# 极值原理, 唯一性与稳定性习题参考解答
在以下各题中, 如不加特别说明, 均假设 $\displaystyle \varOmega$ 是 $\displaystyle \mathbb\{R\}^n$ 中的有界连通区域.
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1、 试证 Dirichlet 外问题解的稳定性.
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / Dirichlet 外问题为
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \Delta\_nu=0, x\in\varOmega,\\\\ u|\_\{\partial\varOmega\}=\varphi, \lim\_\{|x|\to+\infty\}u=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\}
其中 $\displaystyle \varOmega$ 为有界闭区域. 对 $\displaystyle \forall\ \varepsilon > 0, \exists\ \delta=\varepsilon > 0$, 只要 $\displaystyle \max\_\{\partial\varOmega\}|\varphi| < \delta$, 则对 $\displaystyle \forall\ x\_0\in\varOmega$, 由 $\displaystyle \lim\_\{|x|\to+\infty\}u=0$ 知
\begin\{aligned\} \exists\ R > |x\_0|,\mathrm\{ s.t.\} \forall\ x: |x|\geq R, |u(x)| < \varepsilon; B\_R(0)\subset \overline\{\varOmega^c\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在 $\displaystyle \varOmega \cap K$ 上对调和函数 $\displaystyle u$ 应用极值原理知
\begin\{aligned\} -\varepsilon < \min\_\{\partial(\varOmega\cap K)\}u =\min\_\{\overline\{\varOmega\cap K\}\}u \leq u(x\_0) \leq \max\_\{\overline\{\varOmega\cap K\}\}u =\max\_\{\partial(\varOmega\cap K)\}u < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \sup\_\varOmega |u|\leq \varepsilon$. 跟锦数学微信公众号. [在线资料](
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2、 设 $\displaystyle u\in C^3(\varOmega)\cap C^1(\bar\{\varOmega\})$, $\displaystyle u$ 在 $\displaystyle \varOmega$ 中调和, 则 $\displaystyle |\mathrm\{ grad\} u|$ 必在 $\displaystyle \varOmega$ 的边界上取到极值.
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由习题 6-1 题 10 (2) 知 $\displaystyle v=|\mathrm\{ grad\} u|^2$ 在 $\displaystyle \varOmega$ 上是下调和的. 而
\begin\{aligned\} \max\_\{\bar\{\varOmega\}\}v=\max\_\{\partial \varOmega\}v. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle v$ 的最大值必在边界上达到, $\displaystyle |\mathrm\{ grad\} u|=\sqrt\{v\}$ 必在 $\displaystyle \varOmega$ 的边界上取到最大值, 为极值. 跟锦数学微信公众号. [在线资料](
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3、 设 $\displaystyle u\in C^2(\varOmega)\cap C^1(\bar\{\varOmega\})$, 且满足方程
\begin\{aligned\} \sum\_\{i,j=1\}^n a\_\{ij\}(x)\frac\{\partial^2u\}\{\partial x\_j\partial x\_j\} +\sum\_\{i=1\}^n b\_i(x)\frac\{\partial u\}\{\partial x\_i\} +c(x)u=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle c(x\_1,\cdots,x\_n) < 0$, $\displaystyle b\_i,c\in C(\bar\{\varOmega\})$, 矩阵 $\displaystyle (a\_\{ij\})\_\{n\times n\}$ 是正定的, 即存在常数 $\displaystyle \alpha > 0$, 使得对任意的 $\displaystyle \xi\_i\in\mathbb\{R\}\ (i=1,2,\cdots,n)$, 有
\begin\{aligned\} \sum\_\{i,j=1\}^n a\_\{ij\}\xi\_i\xi\_j\geq \alpha\sum\_\{i=1\}^n \xi\_i^2, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle u$ 除恒等于常数外不能在 $\displaystyle \varOmega$ 的内点达到非负最大值和非正最小值.
