6.2极值原理, 最大模估计, 唯一性和稳定性习题参考解答

zhangzujin

# 极值原理, 最大模估计, 唯一性和稳定性习题参考解答 在以下各题中, 如不加特别说明, 均假设区域 $\displaystyle Q\_T=\left\\{(x,t); 0 < x < l, 0 < t\leq T\right\\}$, $\displaystyle \varGamma\_T$ 是 $\displaystyle Q\_T$ 的抛物边界. --- 1、 设 \begin\{aligned\} \mathcal\{L\} u=u\_t-u\_\{xx\}+f(u\_x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 试证明: 当 $\displaystyle f$ 为凸 (或凹) 函数是, 对于算子 $\displaystyle \mathcal\{L\}$, 比较原理成立, 即设 $\displaystyle u,v\in C^\{2,1\}(Q\_T)\cap C(\bar\{Q\}\_T)$, 当 $\displaystyle \mathcal\{L\} u\leq \mathcal\{L\} v, u|\_\{\varGamma\_T\}\leq v|\_\{\varGamma\_T\}$ 时, 在 $\displaystyle Q\_T$ 上有 $\displaystyle u(x,t)\leq v(x,t)$. [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 $\displaystyle w=u-v$, 则 \begin\{aligned\} w\_t-w\_\{xx\}&=(u\_t-u\_\{xx\})-(v\_t-v\_\{xx\}) =\left\[\mathcal\{L\} u-f(u\_x)\right\]-\left\[\mathcal\{L\} v-f(v\_x)\right\]\\\\ &\leq f(v\_x)-f(u\_x) \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} f'(\xi\_x)(v\_x-u\_x). \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\} w\_t-w\_\{xx\}+c(x,t)w\_x=0, c(x,t)=-f'(\xi\_x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 若 $\displaystyle f$ 凸, 则 $\displaystyle f'\nearrow$, \begin\{aligned\} f'\left( \min\left\\{\begin\{array\}\{c\}\min\_\{0\leq x\leq l\atop 0\leq t\leq T\}u\_x(x,t),\\\\ \min\_\{0\leq x\leq l\atop 0\leq t\leq T\}v\_x(x,t)\end\{array\}\right\\}\right) \leq f'(\xi\_x)\leq f'\left( \max\left\\{\begin\{array\}\{c\}\max\_\{0\leq x\leq l\atop 0\leq t\leq T\}u\_x(x,t),\\\\ \max\_\{0\leq x\leq l\atop 0\leq t\leq T\}v\_x(x,t)\end\{array\}\right\\}\right); \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 若 $\displaystyle f$ 凹, 则 $\displaystyle f'\searrow$, \begin\{aligned\} f'\left( \min\left\\{\begin\{array\}\{c\}\min\_\{0\leq x\leq l\atop 0\leq t\leq T\}u\_x(x,t),\\\\ \min\_\{0\leq x\leq l\atop 0\leq t\leq T\}v\_x(x,t)\end\{array\}\right\\}\right) \geq f'(\xi\_x)\geq f'\left( \max\left\\{\begin\{array\}\{c\}\max\_\{0\leq x\leq l\atop 0\leq t\leq T\}u\_x(x,t),\\\\ \max\_\{0\leq x\leq l\atop 0\leq t\leq T\}v\_x(x,t)\end\{array\}\right\\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 不论何种情形, $\displaystyle c(x,t)$ 都有负的下界. 而由热传导方程的弱极值原理知 \begin\{aligned\} w|\_\{\varGamma\_T\}\leq 0\Rightarrow w|\_\{Q\_T\}\leq 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)/ --- 2、 设 \begin\{aligned\} \mathcal\{L\} u=u\_t-u\_\{xx\}+u^3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 证明对于算子 $\displaystyle \mathcal\{L\}$, 比较原理成立. [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 $\displaystyle \mathcal\{L\} u\leq \mathcal\{L\} v, u|\_\{\varGamma\_T\}\leq v|\_\{\varGamma\_T\}$, $\displaystyle w=u-v$, 则 \begin\{aligned\} w\_t-w\_\{xx\}=&(u\_t-u\_\{xx\})-(v\_t-v\_\{xx\}) =(\mathcal\{L\} u-u^3)-(\mathcal\{L\} v-v^3)\\\\ \leq&v^3-u^3=-(u^2+uv+v^2)w. \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\} &w\_t-w\_\{xx\}+c(x,t)w\leq 0,\\\\ &c(x,t)=u^2+uv+v^2=\left(u-\frac\{v\}\{2\}\right)^2+\frac\{3v^2\}\{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\} 由热传导方程的弱极值原理知 \begin\{aligned\} w|\_\{\varGamma\_T\}\leq 0\Rightarrow w|\_\{Q\_T\}\leq 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)/ --- 3、 设 $\displaystyle u\in C^\{1,0\}(\bar\{Q\}\_T)\cap C^\{2,1\}(Q\_T)$ 满足定解问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t-u\_\{xx\}=0, (x,t)\in Q\_T,\\\\ u(x,0)=0, 0\leq x\leq l,\\\\ \left.\left\[\frac\{\partial u\}\{\partial x\}+h(u\_0-u)\right\]\right|\_\{x=0\}=0, 0\leq t\leq T,\\\\ u|\_\{x=l\}=0, 0\leq t\leq T, \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 $\displaystyle h, u\_0$ 为正常数. 