PDE第一次作业

第一题

证明：

（1）

$$\begin{align*} \int_{\Omega}u_{x_{i}}v dx &= -\int_{\Omega} u\cdot v_{x_{i}} dx + \int_{\Omega}\nabla_{x_{i}}(uv) dx \\ &= -\int_{\Omega} u\cdot v_{x_{i}} dx + \int_{\partial\Omega}uv\cdot\vec{n}_{i} dS \end{align*}$$

（2）

$$\begin{align*} \int_{\Omega}\Delta u dx=&\int_{\Omega} \nabla \cdot(\nabla u) dx=\int_{\partial\Omega}\nabla u\cdot\vec{n} dS \\ =&\int_{\partial\Omega}\frac{\partial u}{\partial\overrightarrow{n}} dS \end{align*}$$

（3）

$$\begin{align*} \int_{\Omega}\nabla u\cdot\nabla v dx = & - \int_{\Omega} u\Delta v dx + \int_{\Omega}\nabla\left(u\cdot\nabla v\right)dx \\ =& - \int_{\Omega} u\Delta v dx + \int_{\partial\Omega}u \nabla v\cdot\vec{n} dS \\ =& - \int_{\Omega} u\Delta v dx + \int_{\partial\Omega}u\frac{\partial v}{\partial\overrightarrow{n}}dS \end{align*}$$

（4）

$$\begin{align*} \int_{\Omega}u\Delta v-v\Delta u dx=&\int_{\Omega}u\Delta v dx-\int_{\Omega}v\Delta u dx \\ =&\left(\int_{\partial\Omega}u\frac{\partial v}{\partial\overrightarrow{n}}dS-\int_{\Omega}\nabla u\cdot \nabla v dx\right)-\left(\int_{\partial\Omega}v \frac{\partial u}{\partial\overrightarrow{n}}dS-\int_{\Omega}\nabla u\cdot\nabla v dx\right) \\ =&\int_{\partial\Omega}\left(u\frac{\partial V}{\partial\vec{n}}-v \frac{\partial U}{\partial\vec{n}}\right)dS \end{align*}$$

第二题

T12

法1

（Fzk同学先教会了我T13的解法，我仿照T13的思路凑出了这个解法）

解： 由分部积分，

$$\begin{equation*} \int_{0}^{1}(2x+1)y^{\prime}dx+\int_{0}^{1}2ydx=(2x+1)y|_{0}^{1}=3\cdot y(1)-y(0) = -y(0) \end{equation*}$$

可得，

$$\begin{equation*} -\int_{0}^{1}2ydx=y(0)+\int_{0}^{1}(2x+1)y^{\prime}dx \end{equation*}$$

代入得，

$$\begin{align*} J(y)&=\frac{1}{2}\int_{0}^{1}(y^{\prime})^{2}dx-2\int_{0}^{1}ydx-y(0)\\ &=\frac{1}{2}\int_{0}^{1}(y^{\prime})^{2}dx+\int_{0}^{1}(2x+1)y^{\prime}dx+\frac{1}{2}\int_{0}^{1}(2x+1)^{2}dx -\frac{1}{2}\int_{0}^{1}(2x+1)^{2}dx\\ &=\frac{1}{2}\int_{0}^{1}(y^{\prime}+(2x+1))^{2}dx-\frac{13}{6} \\ &\geq-\frac{13}{6} \end{align*}$$

于是，

$$\begin{equation*} \text{等号成立} \Leftrightarrow \begin{cases} y^{\prime}=-2x-1\\ y(1)=0 \end{cases}, \forall x \in [0,1] \Leftrightarrow \begin{cases} y=-x^{2}-x+c\\ -1-1+c=0 \end{cases} \Leftrightarrow u = -x^{2}-x+2 \end{equation*}$$

法2

（解法 from Yjy同学）

T13

法1

（解法 from Fzk同学）

解：
首先，有分部积分，

$$\begin{equation*} \int_{0}^{1}y y^{\prime} dx=\frac{1}{2} y^{2} |_{0}^{1} = \frac{1}{2} [y^{2}(1)-y^{2}(0)] \end{equation*}$$

代入得，

$$\begin{align*} J(y)&=\frac{1}{2}\int_{0}^{1}(y^{2}+y^{2}) dx+\frac{1}{2}[y^{2}(0)+y^{2}(1)]-2y(0)\\ &=\frac{1}{2}\int_{0}^{1}(y+y^{\prime})^{2}-2yy^{\prime} dx+\frac{1}{2}[y^{2}(0)+y^{2}(1)]-2y(0)\\ &=\frac{1}{2}\int_{0}^{1}(y+y^{\prime})^{2} dx-\int_{0}^{1}yy^{\prime}dx+\frac{1}{2}[y^{2}(0)+y^{2}(1)]-2y(0)\\ &=\frac{1}{2}\int_{0}^{1}(y+y^{\prime})^{2}dx+y^{2}(0)-2y(0)\\ &=\frac{1}{2}\int_{0}^{1}(y+y^{\prime})^{2}dx+(y(0)-1)^{2}+1\\ &\geq 1 \end{align*}$$

