# 能量积分, 唯一性和稳定性习题参考解答
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1、 证明波动方程
\begin\{aligned\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\})=f(x,y,t) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的自由项 $\displaystyle f$ 在 $\displaystyle L^2(K)$ 意义下作微小改变时, 对应的 Cauchy 问题的解 $\displaystyle u(x,y,t)$ 在 $\displaystyle L^2(K)$ 意义下的改变也是微小的, 其中 $\displaystyle K$ 由 (3.17) 式定义.
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle K: (x-x\_0)^2+(y-y\_0)^2\leq a^2(t\_0-t)^2$. 我们让初值为 $\displaystyle 0$, 而 $\displaystyle E(\varOmega\_0)=E\_1(\varOmega\_0)=0$. 设
\begin\{aligned\} E(\varOmega\_t)=\frac\{1\}\{2\}\iint\_\{\varOmega\_t\} u^2\mathrm\{ d\} x\mathrm\{ d\} y =\frac\{1\}\{2\}\int\_0^\{a(t\_0-t)\}\mathrm\{ d\} r \int\_0^\{2\pi r\}u^2\mathrm\{ d\} s, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} E'(\varOmega\_t)&=\int\_0^\{a(t\_0-t)\} \mathrm\{ d\} r\int\_0^\{2\pi r\} uu\_t\mathrm\{ d\} s -\frac\{a\}\{2\}\int\_0^\{2\pi a(t\_0-t)\}u^2|\_\{r=a(t\_0-r)\}\mathrm\{ d\} s\\\\ &=\iint\_\{\varOmega\_t\} uu\_t\mathrm\{ d\} x\mathrm\{ d\} y -\frac\{a\}\{2\}\int\_\{\varGamma\_t\} u^2\mathrm\{ d\} s\\\\ &\leq \iint\_\{\varOmega\_t\}\frac\{u^2+u\_t^2\}\{2\}\mathrm\{ d\} x\mathrm\{ d\} y \leq E(\varOmega\_t)+E\_1(\varOmega\_t).\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle E\_1(\varOmega\_t)$, 类似书定理 5.10 的证明知
\begin\{aligned\} E\_1'(\varOmega\_t)&\leq \iint\_\{\varOmega\_t\} u\_tf\mathrm\{ d\} x\mathrm\{ d\} y \leq \iint\_\{\varOmega\_t\} \frac\{u\_t^2+f^2\}\{2\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &\leq E\_1(\varOmega\_t)+F(t), F(t)=\frac\{1\}\{2\}\iint\_\{\varOmega\_t\}f^2\mathrm\{ d\} x\mathrm\{ d\} y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 Gronwall 不等式知
\begin\{aligned\} E\_1(\varOmega\_t)\leq \int\_0^t F(s)\mathrm\{ d\} s\cdot \mathrm\{e\}^t =\frac\{\mathrm\{e\}^t\}\{2\}\iiint\_K f^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 (I) 变为
\begin\{aligned\} E'(\varOmega\_t)\leq E(\varOmega\_t)+\frac\{\mathrm\{e\}^t\}\{2\}\iiint\_K f^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 Gronwall 不等式知
\begin\{aligned\} E(\varOmega\_t)\leq \iiint\_K f^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} t \cdot \frac\{\mathrm\{e\}^t-1\}\{2\}\cdot \mathrm\{e\}^t \leq \frac\{\mathrm\{e\}^\{2t\}\}\{2\}\iiint\_K f^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
关于 $\displaystyle t$ 在 $\displaystyle [0,t\_0]$ 上积分得
\begin\{aligned\} \iiint\_K u^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} t \leq 2\cdot\frac\{\mathrm\{e\}^\{2t\}-1\}\{4\}\iiint\_K f^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} t \leq \mathrm\{e\}^\{2t\}\iiint\_K f^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ \delta=\frac\{\sqrt\{\varepsilon\}\}\{\mathrm\{e\}^T\} > 0,\mathrm\{ s.t.\} \iiint\_K f^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} t < \delta\Rightarrow \iiint\_K u^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} t < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
2、 利用能量积分法, 证明弦振动方程的混合问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2u\_\{xx\}=f(x,t), 0 < x < l, t > 0,\\\\ u|\_\{t=0\}=\varphi(x), u\_t|\_\{t=0\}=\psi(x), 0\leq x\leq l,\\\\ (u\_x-\sigma u)|\_\{x=0\}=0, u|\_\{x=l\}=0, t\geq 0\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
解的唯一性, 其中 $\displaystyle \sigma$ 为正常数.
