3.1 第2次作业评讲

3.1.1 概念和计算部分

解 (1) 错。左导数的定义为$$\begin{array}{}\text{(1)}& f_{-}^{\prime }(x_0)=\underset{x\to x_0^{-}}{lim}\frac{f(x)-f(x_0)}{x-x_0}\end{array}$$ 方向导数的定义为$$\begin{array}{}\text{(2)}& \frac{\partial f}{\partial \mathit{v}}(x_0)=\underset{t\to 0^{+}}{lim}\frac{f(x_0-t)-f(x_0)}{t}\stackrel{t=x_0-x}{=}-\underset{x\to x_0^{-}}{lim}\frac{f(x)-f(x_0)}{x-x_0}\end{array}$$

(2) 错。如果这个坐标轴上的基底向量不是单位向量，那么偏导数就不是方向导数。

(3) 错。如果自变量空间中没有内积，那么仍然可以定义函数的微分和偏导数，但是不能定义梯度；当自变量空间中有内积时，只有坐标系对应的基底向量是该内积下的一组单位正交基底向量，这个说法才是对的。 $◻$

解 (4) AB。由于$xy=\mathcal{O}(x^2+y^2)$，$f(x,y)=\mathcal{O}\left(\sqrt{x^2+y^2}\right)=o(1)$，故$f$在$(0,0)$连续。计算可得$$\begin{array}{}\text{(6)}& \frac{\partial f}{\partial x}=\frac{y^3}{(x^2+y^2{)}^{3/2}},\frac{\partial f}{\partial x}(x,kx)=\frac{k^3}{(1+k^2{)}^{3/2}}\ne \mathrm{const}\end{array}$$ 故$\frac{\partial f}{\partial x}$在$(0,0)$处不连续。同理$\frac{\partial f}{\partial y}$亦在$(0,0)$处不连续，故$f$在$(0,0)$处不存在偏导数。

注意到$f(x,kx)=\frac{kx}{\sqrt{1+k^2}}$，故$f$沿向量${\mathit{v}}_k:=(1,k)$的导数为$\frac{k}{\sqrt{1+k^2}}$。由于$2{\mathit{v}}_1={\mathit{v}}_0+{\mathit{v}}_2$，而$$\begin{array}{}\text{(7)}& 2\times \frac{1}{\sqrt{1+1^2}}\ne 0+\frac{2}{\sqrt{1+2^2}}⟺\frac{\partial f}{\partial ({\mathit{v}}_0+{\mathit{v}}_2)}\ne \frac{\partial f}{\partial {\mathit{v}}_0}+\frac{\partial f}{\partial {\mathit{v}}_2}\end{array}$$ 故$f$在$(0,0)$处不可微。

(5) ABD。由于$f(x,y)=\mathcal{O}(x^2+y^2)$，故$f$连续、可微、偏导数存在。计算可得$$\begin{array}{}\text{(8)}& \frac{\partial f}{\partial x}=\{\begin{array}{ll}2x(\sin \frac{1}{x^2+y^2}-\frac{1}{x^2+y^2}\cos \frac{1}{x^2+y^2}),& (x,y)\ne (0,0)\\ 0,& (x,y)=(0,0)\end{array}\end{array}$$ 然而$$\begin{array}{}\text{(9)}& \frac{\partial f}{\partial x}(x,x)=2x\sin \frac{1}{2x^2}-\frac{1}{x}\cos \frac{1}{2x^2}\nrightarrow 0,x\to 0\end{array}$$ 故$\frac{\partial f}{\partial x}$在$(0,0)$处不连续。同理$\frac{\partial f}{\partial y}$亦在$(0,0)$处不连续，故$f$在$(0,0)$处偏导数不连续。

(6) B。由于$f(x,k{x}^3)=k\ne \mathrm{const}$，故$f$不连续，自然不可微。设$\mathit{v}=(\cos \theta ,\sin \theta {)}^{\mathrm{T}}$，计算可得$$\begin{array}{}\text{(10)}& \frac{\partial f}{\partial \mathit{v}}(0,0)=\underset{t\to 0}{lim}\frac{f(t\cos \theta ,t\sin \theta )-f(0,0)}{t}=\underset{t\to 0}{lim}{t}^2\frac{{\cos }^3\theta }{\sin \theta }=0\end{array}$$ 故$f$在$(0,0)$处沿所有方向都有方向导数。 $◻$