于是有，

$$\begin{equation*} \text{等号成立} \Leftrightarrow \begin{cases} y + y^{\prime}=0 \\ y(0)=1 \end{cases}, \forall x \in [0,1] \Leftrightarrow \left.\left\{ \begin{array}{l} {u^{\prime}+u=0}\\ {u(0)=1}\\ \end{array} \right.\right. \Leftrightarrow u=e^{-x} \end{equation*}$$

法2

（解法 from Yjy同学）

T14

第一问

证明： 令

$$\begin{equation*} j(\epsilon)=J(u+\epsilon v), \quad v\in M. \end{equation*}$$

则有，

$$\begin{align*} j(\epsilon)=&J(u+\epsilon v) \\ =&\frac{1}{2}\int_{\Omega}\left(\left|\nabla(u+\epsilon v)\right|^{2}+\left(u+\epsilon v\right)^{2}\right)dx+\frac{1}{2}\int_{\partial\Omega}\alpha(x)\cdot\left(u+\epsilon v\right)^{2}ds\\ &-\int_{\Omega}f(u+\epsilon v)dx-\int_{\partial\Omega}g(u+\epsilon v)ds\\ =&\frac{1}{2}\int_{\Omega}\left(|\nabla u|^{2}+\epsilon^{2}|\nabla v|^{2}+2\epsilon\cdot\nabla u\cdot\nabla v+u^{2}+2\epsilon uv+v^{2}\right)dx\\ &+\frac{1}{2}\int_{\partial\Omega}\alpha(x)\left(u^{2}+2\epsilon\cdot uv+\epsilon^{2}v^{2}\right)ds-\int_{\Omega}f(u+\epsilon v)dx-\int_{\partial\Omega}g(u+\epsilon v)ds \end{align*}$$

对 $\epsilon$ 求导，可得，

$$\begin{align*} j^{\prime}(\epsilon)=&\int_{\Omega}\left(\epsilon|\nabla v|^{2}+\nabla u\cdot\nabla v +uv\right)dx+\int_{\partial\Omega}\alpha(x)\left(uv+\epsilon v^{2}\right)ds\\ &-\int_{\Omega}fv dx-\int_{\partial\Omega}gv ds \end{align*}$$

于是，

$$\begin{align*} \text{问题1} &\Leftrightarrow j^{\prime}(0)=0 \\ &\Leftrightarrow \int_{\Omega}\left(\nabla u\cdot\nabla v+uv-fv\right)dx+\int_{\partial\Omega}(\alpha(x)uv-gv)ds=0 , \forall v\in M \\ &\Leftrightarrow \text{问题2} \end{align*}$$

第二问

证明： 从第一问可得：$\text{问题1} \Leftrightarrow \text{问题2}$

于是只需证：
$\text{问题2} \Leftrightarrow \text{问题3}$

由 $u \in C^2(\Omega) \cup C^1(\overline{\Omega})$，使用分部积分以及 Gauss-Green 公式，

$$\begin{align*} &\int_{\Omega}\left(\nabla u\cdot\nabla v+v\cdot\Delta u\right)dx=\int_{\Omega}\nabla\left(v\cdot\nabla u\right)dx=\int_{\partial\Omega}v\cdot\nabla u\cdot\vec{n} ds \\ \Rightarrow &\int_{\Omega}\nabla u\cdot\nabla v dx=\int_{\partial\Omega}v\frac{\partial u}{\partial\vec{n}} ds-\int_{\Omega}v\Delta u dx \end{align*}$$

于是，

$$\begin{align*} \text{问题2}&\Leftrightarrow \int_{\Omega}\left(\nabla u\cdot\nabla v+uv-fv\right)dx+\int_{\partial\Omega}\left(\alpha(x)uv-gv\right)ds=0 , \forall v\in M .\\ &\Leftrightarrow \int_{\Omega}\left(u-\Delta u-f\right)v dx+\int_{\partial\Omega}\left(\frac{\partial u}{\partial \vec{n}}+\alpha(x)u-g\right)v ds , \forall v\in M .\\ &\Leftrightarrow \begin{cases} -\Delta u+u=f&, \forall x\in\Omega\\ \frac{\partial u}{\partial \vec{n}}+\alpha(x)u=g &, \forall x\in\partial\Omega \end{cases} \\ &\Leftrightarrow\text{问题3} \end{align*}$$