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \bar\{u\}$ 也是方程的解, 则令 $\displaystyle v=u-\bar\{u\}$ 后,
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} v\_\{tt\}-a^2v\_\{xx\}=0, 0 < x < l, t > 0,\\\\ v|\_\{t=0\}=0, v\_t|\_\{t=0\}=0, 0\leq x\leq l,\\\\ (v\_x-\sigma v)|\_\{x=0\}=0, v|\_\{x=l\}=0, t\geq 0,\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设
\begin\{aligned\} E\_1(t)=\frac\{1\}\{2\}\int\_0^l (v\_t^2+a^2v\_x^2)\mathrm\{ d\} x+\frac\{a^2\sigma\}\{2\}v^2(0,t), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} E\_1'(t)&=\int\_0^l (v\_tv\_\{tt\}+a^2v\_xv\_\{xt\})\mathrm\{ d\} x+a^2\sigma v(0,t)v\_t(0,t)\\\\ &\stackrel\{v\_\{tt\}=a^2v\_\{xx\}\}\{=\}a^2(v\_xv\_t)|\_\{x=0\}^\{x=l\} +a^2\sigma v(0,t)v\_t(0,t)\\\\ &\stackrel\{v|\_\{x=l\}=0\Rightarrow v\_t|\_\{x=l\}=0\}\{=\} -a^2v\_x(0,t)v\_t(0,t)+a^2\sigma v(0,t)v\_t(0,t) =0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} E\_1(t)=E\_1(0)=0\Rightarrow v\_t=v\_x=0\Rightarrow v\equiv C\stackrel\{v|\_\{x=l\}=0\}\{\Rightarrow\}v\equiv 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
3、 设 $\displaystyle u$ 是带一阶耗散项的波动方程的混合问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2\Delta u+\alpha u\_t=0, (x\_1,\cdots,x\_n)\in \varOmega, t\geq 0,\\\\ u|\_\{t=0\}=\varphi(x\_1,\cdots,x\_n), u\_t|\_\{t=0\}=\psi(x\_1,\cdots,x\_n), (x\_1,\cdots,x\_n)\bar\{\\}varOmega,\\\\ u|\_\{\partial\varOmega\times [0,\infty)\}=0 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的解, 其中 $\displaystyle a > 0, \alpha > 0$ 为常数, $\displaystyle \varOmega\subset \mathbb\{R\}^n$ 为具有光滑边界 $\displaystyle \partial\varOmega$ 的有界区域.
(1)、 试证明其能量积分
\begin\{aligned\} E(t)=\frac\{1\}\{2\}\int\_\{\varOmega\} (u\_t^2+a^2|\nabla u|^2)\mathrm\{ d\} x\_1\cdots \mathrm\{ d\} x\_n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
随时间增加而不增加.
(2)、 证明该问题解的唯一性.