解 计算可知$$\begin{array}{}\text{(12)}& \begin{array}{rl}\frac{\partial g}{\partial x}& =3x^{2}f(xy,\frac{y}{x})+x^3[y{f}_u(xy,\frac{y}{x})-\frac{y}{x^2}{f}_v(xy,\frac{y}{x})]\\ \frac{\partial g}{\partial y}& =x^3[x{f}_u(xy,\frac{y}{x})+\frac{1}{x}{f}_v(xy,\frac{y}{x})]\end{array}\end{array}$$ 代入$(x,y)=(1,1)$可得$$\begin{array}{}\text{(13)}& g_x(1,1)=10,g_y(1,1)=9\end{array}$$ 根据可微（切空间）的定义可得切线方程为$$\begin{array}{}\text{(14)}& 0=\mathrm{d}g(1,1)=g_x(1,1)\mathrm{d}x+g_y(1,1)\mathrm{d}y=10(x-1)+9(y-1)\end{array}$$ $◻$

3.1.2 解答和证明部分

注 大家在使用大O、小o记号时一定要注意：确保你的记号与其他变量无关！不然，我们可以这么“证明”若$f(x,y)$分别关于$x$和$y$连续，则$f$关于$(x,y)$二元连续：$$\begin{array}{}\text{(20)}& f(x,y)=f(x,0)+o(1)=f(0,0)+o(1)+o(1)=f(0,0)+o(1)\end{array}$$ 我推荐大家采用$o_x(h)$这种写法，强调这里关于$h$的小o可能与$x$有关。

注¹ 本题(3)问的难点在于如何理解$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=kz$，它的意思是：$$\begin{array}{}\text{(28)}& x\frac{\partial z}{\partial 1}(x,y)+y\frac{\partial z}{\partial 2}(x,y)=kz(x,y),\mathrm{\forall }(x,y)\ne (0,0)\end{array}$$

注² 本题(1)问是验证直线为$z$的等高线。若是证明$z$的等高线是直线，即求出方程的解，难度更大。求解一阶线性偏微分方程有专门的方法，称为特征线法，参见第4.2.8节。对于本题而言，特征线为$$\begin{array}{}\text{(29)}& \frac{\mathrm{d}x}{a}=\frac{\mathrm{d}y}{b}⟹bx-ay=h\end{array}$$ 作换元$(x,y)\mapsto (\xi ,\eta )=(bx-ay,y)$，计算可得$$\begin{array}{}\text{(30)}& \left(\begin{array}{cc}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}\end{array}\right)=\left(\begin{array}{cc}\frac{\partial }{\partial \xi }& \frac{\partial }{\partial \eta }\end{array}\right)\left(\begin{array}{cc}\frac{\partial \xi }{\partial x}& \frac{\partial \xi }{\partial y}\\ \frac{\partial \eta }{\partial x}& \frac{\partial \eta }{\partial y}\end{array}\right)=\left(\begin{array}{cc}\frac{\partial }{\partial \xi }& \frac{\partial }{\partial \eta }\end{array}\right)\left(\begin{array}{cc}b& -a\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}b\frac{\partial }{\partial \xi }& -a\frac{\partial }{\partial \xi }+\frac{\partial }{\partial \eta }\end{array}\right)\end{array}$$ 代入原式可得$$\begin{array}{}\text{(31)}& \frac{\partial z}{\partial \eta }=0⟹z=f(\xi )=f(bx-ay)\end{array}$$ 所以$bx-ay=C$即为$z$的等高线。这实际上是原一阶线性偏微分方程的首次积分。

解 (1) 计算可得$$\begin{array}{}\text{(33)}& \begin{array}{rl}\left(\begin{array}{cc}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}\end{array}\right)& =\left(\begin{array}{cc}\frac{\partial }{\partial u}& \frac{\partial }{\partial v}\end{array}\right)\left(\begin{array}{cc}\frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}\end{array}\right)=\left(\begin{array}{cc}\frac{\partial }{\partial u}& \frac{\partial }{\partial v}\end{array}\right)\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}\frac{\partial }{\partial u}+\frac{\partial }{\partial v}& \frac{\partial }{\partial v}\end{array}\right)\\ \left(\begin{array}{cc}\frac{{\partial }^2}{\partial x^2}& \frac{{\partial }^2}{\partial x\partial y}\\ \frac{{\partial }^2}{\partial y\partial x}& \frac{{\partial }^2}{\partial y^2}\end{array}\right)& =\left(\begin{array}{c}\frac{\partial }{\partial u}+\frac{\partial }{\partial v}\\ \frac{\partial }{\partial v}\end{array}\right)\left(\begin{array}{cc}\frac{\partial }{\partial u}+\frac{\partial }{\partial v}& \frac{\partial }{\partial v}\end{array}\right)=\left(\begin{array}{cc}\frac{{\partial }^2}{\partial u^2}+2\frac{{\partial }^2}{\partial u\partial v}+\frac{{\partial }^2}{\partial v^2}& \frac{{\partial }^2}{\partial u\partial v}+\frac{{\partial }^2}{\partial v^2}\\ \frac{{\partial }^2}{\partial u\partial v}+\frac{{\partial }^2}{\partial v^2}& \frac{{\partial }^2}{\partial v^2}\end{array}\right)\end{array}\end{array}$$ 首先作因变量换元$z\mapsto w=x+y+z$，计算可得$$\begin{array}{}\text{(34)}& w_{xx}-2w_{xy}+w_{yy}+w_x-w_y=0\end{array}$$ 再作自变量换元$(x,y)\mapsto (u,v)=(x,x+y)$，计算可得$$\begin{array}{}\text{(35)}& \underset{w_{xx}}{\underbrace{w_{uu}+2w_{uv}+w_{vv}}}-2\underset{w_{xy}}{\underbrace{(w_{uv}+w_{vv})}}+\underset{w_{yy}}{\underbrace{w_{vv}}}+\underset{w_x}{\underbrace{w_u+w_v}}-\underset{w_y}{\underbrace{w_v}}=w_{uu}+w_u=0\end{array}$$