第三题

T16

第一问

证明： 由

$$\begin{equation*} \frac{\partial u}{\partial t}=\frac{\partial u}{\partial\xi}\cdot\frac{\partial\xi}{\partial t}+\frac{\partial u}{\partial\eta}\cdot\frac{\partial\eta}{\partial t}=\frac{\partial u}{\partial \xi}(-\alpha)+\frac{\partial u}{\partial \eta}(\alpha) \end{equation*}$$

有，

$$\begin{align*} \frac{\partial^2 u}{\partial t^{2}}&=\frac{\partial}{\partial t}\left(-a\frac{\partial u}{\partial\xi}+a\frac{\partial u}{\partial\eta}\right)\\ &=(-a)\left(\frac{\partial^2 u}{\partial\xi^{2}}\frac{\partial\xi}{\partial t}+\frac{\partial^2 u}{\partial\eta\partial\xi}\frac{\partial\eta}{\partial t}\right)+a\left(\frac{\partial^2 u}{\partial\eta^{2}}\frac{\partial\eta}{\partial t}+\frac{\partial^2 u}{\partial\xi\partial\eta}\frac{\partial\xi}{\partial t}\right)\\ &=(-a)\left[\frac{\partial^2 u}{\partial\xi^{2}}(-a)+\frac{\partial^2 u}{\partial\eta\partial\xi}(a)\right]+a\left[\frac{\partial^2 u}{\partial\eta^{2}}(a)+\frac{\partial^2 u}{\partial\xi\partial\eta}(-a)\right]\\ &=a^2 \left(\frac{\partial^2 u}{\partial\xi^{2}}-2\frac{\partial^2 u}{\partial\eta\partial\xi}+\frac{\partial^2 u}{\partial\eta^{2}}\right) \end{align*}$$

又

$$\begin{equation*} \frac{\partial^2 u}{\partial x}=\frac{\partial^2 u}{\partial\xi}\cdot\frac{\partial\xi}{\partial x}+\frac{\partial^2 u}{\partial \eta}\cdot\frac{\partial\eta}{\partial x}=\frac{\partial u}{\partial\xi}+\frac{\partial u}{\partial\eta} \end{equation*}$$

得

$$\begin{align*} \frac{\partial^2 u}{\partial x^{2}}&=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial\xi}+\frac{\partial u}{\partial\eta}\right)\\ &=\left(\frac{\partial^2 u}{\partial\xi^{2}}\cdot\frac{\partial\xi}{\partial x}+\frac{\partial^2 u}{\partial\eta\partial\xi}\cdot\frac{\partial\eta}{\partial x}\right)+\left(\frac{\partial^2 u}{\partial\eta^{2}}\cdot\frac{\partial\eta}{\partial x}+\frac{\partial^2 u}{\partial\eta \partial \xi}\cdot\frac{\partial\xi}{\partial x}\right)\\ &=\frac{\partial^2 u}{\partial\xi^{2}}+2\frac{\partial^2 u}{\partial\eta\partial\xi}+\frac{\partial^2 u}{\partial\eta^{2}} \end{align*}$$

代入 $u_{tt} - a^2 \cdot u_{xx} = 0$，有：

$$\begin{align*} a^2 \left(\frac{\partial^2 u}{\partial\xi_{0}^{2}}-2\frac{\partial^2 u}{\partial\eta\partial\xi}+\frac{\partial^2 u}{\partial\eta^{2}}\right)- a^{2}\left(\frac{\partial^2 u}{\partial\xi^{2}}+2\frac{\partial^2 u}{\partial\eta\partial\xi}+\frac{\partial^2 u}{\partial\eta^{2}}\right)=0 \Rightarrow\frac{\partial^2 u}{\partial\eta\partial\xi}=0 \end{align*}$$

于是可得通解，

$$\begin{align*} \int\frac{\partial^2 u}{\partial\eta\partial\xi}d\eta=F_{1}(\xi)&\Rightarrow u= \int F_{1}(\xi)d\xi+G(\eta)=F(\xi)+G(\eta)\\ &\Rightarrow u(x,t)=F(x-at)+G(x+at) \end{align*}$$

第二问

证明： 首先，

$$\begin{align*} &\frac{\partial u}{\partial t}=\frac{\partial u}{\partial\xi}\cdot\frac{\partial\xi}{\partial t}+\frac{\partial u}{\partial\tau}\cdot\frac{\partial\tau}{\partial t}=\frac{\partial u}{\partial\xi}(-\alpha)+\frac{\partial u}{\partial\tau}\\ &\frac{\partial u}{\partial x}=\frac{\partial u}{\partial\xi}\cdot\frac{\partial\xi}{\partial x}+\frac{\partial u}{\partial\tau}\cdot\frac{\partial\tau}{\partial x}=\frac{\partial u}{\partial\xi}\\ &\frac{\partial^2 u}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{\partial^2 u}{\partial\xi}\right)=\frac{\partial^2 u}{\partial\xi^{2}}\cdot\frac{\partial\xi}{\partial x}+\frac{\partial^2 u}{\partial\tau\partial\xi}\cdot\frac{\partial\tau}{\partial x}=\frac{\partial^{2}u}{\partial\xi^{2}} \end{align*}$$

