第十六周作业

本次作业题包括习题4.2第3题和习题4.3第4题。

习题4.2

3

设$\varphi(x_1,\ldots,x_n)=0$给出方程$\displaystyle\sum_{i,j=1}^na_{ij}\frac{\partial^2 u}{\partial x_i\partial x_j}+\sum_{i=1}^nb_{i}\frac{\partial u}{\partial x_i}+cu-f=0$的特征曲面，即成立$\displaystyle\sum_{i,j=1}^na_{ij}\frac{\partial\varphi}{\partial x_i}\frac{\partial\varphi}{\partial x_j}=0$。作可逆坐标变换$x_i=f_i(y_1,\ldots,y_n)$，$y_i=g_i(x_1,\ldots,x_n)$，则有$\frac{\partial u}{\partial x_i}=\displaystyle\sum_{k=1}^n\frac{\partial u}{\partial y_k}\frac{\partial g_k}{\partial x_i}$，$\frac{\partial^2 u}{\partial x_i\partial x_j}=\displaystyle\sum_{k,l=1}^n\frac{\partial^2 u}{\partial y_k\partial y_l}\frac{\partial g_k}{\partial x_i}\frac{\partial g_l}{\partial x_j}+\sum_{k=1}^n\frac{\partial u}{\partial y_k}\frac{\partial^2 g_k}{\partial x_i\partial x_j}$，于是方程化为$\displaystyle\sum_{i,j=1}^na_{ij}\sum_{k,l=1}^n\frac{\partial^2 u}{\partial y_k\partial y_l}\frac{\partial g_k}{\partial x_i}\frac{\partial g_l}{\partial x_j}+\cdots=0$，即$\displaystyle\sum_{k,l=1}^n\hat{a}_{kl}\frac{\partial^2 u}{\partial y_k\partial y_l}+\cdots=\displaystyle\sum_{k,l=1}^n(\sum_{i,j=1}^na_{ij}\frac{\partial g_k}{\partial x_i}\frac{\partial g_l}{\partial x_j})\frac{\partial^2 u}{\partial y_k\partial y_l}+\cdots=0$，而原方程的特征曲面化为$\hat{\varphi}(y_1,\ldots,y_n)=\varphi(f_1(y_1,\ldots,y_n),\ldots,f_n(y_1,\ldots,y_n))=0$。

$\begin{align}\displaystyle\sum_{k,l=1}^n\hat{a}_{kl}\frac{\partial\hat{\varphi}}{\partial y_k}\frac{\partial\hat{\varphi}}{\partial y_l}&=\sum_{k,l=1}^n(\sum_{i,j=1}^na_{ij}\frac{\partial g_k}{\partial x_i}\frac{\partial g_l}{\partial x_j})(\sum_{r=1}^n\frac{\partial\varphi}{\partial x_r}\frac{\partial f_r}{\partial y_k})(\sum_{s=1}^n\frac{\partial\varphi}{\partial x_s}\frac{\partial f_s}{\partial y_l})\\&=\sum_{i,j=1}^na_{ij}\sum_{r,s=1}^n\frac{\partial\varphi}{\partial x_r}\frac{\partial\varphi}{\partial x_s}(\sum_{k=1}^n\frac{\partial f_r}{\partial y_k}\frac{\partial g_k}{\partial x_i})(\sum_{l=1}^n\frac{\partial f_s}{\partial y_l}\frac{\partial g_l}{\partial x_j})\\&=\sum_{i,j=1}^na_{ij}\sum_{r,s=1}^n\frac{\partial\varphi}{\partial x_r}\frac{\partial\varphi}{\partial x_s}\delta_{ri}\delta_{sj}\\&=\sum_{i,j=1}^na_{ij}\frac{\partial\varphi}{\partial x_i}\frac{\partial\varphi}{\partial x_j}\\&=0\end{align}$

这表明$\hat{\varphi}(y_1,\ldots,y_n)=0$给出了新方程的特征曲面。

习题4.3

4

反设存在$M_0\in S$使$\displaystyle\lim_{\Omega_1\ni M\to M_0}\begin{pmatrix}u_{xx}&u_{xy}\\u_{yx}&u_{yy}\end{pmatrix}\neq\lim_{\Omega_2\ni M\to M_0}\begin{pmatrix}u_{xx}&u_{xy}\\u_{yx}&u_{yy}\end{pmatrix}$，则对$M_0$在$S$的某个邻域内的$M_1\in S\cap B_r(M_0)$都成立$\displaystyle\lim_{\Omega_1\ni M\to M_1}\begin{pmatrix}u_{xx}&u_{xy}\\u_{yx}&u_{yy}\end{pmatrix}\neq\lim_{\Omega_2\ni M\to M_1}\begin{pmatrix}u_{xx}&u_{xy}\\u_{yx}&u_{yy}\end{pmatrix}$，于是这邻域为调和方程的弱间断线。然而调和方程没有弱间断线，这个矛盾表明$u$的二阶偏导数在$S$也连续。