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
(1)、
\begin\{aligned\} E'(t)&=\int\_\varOmega (u\_tu\_\{tt\}+a^2\nabla u\nabla u\_t)\mathrm\{ d\} x\\\\ &\stackrel\{u\_\{tt\}=a^2\Delta u-\alpha u\_t\}\{=\} \int\_\{\varOmega\}(a^2u\_t\Delta u+a^2\nabla u\nabla u\_t-\alpha u\_t^2)\mathrm\{ d\} x\\\\ &=a^2\int\_\varOmega \mathrm\{ div\}(u\_t\nabla u)\mathrm\{ d\} x-\alpha^2\int\_\varOmega u\_t^2\mathrm\{ d\} x\leq a^2\int\_\{\partial\varOmega\}u\_t\frac\{\partial u\}\{\partial \nu\}\mathrm\{ d\} S\\\\ &\stackrel\{u|\_\{\partial\varOmega\times[0,\infty)\}=0\Rightarrow u\_t|\_\{\partial\varOmega\times[0,\infty)\}=0\}\{=\}0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle E(t)\searrow$.
(2)、 设 $\displaystyle \bar\{u\}$ 也是方程的解, 则令 $\displaystyle v=u-\bar\{u\}$,
\begin\{aligned\} E(t)=\frac\{1\}\{2\}\int\_\varOmega (v\_t^2+a^2|\nabla v|^2)\mathrm\{ d\} x \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
后, $\displaystyle E(0)=0$. 由第 1 步知
\begin\{aligned\} &E(t)\leq 0\Rightarrow E(t)=0\Rightarrow v\_t=0, \nabla v=0\Rightarrow v\equiv C \stackrel\{u|\_\{\partial\varOmega\times[0,\infty)\}=0\}\{\Rightarrow\}v\equiv 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
4、 设 $\displaystyle u$ 为混合问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-\sum\_\{i=1\}^n \frac\{\partial\}\{\partial x\_i\}\left(p\_i(x)u\_\{x\_i\}\right) +c^2u=0, (x\_1,\cdots,x\_n)\in\varOmega, t > 0,\\\\ u|\_\{t=0\}=\varphi(x\_1,\cdots,x\_n), u\_t|\_\{t=0\}=\psi(x\_1,\cdots,x\_n), (x\_1,\cdots,x\_n)\in \bar\varOmega,\\\\ u|\_\{\partial\varOmega\times [0,\infty)\}=0 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的解, 其中 $\displaystyle \varOmega\subset \mathbb\{R\}^n$ 为具有光滑边界的有界域, $\displaystyle p\_1,\cdots,p\_n\geq a^2 > 0$ 为适当光滑的函数, $\displaystyle a,c$ 为非零常数. 试证明: 存在常数 $\displaystyle M > 0)$, 使得对任意的 $\displaystyle t > 0$, 有
\begin\{aligned\} \int\_\varOmega \left(u^2+|\nabla u|^2+u\_t^2\right)\mathrm\{ d\} x\_1\cdots \mathrm\{ d\} x\_n \leq M\int\_\{\varOmega\}\left(\varphi^2+|\nabla \varphi|^2+\psi^2\right)\mathrm\{ d\} x\_1\cdots \mathrm\{ d\} x\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 方程两边乘以 $\displaystyle u\_t$, 由 $\displaystyle \int\_\varOmega u\_\{tt\}u\_t\mathrm\{ d\} t=\frac\{1\}\{2\}\frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} t\}\int\_\varOmega u\_t^2\mathrm\{ d\} x$,