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(1)、 类似于球上的 Hopf 引理, 我们有如下结论. 设
(1-1)、 $\displaystyle K$ 是以原点为中心半径为 $\displaystyle R$ 的球;
(1-2)、 $\displaystyle u\in C^2(K)\cap C(\bar\{K\})$ 满足
\begin\{aligned\} \sum\_\{i,j\}a\_\{ij\}u\_\{x\_ix\_j\}+\sum\_i b\_iu\_\{x\_i\}+cu=0, c\leq 0, c\mbox\{有界\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且在 $\displaystyle S=\partial K$ 上某点 $\displaystyle M\_0(x\_0)$ 处取得非正最小 (非负最大) 值, 且
\begin\{aligned\} \forall\ M(x)\in K, u(M) > u(M\_0)\left(u(M) < u(M\_0)\right); \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-3)、 在 $\displaystyle M\_0$ 处, $\displaystyle \mu$ 是与球的内法线方向成锐角的方向向量, 且 $\displaystyle \frac\{\partial u\}\{\partial \mu\}$ 存在.
则 $\displaystyle \frac\{\partial u\}\{\partial \mu\} > 0\ ( < 0)$.
(2)、 用反证法证明题目. 若不恒等于常数的 $\displaystyle u$ 在 $\displaystyle \varOmega$ 的某个内点 $\displaystyle x\_0$ 处取得非负最大值. 设
\begin\{aligned\} \varOmega^-=\left\\{x\in\varOmega; u(x) < M\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle x\_0\in\partial\varOmega^-$. 由 $\displaystyle x\_0\in \varOmega\cap \partial\varOmega^-$ 及 $\displaystyle u\in C^2(\varOmega)$ 知 $\displaystyle \varOmega^-$ 在 $\displaystyle x\_0$ 处具有内部球条件. 由第 1 步知 $\displaystyle \frac\{\partial u\}\{\partial \nu\}(x\_0) < 0$. 这与 $\displaystyle u$ 在 $\displaystyle x\_0\in\varOmega$ 处取得最大值 $\displaystyle \Rightarrow \nabla u(x\_0)=0$ 矛盾. 故有结论. 同理可证 $\displaystyle u$ 除恒等于常数外不能在 $\displaystyle \varOmega$ 的内点达到非正最小值.
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4、 设 $\displaystyle u\in C^2(\varOmega)\cap C(\bar\{\varOmega\})$, 试证问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \Delta\_nu+|\nabla u|^4=u^3, (x\_1,\cdots,x\_n)\in\varOmega\subset \mathbb\{R\}^n,\\\\ u|\_\{\partial\varOmega\}=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\}
只有零解.
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 用反证法. 若 $\displaystyle \exists\ x\_0\in\varOmega,\mathrm\{ s.t.\} u(x\_0)\neq 0$.
(1)、 若 $\displaystyle u(x\_0) > 0$, 则 $\displaystyle u$ 在 $\displaystyle \bar\{\varOmega\}$ 上的正的最大值在 $\displaystyle \varOmega$ 内某点 $\displaystyle x\_1$ 处达到. 由极值条件知在 $\displaystyle x\_1$ 处,
\begin\{aligned\} \Delta\_nu \leq 0, \nabla u=0\Rightarrow u^3=\Delta\_nu+|\nabla u|^4\leq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与 $\displaystyle u(x\_1)\geq u(x\_0) > 0$ 矛盾.
(2)、 若 $\displaystyle u(x\_0) < 0$, 则 $\displaystyle u$ 在 $\displaystyle \bar\{\varOmega\}$ 上的负的最小值在 $\displaystyle \varOmega$ 内某点 $\displaystyle x\_2$ 处达到. 由极值条件知在 $\displaystyle x\_2$ 处,
\begin\{aligned\} \Delta\_nu \geq 0, \nabla u=0\Rightarrow u^3=\Delta\_nu+|\nabla u|^4\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与 $\displaystyle u(x\_2)\leq u(x\_0) < 0$ 矛盾.