代入原式，

$$\begin{align*} u_{t}+\alpha u_{x}=a^2 u_{xx} \Leftrightarrow& (-\alpha) u_{\xi}+u_{\tau}+\alpha (u_{\xi} )=a^{2} u_{\xi\xi}\\ \Leftrightarrow& u_{\tau}= a^2 u_{\xi\xi} \end{align*}$$

T17

证明： 首先，

$$\begin{align*} \frac{\partial\widetilde{u}}{\partial x_{i}}&=\frac{\partial\widetilde{u}}{\partial r}\cdot\frac{\partial r}{\partial x_{i}}=\widetilde{u}^{\prime}\cdot\frac{x_{i}}{r}\\ \frac{\partial^{2}\widetilde{u}}{\partial x_{i}^{2}}&=\frac{\partial}{\partial x_{i}}\left(\widetilde{u}^{\prime}\cdot\frac{x_{i}}{r}\right)=\frac{\partial}{\partial r}\left(\widetilde{u}^{\prime}\cdot\frac{x_{i}}{r}\right)\cdot\frac{\partial r}{\partial x_{i}}\\ &=\left(\widetilde{u}^{\prime\prime}\cdot\frac{x_{i}}{r}+\widetilde{u}^{\prime}\cdot\frac{r\cdot\frac{r}{x_i} - x_{i}}{r^{2}}\right)\cdot\frac{x_{i}}{r}\\ &=\widetilde{u}^{\prime\prime}\cdot\frac{x_{i}^{2}}{r^{2}}+\widetilde{u}^{\prime}\cdot\frac{r^2 - x_{i}^{2}}{r^{3}} \end{align*}$$

因此，

$$\begin{align*} 0 &=\Delta u=\left(\sum_{i=1}^{n}\frac{\partial^{2}}{\partial x_{i}^{2}}\right)u=\left(\sum_{i=1}^{n}\frac{\partial^{2}}{\partial x_{i}^{2}}\right)\widetilde{u}\\ &=\widetilde{u}^{\prime\prime}\cdot\sum_{i=1}^{n}\frac{x_{i}^{2}}{r^{2}}+\widetilde{u}^{\prime}\cdot\sum_{i=1}^{n}\frac{r^2 - x_{i}^{2}}{r^{3}}\\ &=\widetilde{u}^{\prime\prime}+\widetilde{u}^{\prime}\cdot\frac{n-1}{r} \end{align*}$$

得到 $\widetilde{u}(r)$ 对应得常微分方程，

$$\begin{equation*} \widetilde{u}^{\prime\prime}+\frac{n-1}{r}\widetilde{u}^{\prime}=0 \end{equation*}$$

第四题

解：
首先，考虑二阶项系数构成的矩阵之特征值，设特征值为 $\lambda$，则有，

$$\begin{align*} \begin{vmatrix} A-\lambda&B\\ B&C-\lambda \end{vmatrix} &=(A-\lambda)(C-\lambda)-B^{2}\\ &=\lambda^{2}-(A+C)+(AC-B^{2}) \end{align*}$$

由韦达定理，

$$\begin{equation*} \begin{cases}\lambda_{1}+\lambda_{2}=A+C\\\lambda_{1}\lambda_{2}=AC-B^{2}\end{cases} \end{equation*}$$

上述方程的判别式为，

$$\begin{align*} \Delta&=(A+C)^{2}-4(AC-B^{2})\\ &=(A-C)^{2}+4B^{2}\geq0 \end{align*}$$

因此，$\lambda_{1},\lambda_{2}$ 都是实根。

情形一：两特征值全负

此时有，

$$\begin{equation*} \begin{cases}A+C<0\\AC-B^{2}>0\end{cases} \end{equation*}$$

在此情况下，上述二阶偏微分方程是椭圆型方程。

情形二：两特征值一0一负

此时有，

$$\begin{equation*} \begin{cases}A+C<0\\AC-B^{2}=0\end{cases} \end{equation*}$$

在此情况下，上述二阶偏微分方程是抛物型方程。

情形三：两特征值一正一负

此时有，

$$\begin{equation*} AC-B^{2}<0 \end{equation*}$$

在此情况下，上述二阶偏微分方程是双曲型方程。

情形四：以上均不是

此时类型不为以上三种。