\begin\{aligned\} &-\int\_\varOmega \sum\_i \frac\{\partial\}\{\partial x\_i\}\left(p\_i(x)u\_\{x\_i\}\right)u\_t\mathrm\{ d\} x =\int\_\varOmega \sum\_i p\_i(x) u\_\{x\_i\}u\_\{x\_it\}\mathrm\{ d\} x\\\\ =&\int\_\varOmega \sum\_i p\_i(x)\frac\{1\}\{2\}\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}u\_\{x\_i\}^2\mathrm\{ d\} x =\frac\{1\}\{2\}\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\int\_\varOmega \sum\_i p\_i(x)u\_\{x\_i\}^2\mathrm\{ d\} x \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \frac\{1\}\{2\}\frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} t\}\int\_\varOmega u\_t^2\mathrm\{ d\} x +\frac\{1\}\{2\}\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\int\_\varOmega \sum\_i p\_i(x)u\_\{x\_i\}^2\mathrm\{ d\} x +\frac\{c^2\}\{2\}\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\int\_\varOmega u^2\mathrm\{ d\} x=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} &\int\_\varOmega u\_t^2\mathrm\{ d\} x +\int\_\varOmega \sum\_i p\_i(x)u\_\{x\_i\}^2\mathrm\{ d\} x +c^2\int\_\varOmega u^2\mathrm\{ d\} x\\\\ =&\int\_\varOmega \psi^2\mathrm\{ d\} x +\int\_\varOmega \sum\_i p\_i(x)\varphi\_\{x\_i\}^2\mathrm\{ d\} x +c^2\int\_\varOmega \varphi^2\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle A=\min\left\\{1,a^2,c^2\right\\}, B=\max\left\\{1,\max\_\{1\leq i\leq n\atop x\in\varOmega\}p\_i(x), c^2\right\\}$, 则
\begin\{aligned\} A\int\_\varOmega \left(u^2+|\nabla u|^2+u\_t^2\right)\mathrm\{ d\} x\_1\cdots \mathrm\{ d\} x\_n \leq B\int\_\{\varOmega\}\left(\varphi^2+|\nabla \varphi|^2+\psi^2\right)\mathrm\{ d\} x\_1\cdots \mathrm\{ d\} x\_n. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle M=\frac\{B\}\{A\}$ 即知结论成立. 跟锦数学微信公众号. [在线资料](
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---
5、 设有界区域 $\displaystyle \varOmega\subset\mathbb\{R\}^n$ 的边界由 $\displaystyle \varGamma\_0, \varGamma\_1$ 两部分组成, $\displaystyle u$ 为混合问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-\sum\_\{i=1\}^n \frac\{\partial\}\{\partial x\_i\}\left(p\_i(x)u\_\{x\_i\}\right) +c^2u=0, (x\_1,\cdots,x\_n)\in\varOmega, t > 0,\\\\ u|\_\{t=0\}=\varphi(x\_1,\cdots,x\_n), u\_t|\_\{t=0\}=\psi(x\_1,\cdots,x\_n), (x\_1,\cdots,x\_n)\in \bar\varOmega,\\\\ u|\_\{\varGamma\_0\times [0,\infty)\}=0, \left.\left(\frac\{\partial u\}\{\partial \nu\}+\sigma \frac\{\partial u\}\{\partial t\}\right)\right|\_\{\varGamma\_1\times [0,\infty)\}=0 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的解, 其中 $\displaystyle \sigma > 0$ 为常数. 试证明其总能量
\begin\{aligned\} E(t)=\frac\{1\}\{2\}\int\_\varOmega (u\_t^2+a^2|\nabla u|^2)\mathrm\{ d\} x\_1\cdots \mathrm\{ d\} x\_n \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
随时间增加而减少, 由此证明上述问题解的唯一性.