不论何种情形, 都得到矛盾. 故有结论. 跟锦数学微信公众号. [在线资料](
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5、 设 $\displaystyle u(x,y)$ 在 $\displaystyle x^2+y^2\leq 1$ 上连续, 且满足
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \Delta\_2u+a\_1(x,y)u\_x+b\_1(x,y)u\_y=0, x^2+y^2 < 1, x < 0,\\\\ \Delta\_2u+a\_2(x,y)u\_x+b\_2(x,y)u\_y=0, x^2+y^2 < 1, x > 0,\\\\ k\_1u\_x(0^-,y)=k\_2u\_x(0^+,y), \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\}
其中 $\displaystyle a\_i(x,y), b\_i(x,y)$ 分别在对应的区域中有界连续, $\displaystyle k\_i > 0$. 试证: 若在圆周 $\displaystyle x^2+y^2=1$ 上 $\displaystyle u(x,y)\geq 0$, 则在圆 $\displaystyle x^2+y^2 < 1$ 内 $\displaystyle u(x,y)\geq 0$.
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 用反证法. 若
\begin\{aligned\} \exists\ (x\_0,y\_0)\in B\_1(0),\mathrm\{ s.t.\} u(x\_0,y\_0) < 0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle u$ 在 $\displaystyle \overline\{B\_1(0)\}$ 上的负的最小值 $\displaystyle m$ 只能在 $\displaystyle \varOmega$ 内某 $\displaystyle (x\_1,y\_1)$ 处达到. 设
\begin\{aligned\} \varOmega^+=\left\\{(x,y)\in B\_1(0); u(x,y) > m\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle (x\_1,y\_1)\in \partial\varOmega^+$.
(1)、 若 $\displaystyle x\_1\neq 0$, 则不妨设 $\displaystyle x\_1 > 0$, 在 $\displaystyle \partial\varOmega^+$ 上点 $\displaystyle (x\_1,y\_1)$ 处应用 Hopf 引理知 $\displaystyle \frac\{\partial u\}\{\partial \nu\}(x\_1,y\_1) < 0$. 这与极值条件矛盾.
(2)、 若 $\displaystyle x\_1=0$, $\displaystyle \varOmega^+$ 在 $\displaystyle x=0$ 的同一侧, 不妨设 $\displaystyle \varOmega^+$ 在 $\displaystyle x > 0$ 上, 则在 $\displaystyle \partial \varOmega^+$ 上点 $\displaystyle (x\_1=0,y\_1)$ 处应用 Hopf 引理知
\begin\{aligned\} \frac\{\partial u\}\{\partial \nu\}(x\_1=0^+,y\_1) < 0 \Rightarrow u\_x(x\_1=0^+,y\_1) > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由边界条件知 $\displaystyle u\_x(x\_1=0^-,y\_1) > 0$. 这与
\begin\{aligned\} x < 0\Rightarrow (x,y)\not\in \varOmega^+\Rightarrow u(x,y)=m\Rightarrow u\_x(x\_1=0^-,y\_1)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
矛盾.
(3)、 若 $\displaystyle x\_1=0$, $\displaystyle \varOmega^+$ 与 $\displaystyle x < 0, x > 0$ 都相交, 则 $\displaystyle \partial\varOmega^+\cap \left\\{x > 0\right\\}\neq 0$. 回到第 1 种情况, 亦得矛盾.
综上, 不论何种情形, 都得矛盾. 故有结论. 跟锦数学微信公众号. [在线资料](
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6、 设 $\displaystyle u\in C^2(\varOmega)\cap C(\bar\{\varOmega\})$, 当 $\displaystyle x\in \varOmega$ 时满足 $\displaystyle \Delta\_nu-u^\{2m\}=0$, $\displaystyle m$ 为正整数, 试证: $\displaystyle u$ 在 $\displaystyle \varOmega$ 内不可能达到最大值, 除非 $\displaystyle u\equiv 0$.
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle \Delta\_nu=u^\{2m\}\geq 0$ 知 $\displaystyle u$ 下调和.
(1)、 若 $\displaystyle u$ 不是常数, 则由强极值原理知 $\displaystyle u$ 不能在 $\displaystyle \varOmega$ 内达到最大值.