(2) 换元后的方程相当于$w$关于$u$的常微分方程，但要注意任意常数可以与$v$有关，解得$$\begin{array}{}\text{(36)}& w(u,v)=c_1(v){\mathrm{e}}^{-u}+c_2(v)=c_1(x+y){\mathrm{e}}^{-x}+c_2(x+y)\end{array}$$ 代入$z=w-(x+y)$，将$-(x+y)$并入$c_2(x+y)$，得$$\begin{array}{}\text{(37)}& z(x,y)=c_1(x+y){\mathrm{e}}^{-x}+c_2(x+y)\end{array}$$ 其中$c_1,c_2\in {\mathcal{C}}^2$。

(3) 这是一个二阶线性偏微分方程，换元法是为了将其化为标准形，参见第4.2.9节。关注二阶项的系数$a_{11}{w}_{xx}+2a_{12}{w}_{xy}+a_{13}{w}_{yy}$，判别式$\mathrm{\Delta }=a_{12}^2-a_{11}{a}_{22}=0$，故这是一个抛物型方程，其特征线为$$\begin{array}{}\text{(38)}& \mathrm{d}{x}^2+2\mathrm{d}x\mathrm{d}y+\mathrm{d}{y}^2=0⟹\mathrm{d}(x+y)=0\end{array}$$ 由此解得唯一一个首次积分$x+y=h$。任选与此无关的变量（如$x$）构成换元$(x,y)\mapsto (u,v)=(x,x+y)$，使得原方程化为抛物型方程的标准式。 $◻$

解 (1) 计算可得$$\begin{array}{}\text{(39)}& \begin{array}{rl}\left(\begin{array}{cc}\frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }\end{array}\right)& =\left(\begin{array}{cc}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}\end{array}\right)\left(\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }\end{array}\right)=\left(\begin{array}{cc}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}\end{array}\right)\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right)\\ & =\left(\begin{array}{cc}\cos \theta \frac{\partial }{\partial x}+\sin \theta \frac{\partial }{\partial y}& -r\sin \theta \frac{\partial }{\partial x}+r\cos \theta \frac{\partial }{\partial y}\end{array}\right)\end{array}\end{array}$$ 故$u_r$可视作$u$沿半径方向的方向导数；但由于$r$的存在，$u_{\theta }$不是$u$沿角度方向的方向导数。

(2) 梯度与方向导数的关系为$$\begin{array}{}\text{(40)}& \frac{\partial u}{\partial \mathit{v}}=⟨\mathrm{\nabla }u,\mathit{v}⟩\end{array}$$ 分别取$\mathit{v}=\hat{r}=(\cos \theta ,\sin \theta )$和$\mathit{v}=\hat{\theta }=(-\sin \theta ,\cos \theta )$，则$$\begin{array}{}\text{(41)}& ⟨\mathrm{\nabla }u,\hat{r}⟩=u_r,⟨\mathrm{\nabla }u,\hat{\theta }⟩=\frac{1}{r}{u}_{\theta }\end{array}$$ 由于$\hat{r}\perp \hat{\theta }$，故$$\begin{array}{}\text{(42)}& \mathrm{\nabla }u=⟨\mathrm{\nabla }u,\hat{r}⟩\hat{r}+⟨\mathrm{\nabla }u,\hat{\theta }⟩\hat{\theta }=u_r\hat{r}+\frac{1}{r}{u}_{\theta }\hat{\theta }\end{array}$$