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} E'(t)&=\int\_\varOmega(u\_tu\_\{tt\}+a^2\nabla u\nabla u\_t)\mathrm\{ d\} x =\int\_\varOmega (a^u\_t\Delta u+a^2\nabla u\nabla u\_t)\mathrm\{ d\} x\\\\ &=a^2\int\_\varOmega \mathrm\{ div\}(u\_t\nabla u)\mathrm\{ d\} x =a^2\left\[\int\_\{\varGamma\_0\} u\_t\frac\{\partial u\}\{\partial \nu\}\mathrm\{ d\} S +\int\_\{\varGamma\_1\} u\_t\frac\{\partial u\}\{\partial \nu\}\mathrm\{ d\} S\right\]\\\\ &\stackrel\{u|\_\{\varGamma\_0\}=0\Rightarrow u\_t|\_\{\varGamma\_0\}=0\}\{=\} a^2\int\_\{\varGamma\_1\} u\_t(-\sigma^2u\_t)\mathrm\{ d\} S\leq 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle E(t)\searrow$. 设 $\displaystyle \bar\{u\}$ 也是方程的解, 则令
\begin\{aligned\} v=u-\bar\{u\}, E(t)=\frac\{1\}\{2\}\int\_\varOmega (v\_t^2+a^2|\nabla v|^2)\mathrm\{ d\} x \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
后, $\displaystyle E(0)=0$, 而
\begin\{aligned\} E(t)\leq E(0)=0\Rightarrow E(t)=0\Rightarrow v\_t=0, \nabla v=0\Rightarrow v\equiv C \stackrel\{v|\_\{t=0\}=0\}\{\Rightarrow\}v\equiv 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
6、 设 $\displaystyle \varOmega=\left\\{(x,y)\in\mathbb\{R\}\times \left\\{y > 0\right\\}\right\\}$, $\displaystyle u$ 为无界区域 $\displaystyle \varOmega$ 中的混合问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\})=0, (x,y)\in\varOmega, t > 0,\\\\ u|\_\{t=0\}=\varphi(x,y), u\_t|\_\{t=0\}=\psi(x,y), (x,y)\in\bar\varOmega,\\\\ u|\_\{y=0\}=0 \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的解, 对任意 $\displaystyle (\xi,\eta,T)\in\varOmega\times (0,\infty)$, 记 $\displaystyle \varOmega\_r$ 为过该点的特征锥与平面 $\displaystyle t=\tau$ 的截面, 试证明能量积分
\begin\{aligned\} E(t)=\frac\{1\}\{2\}\iint\_\{\varOmega\_r\cap \left\\{y > 0\right\\}\} \left\[u\_t+a^2(u\_x^2+u\_y^2)\right\]\mathrm\{ d\} x\mathrm\{ d\} y \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
随时间增加而减少 $\displaystyle (t\leq T)$, 由此证明上述问题解的唯一性.
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 当特征锥与 $\displaystyle \left\\{y=0\right\\}$ 无交时, 由书定理 5.10 知 $\displaystyle E(t)\searrow$. 当特征锥与 $\displaystyle \left\\{y=0\right\\}$ 有交时, 若 $\displaystyle t\leq T-\frac\{\eta\}\{a\}$, 则 $\displaystyle \varOmega\_t\cap\left\\{y=0\right\\}$ 无交, 而由书定理 5.10 知 $\displaystyle E(t)\searrow$. 我们只要证明当 $\displaystyle 0 < t < T-\frac\{\eta\}\{a\}$ 时, $\displaystyle E(t)$ 递减即可. 此时, $\displaystyle \varOmega\_t\cap \left\\{y=0\right\\}$ 是一顶’安全帽‘:
\begin\{aligned\} E(t)=&\frac\{1\}\{2\}\int\_0^\{a(T-t)\}\mathrm\{ d\} r \int\_\{-\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\}^\{\pi+\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\} \left\[u\_t^2+a^2|\nabla u|^2\right\]r\mathrm\{ d\} \theta\\\\ &+\frac\{1\}\{2\}\int\_\{\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}-\pi\}^\{-\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\} \mathrm\{ d\}\theta \int\_0^\{-\frac\{\eta\}\{\sin\theta\}\} \left\[u\_t^2+a^2|\nabla u|^2\right\]r\mathrm\{ d\} r. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再求 $\displaystyle E'(t)$ 时, 关于被积函数的导数部分