(2)、 若 $\displaystyle u$ 是常数, 则 $\displaystyle u^\{2m\}=\Delta\_n u=0\Rightarrow u\equiv 0$.
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7、 举例说明: 方程 $\displaystyle u\_\{xx\}+u\_\{yy\}+cu=0\ (c > 0)$ 不成立最大 (小) 值原理.
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle u(x,y)=\sin x\sin y, (x,y)\in \varOmega=(-\pi,\pi)^2$, 则
\begin\{aligned\} u\_\{xx\}+u\_\{yy\}+2u=0, u|\_\{\partial\varOmega\}=0, u\left(\frac\{\pi\}\{2\},\frac\{\pi\}\{2\}\right)=1, u\left(-\frac\{\pi\}\{2\},\frac\{\pi\}\{2\}\right)=-1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故最值原理不成立. 跟锦数学微信公众号. [在线资料](
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8、 举例说明: 一维波动方程 $\displaystyle u\_\{tt\}-u\_\{xx\}=0$ 不成立最大 (小) 值原理.
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle u(x,t)=\pm (x^2+(t-1)^2), (x,t)\in \varOmega=(-1,1)\times(0,2)$, 则 $\displaystyle u\_\{tt\}-u\_\{xx\}=0$, 但 $\displaystyle u$ 在 $\displaystyle \varOmega$ 上的最小/大值只能在 $\displaystyle (0,1)$ 处取得. 跟锦数学微信公众号. [在线资料](
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9、 假设 $\displaystyle \varOmega\subset\mathbb\{R\}^n$ 是一个有界开集, $\displaystyle u\_i(x)\in C^2(\varOmega)\cap C^1(\bar\{\varOmega\})\ (i=1,2)$ 满足定解问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} -\Delta u\_i+|Du\_i|^2=f\_i(x), x\in\varOmega,\\\\ u=g\_i(x), x\in\partial\varOmega. \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\}
如果 $\displaystyle f\_1(x)\geq f\_2(x)$, $\displaystyle g\_1(x)\geq g\_2(x)$, 则
\begin\{aligned\} u\_1(x)\geq u\_2(x), x\in\bar\varOmega. \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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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle v=u\_1-u\_2$, 则
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} -\Delta v+D(u\_1+u\_2)\cdot Dv=f\equiv f\_1-f\_2\geq 0, x\in \varOmega,\\\\ v=g\equiv g\_1-g\_2, x\in \partial\varOmega.\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\}
由弱极值原理 (书第 183 页推论 3 第 2 条) 知
\begin\{aligned\} \min\_\{\bar\{\varOmega\}\}v=\min\_\{\partial\varOmega\}v\geq 0 \Rightarrow u\_1(x)\geq u\_2(x), x\in\bar\varOmega. \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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10、 设 $\displaystyle u\in C^2(\varOmega)\cap C(\overline\{\varOmega\})$ 是定解问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}-\Delta\_nu+\sin (u\_\{x\_1\}+2u\_\{x\_n\})+c(x)u^3=f^3(x), x\in \varOmega,\\\\ u|\_\{\partial\varOmega\}=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\}
的一个解, 证明: 若 $\displaystyle c(x)\geq c\_0 > 0$, 则
\begin\{aligned\} \max\_\{\overline\{\varOmega\}\}|u(x)|\leq c\_0^\{-\frac\{1\}\{3\}\}\max\_\{\bar\{\varOmega\}\}|f(x)|. \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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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle F=\max\_\{\bar\{\varOmega\}\} |f|$. 由 $\displaystyle u|\_\{\partial\varOmega\}=0$ 知 $\displaystyle \max\_\{\bar\varOmega\}u\geq 0\geq \min\_\{\bar\{\varOmega\}\}u$.