(3) 容易发现$$\begin{array}{}\text{(43)}& \parallel \mathrm{\nabla }u{\parallel }^2=(u_x{)}^2+(u_y{)}^2=(u_r{)}^2+{\left(\frac{1}{r}{u}_{\theta }\right)}^2\end{array}$$ 这个等式的直观解释就是梯度与具体的坐标系无关。

(4) 计算可得$$\begin{array}{}\text{(44)}& \begin{array}{rl}\left(\begin{array}{cc}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}\end{array}\right)& =\left(\begin{array}{cc}\frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }\end{array}\right){\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right)}^{-1}=\left(\begin{array}{cc}\frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\frac{1}{r}\sin \theta & \frac{1}{r}\cos \theta \end{array}\right)\\ & =\left(\begin{array}{cc}\cos \theta \frac{\partial }{\partial r}-\frac{1}{r}\sin \theta \frac{\partial }{\partial \theta }& \sin \theta \frac{\partial }{\partial r}+\frac{1}{r}\cos \theta \frac{\partial }{\partial \theta }\end{array}\right)\end{array}\end{array}$$ 继续计算可得$$\begin{array}{}\text{(45)}& \begin{array}{rl}\frac{{\partial }^2}{\partial x^2}& =\cos \theta \frac{\partial }{\partial r}(\cos \theta \frac{\partial }{\partial r}-\frac{1}{r}\sin \theta \frac{\partial }{\partial \theta })-\frac{1}{r}\sin \theta \frac{\partial }{\partial \theta }(\cos \theta \frac{\partial }{\partial r}-\frac{1}{r}\sin \theta \frac{\partial }{\partial \theta })\\ & ={\cos }^2\theta \frac{{\partial }^2}{\partial r^2}+\frac{2}{r^2}\cos \theta \sin \theta \frac{\partial }{\partial \theta }-\frac{1}{r}\cos \theta \sin \theta \frac{{\partial }^2}{\partial r\partial \theta }+\frac{1}{r}{\sin }^2\theta \frac{{\partial }^2}{\partial r\partial \theta }+\frac{1}{r^2}{\sin }^2\theta \frac{{\partial }^2}{\partial {\theta }^2}\end{array}\end{array}$$ 对于$\frac{{\partial }^2}{\partial y^2}$，将上式中的$\theta$换为$\theta -\frac{\pi }{2}$即可，亦即$$\begin{array}{}\text{(46)}& \frac{{\partial }^2}{\partial y^2}={\sin }^2\theta \frac{{\partial }^2}{\partial r^2}-\frac{2}{r^2}\sin \theta \cos \theta \frac{\partial }{\partial \theta }+\frac{1}{r}\sin \theta \cos \theta \frac{{\partial }^2}{\partial r\partial \theta }+\frac{1}{r}{\cos }^2\theta \frac{{\partial }^2}{\partial r\partial \theta }+\frac{1}{r^2}{\cos }^2\theta \frac{{\partial }^2}{\partial {\theta }^2}\end{array}$$ 两者相加可得$$\begin{array}{}\text{(47)}& \frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}=\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}\end{array}$$ $◻$

注 第(4)问不要想当然地认为$$\begin{array}{}\text{(48)}& \begin{array}{rl}\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}& =\left(\begin{array}{cc}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}\end{array}\right)\left(\begin{array}{c}\frac{\partial }{\partial x}\\ \frac{\partial }{\partial y}\end{array}\right)\ne \left(\begin{array}{cc}\frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }\end{array}\right)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\frac{1}{r}\sin \theta & \frac{1}{r}\cos \theta \end{array}\right)\left(\begin{array}{cc}\cos \theta & -\frac{1}{r}\sin \theta \\ \sin \theta & \frac{1}{r}\cos \theta \end{array}\right)\left(\begin{array}{c}\frac{\partial }{\partial r}\\ \frac{\partial }{\partial \theta }\end{array}\right)\\ & =\left(\begin{array}{cc}\frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& \frac{1}{r^2}\end{array}\right)\left(\begin{array}{c}\frac{\partial }{\partial r}\\ \frac{\partial }{\partial \theta }\end{array}\right)=\frac{{\partial }^2}{\partial r^2}+\frac{1}{r^2}\frac{{\partial }^2}{\partial {\theta }^2}\end{array}\end{array}$$ 原因在于求偏导和乘法不可交换，$\left(\begin{array}{cc}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}\end{array}\right)=\left(\begin{array}{cc}\frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }\end{array}\right)B$是个形式记号，是右边的矩阵系数直接乘在偏导算子的左边，而不是偏导算子作用在右边的矩阵上。