\begin\{aligned\} f(t)&=\iint\_\{\varOmega\_t\cap \left\\{y > 0\right\\}\} \left\[u\_tu\_\{tt\}+a^2\nabla u\nabla u\_t\right\]\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=\iint\_\{\varOmega\_t\cap \left\\{y > 0\right\\}\} \left\[u\_ta^2u\_\{xx\}+a^2\nabla u\nabla u\_t\right\]\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=a^2\iint\_\{\varOmega\_t\cap \left\\{y > 0\right\\}\} \mathrm\{ div\}(u\_t\nabla u)\mathrm\{ d\} x\mathrm\{ d\} y =a^2\int\_\{\partial \left(\varOmega\_t\cap \left\\{y > 0\right\\}\right)\} u\_t\frac\{\partial u\}\{\partial\nu\}\mathrm\{ d\} s\\\\ &=a^2\int\_\{-\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\}^\{\pi+\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\} u\_t\left\[u\_x\cos(\nu,x) +u\_y\cos(\nu,y)\right\]\cdot a(T-t)\mathrm\{ d\} \theta\\\\ &\quad + \int\_\{\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}-\pi\}^\{-\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\} u\_t(-u\_y)\cdot a(T-t)\mathrm\{ d\} \theta\\\\ &\xlongequal[\Rightarrow u\_t|\_\{y=0\}=0]\{u|\_\{y=0\}=0\}a^2\int\_\{-\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\}^\{\pi+\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\} u\_t\left\[\begin\{array\}\{c\} u\_x\cos(\nu,x)\\\\ +u\_y\cos(\nu,y)\end\{array\}\right\]\cdot a(T-t)\mathrm\{ d\} \theta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而关于积分限的导数部分 [记 $\displaystyle H=\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}$]
\begin\{aligned\} g(t)=&-\frac\{a\}\{2\}\int\_\{-\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\}^\{\pi+\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\} \left\[u\_t^2+a^2|\nabla u|^2\right\]|\_\{r=a(T-t)\} a(T-t)\mathrm\{ d\} \theta\\\\ &+\frac\{1\}\{2\}\int\_0^\{a(T-t)\} r\boxed\{\begin\{array\}\{c\} \left\[u\_t^2+a^2|\nabla u|^2\right\]\_\{\theta=\pi+\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\}\\\\ +\left\[u\_t^2+a^2|\nabla u|^2\right\]\_\{\theta=-\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\}\end\{array\} \}H\mathrm\{ d\} r\\\\ &+\frac\{1\}\{2\}\int\_0^\{a(T-t)\}\boxed\{\begin\{array\}\{c\} \left\[u\_t^2+a^2|\nabla u|^2\right\]\_\{\theta=-\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\}\\\\ +\left\[u\_t^2+a^2|\nabla u|^2\right\]\_\{\theta=\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}-\pi\}\end\{array\}\} r\mathrm\{ d\} r\cdot (-H)\\\\ =&-\frac\{a\}\{2\}\int\_\{-\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\}^\{\pi+\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\} \left\[u\_t^2+a^2|\nabla u|^2\right\]|\_\{r=a(T-t)\} a(T-t)\mathrm\{ d\} \theta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} E'(t)=&-\frac\{a\}\{2\}\int\_\{-\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\}^\{\pi+\arcsin \frac\{\eta\}\{\sqrt\{a^2(T-t)^2-\eta^2\}\}\} \left\\{\begin\{array\}\{c\} \left\[u\_t\cos(\nu,x)+u\_x\right\]^2\\\\ +\left\[u\_t\cos(\nu,y)+u\_y\right\]^2\end\{array\}\right\\}a(T-t)\mathrm\{ d\} \theta\leq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle E(t)\searrow$. 这可用来证明解的唯一性. 事实上, 若 $\displaystyle \bar\{u\}$ 也是解, 则令
\begin\{aligned\} v=u-\bar\{u\}, E(t)=E(t)=\frac\{1\}\{2\}\iint\_\{\varOmega\_r\cap \left\\{y > 0\right\\}\} \left\[u\_t+a^2(u\_x^2+u\_y^2)\right\]\mathrm\{ d\} x\mathrm\{ d\} y \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
后, $\displaystyle E(0)=0$,
\begin\{aligned\} E(t)\leq E(0)=0\Rightarrow E(t)=0\Rightarrow v\_t=0, \nabla v=0\stackrel\{v|\_\{t=0\}=0\}\{\Rightarrow\}v\equiv 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-2-11 07:51:53