(1)、 若 $\displaystyle \max\_\{\bar\{\varOmega\}\}u=0$, 则
\begin\{aligned\} \max\_\{\bar\{\varOmega\}\}u=0\leq \frac\{1\}\{c\_0\}F^3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 若 $\displaystyle \max\_\{\bar\{\varOmega\}\}u > 0$, 则 $\displaystyle \max\_\{\bar\{\varOmega\}\}u$ 必在 $\displaystyle \varOmega$ 内某 $\displaystyle x\_0$ 处达到, 而 $\displaystyle \nabla u(x\_0)=0, \Delta u(x\_0)\leq 0$,
\begin\{aligned\} &c\_0\max\_\{\bar\{\varOmega\}\}u^3=c\_0u^3(x\_0) \leq c(x\_0)u^3(x\_0)\\\\ =& f^3(x\_0)+\Delta u(x\_0)-\sin (u\_\{x\_1\}+2u\_\{x\_n\})\leq f^3(x\_0)\leq F^3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 若 $\displaystyle \min\_\{\bar\{\varOmega\}\}u=0$, 则
\begin\{aligned\} \min\_\{\bar\{\varOmega\}\}u=0 \leq \frac\{1\}\{c\_0\}F^3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(4)、 若 $\displaystyle \min\_\{\bar\{\varOmega\}\}u < 0$, 则 $\displaystyle \min\_\{\bar\{\varOmega\}\}u$ 必在 $\displaystyle \varOmega$ 内某 $\displaystyle x\_1$ 处达到, 而 $\displaystyle \nabla u(x\_1)=0, \Delta u(x\_1)\geq 0$,
\begin\{aligned\} &c\_0\min\_\{\bar\{\varOmega\}\}u^3=c\_0u^3(x\_0) \geq c(x\_0)u^3(x\_0)\\\\ =&f^3(x\_0)+\Delta u(x\_0)-\sin (u\_\{x\_1\}+2u\_\{x\_n\}) \geq f^3(x\_0)\geq -F^3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle \forall\ x\in\bar\varOmega$,
\begin\{aligned\} -\frac\{1\}\{c\_0\}F^3\leq \min\_\{\bar\{\varOmega\}\}u^3\leq u(x)\leq \max\_\{\bar\{\varOmega\}\}u^3\leq \frac\{1\}\{c\_0\}F^3. \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)/
---
11、 设 $\displaystyle \varOmega=\left\\{(x,y); x^2+y^2 < 1\right\\}$, $\displaystyle u(x,y)\in C^2(\varOmega)\cap C(\bar\{\varOmega\})$ 是问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}u\_\{xx\}+5u\_\{yy\}+\sin(3u\_x-2u\_y)-1=0,&(x,y)\in \varOmega,\\\\ u=\mathrm\{e\}^\{x+y\},&(x,y)\in\partial\varOmega\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, 证明 $\displaystyle u$ 在 $\displaystyle \bar\{\varOmega\}$ 上的最大值只能在 $\displaystyle \varOmega$ 的边界 $\displaystyle \partial\varOmega$ 上达到, 并求出 $\displaystyle u$ 在 $\displaystyle \bar\{\varOmega\}$ 上的最大值.
[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) / 用反证法. 若 $\displaystyle \max\_\{\bar\{\varOmega\}\}u$ 在 $\displaystyle \varOmega$ 内某 $\displaystyle (x\_\star,y\_\star)$ 处取得, 则由极值条件知
\begin\{aligned\} u\_\{xx\}(x\_\star,y\_\star)\leq 0, u\_\{yy\}(x\_\star,y\_\star)\leq 0, u\_x(x\_\star,y\_\star)=0, u\_y(x\_\star,y\_\star)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
代入题中方程得 $\displaystyle 0=\mbox\{右端\}=\mbox\{左端\}\leq -1$. 这是一个矛盾. 故有结论
\begin\{aligned\} \max\_\{\bar\{\varOmega\}\}u=\max\_\{x^2+y^2=1\}\mathrm\{e\}^\{x+y\}=\max\_\{0\leq \theta\leq 2\pi\}\mathrm\{e\}^\{\cos\theta+\sin \theta\} =\max\_\{0\leq \theta\leq 2\pi\}\mathrm\{e\}^\{\sqrt\{2\}\sin\left(\theta+\frac\{\pi\}\{2\}\right)\} =\mathrm\{e\}^\{\sqrt\{2\}\}. \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 08:05:37
