 2\left(\dfrac{\partial^2 w}{\partial u\partial v} + \dfrac{\partial^2 w}{\partial v^2}\right) + \dfrac{\partial^2 w}{\partial v^2} + \left(\dfrac{\partial w}{\partial u} + \dfrac{\partial w}{\partial v} - 1\right) - \left(\dfrac{\partial w}{\partial v} - 1\right) \\ & = \dfrac{\partial^2w}{\partial u^2} + \dfrac{\partial w}{\partial u} \end{aligned} \]

所以化简为 \(\dfrac{\partial^2w}{\partial u^2} + \dfrac{\partial w}{\partial u} = 0\).

第 9 题

向量值函数 \(\mathbf{Y} = \mathbf{f}(\mathbf{U}), \mathbf{U} = \mathbf{g}(\mathbf{X})\) 均可微, 求复合函数 \(\mathbf{Y} = \mathbf{f} \circ \mathbf{g}(\mathbf{X})\) 的 Jacobi 矩阵和全微分.

(1) \(\begin{cases} y_1 = u_1 + u_2 \\ y_2 = u_1u_2 \\ y_3 = \dfrac{u_2}{u_1},\end{cases}\begin{cases}u_1 = \dfrac{x}{x^2 + y^2}, \\ u_2 = \dfrac{y}{x^2 + y^2};\end{cases}\)

解

(1) 因为

\[ J_{\mathbf{g}} = \begin{pmatrix}\dfrac{\partial u_1}{\partial x} & \dfrac{\partial u_1}{\partial y} \\ \dfrac{\partial u_2}{\partial x} & \dfrac{\partial u_2}{\partial y}\end{pmatrix} = \begin{pmatrix}\dfrac{y^2 - x^2}{x^2 + y^2} & \dfrac{-2xy}{x^2 + y^2} \\ \dfrac{-2xy}{x^2 + y^2} & \dfrac{x^2 - y^2}{x^2 + y^2}\end{pmatrix}, \]
\[ J_{\mathbf{f}} = \begin{pmatrix}\dfrac{\partial y_1}{\partial u_1} & \dfrac{\partial y_1}{\partial u_2} \\ \dfrac{\partial y_2}{\partial u_1} & \dfrac{\partial y_2}{\partial u_2} \\ \dfrac{\partial y_3}{\partial u_1} & \dfrac{\partial y_3}{\partial u_2}\end{pmatrix} = \begin{pmatrix}1 & 1 \\ \dfrac{y}{x^2 + y^2} & \dfrac{x}{x^2 + y^2} \\ -\dfrac{y}{x} & \dfrac{x^2 + y^2}{x}\end{pmatrix}, \]

所以 \(J = J_{\mathbf{f}}J_{\mathbf{g}} = \begin{pmatrix}1 & 1 \\ \dfrac{y}{x^2 + y^2} & \dfrac{x}{x^2 + y^2} \\ -\dfrac{y}{x} & \dfrac{x^2 + y^2}{x}\end{pmatrix}\begin{pmatrix}\dfrac{y^2 - x^2}{x^2 + y^2} & \dfrac{-2xy}{x^2 + y^2} \\ \dfrac{-2xy}{x^2 + y^2} & \dfrac{x^2 - y^2}{x^2 + y^2}\end{pmatrix}\), 且全微分 \(\text{d}{\mathbf{Y}} = J\begin{pmatrix}\text{d}{x} \\ \text{d}{y}\end{pmatrix}\).