证明: (1)、 $\displaystyle 0\leq u(x,t)\leq u\_0, (x,t)\in Q\_T$; (2)、 $\displaystyle u=u\_h(x,t)$ 关于 $\displaystyle h$ 单调递增. [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)、 由热方程的极值原理知 \begin\{aligned\} \max\_\{\bar\{Q\}\_T\}u=\max\_\{\partial\_pQ\_T\}u, \min\_\{\bar\{Q\}\_T\}u=\min\_\{\partial\_pQ\_T\}u.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle \min\_\{0\leq t\leq T\}u(0,t)=u(0,t\_1)$, $\displaystyle \max\_\{0\leq t\leq T\}u(0,t)=u(0,t\_2)$. (1-1)、 若 $\displaystyle u(0,t\_1) < 0$, 则由 $\displaystyle (I)$ 知第 2 式知 \begin\{aligned\} &u(x,t)\geq u(0,t\_1), \forall\ (x,t)\in Q\_T\\\\ \Rightarrow&\frac\{\partial u\}\{\partial x\}(0,t\_1)=\lim\_\{x\to 0^+\} \frac\{u(x,t\_1)-u(0,t\_1)\}\{x\}\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 \begin\{aligned\} \frac\{\partial u\}\{\partial x\}(0,t\_1)=h [u(0,t\_1)-u\_0] < 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. 故 $\displaystyle u(0,t\_1)\geq 0$, $\displaystyle \min\_\{\bar\{Q\}\_T\}u\geq 0$. (1-2)、 若 $\displaystyle u(0,t\_2) > u\_0$, 则由 $\displaystyle (I)$ 知第 1 式知 \begin\{aligned\} &u(x,t)\leq u(0,t\_2), \forall\ (x,t)\in Q\_T\\\\ \Rightarrow&\frac\{\partial u\}\{\partial x\}(0,t\_2)=\lim\_\{x\to 0^+\}\frac\{u(x,t\_2)-u(0,t\_2)\}\{x\}\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 \begin\{aligned\} \frac\{\partial u\}\{\partial x\}(0,t\_2)=h[u(0,t\_2)-u\_0] > 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. 故 $\displaystyle u(0,t\_2)\leq u\_0$, $\displaystyle \max\_\{\bar\{Q\}\_T\}u=u(0,t\_2)\leq u\_0$. (2)、 对 $\displaystyle 0 < h < k$, 设 $\displaystyle v(x,t)=u\_h(x,t)-u\_k(x,t)$, 则 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} v\_t-v\_\{xx\}=0, (x,t)\in Q\_T,\\\\ v(x,0)=0, 0\leq x\leq l,\\\\ \frac\{\partial v\}\{\partial x\}-hv=(h-k)(u\_h-u\_0),\\\\ v|\_\{x=l\}=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\} 由抛物方程的极大模原理知仅需证明 \begin\{aligned\} &\max\_\{\bar\{Q\}\_T\}v=\max\_\{\partial Q\_T\}v\leq 0 \Leftrightarrow v(0,t)\leq 0, 0\leq t\leq T. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 事实上, 设 $\displaystyle \max\_\{0\leq t\leq T\} v(0,t)=v(0,t\_1)$, 则 \begin\{aligned\} hv(0,t\_1)=\frac\{\partial v\}\{\partial x\}(0,t\_1)+(k-h)(u\_h-u\_0) \leq 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)/ --- 4、 设 $\displaystyle u\in C^\{1,0\}(\bar\{Q\}\_T)\cap C^\{2,1\}(Q\_T)$ 且满足定解问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t-u\_\{xx\}=f(x,t), (x,t)\in Q\_T,\\\\ u(x,0)=\varphi(x), 0\leq x\leq l,\\\\ u|\_\{x=0\}=\mu\_1(t), u|\_\{x=l\}=\mu\_2(t), 0\leq t\leq T,\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\} 则有以下估计: \begin\{aligned\} \max\_\{\bar\{Q\}\_T\} |u\_t(x,t)|\leq C\left\[\begin\{array\}\{c\} \left\Vert f\_t\right\Vert \_\{C(\bar\{Q\}\_T)\} +\left\Vert f\right\Vert \_\{C(\bar\{Q\}\_T)\}\\\\ +\left\Vert \varphi''\right\Vert \_\{C([0,l])\} +\left\Vert \mu\_1'\right\Vert \_\{C([0,T])\} +\left\Vert \mu\_2'\right\Vert \_\{C([0,T])\}\end\{array\}\right\], \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle C$ 仅依赖于 $\displaystyle T$. [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 $\displaystyle v(x,t)=u\_t(x,t)$, 则 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} v\_t-a^2v\_\{xx\}=f(x,t), (x,t)\in Q\_T,\\\\ v|\_\{t=0\}=u\_t|\_\{t=0\}=u\_\{xx\}|\_\{t=0\}=\varphi''(x), 0\leq x\leq l,\\\\ v|\_\{x=0\}=u\_t|\_\{x=0\}=\mu\_1'(t), v|\_\{x=l\}=u\_t|\_\{x=l\}=\mu\_2'(t), 0\leq t\leq T.\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\} 由抛物方程的最大模估计即知 \begin\{aligned\} &\max\_\{\bar\{Q\}\_T\}|u\_t|=\max\_\{\bar\{Q\}\_T\}|v|\\\\ \leq&T \max\_\{\bar\{Q\}\_T\}|f\_t| +\max\left\\{ \max\_\{[0,l]\}|\varphi''|, \max\_\{[0,T]\}|\mu\_1'|, \max\_\{[0,T]\}|\mu\_2'| \right\\}\\\\ \leq& C\left\[\begin\{array\}\{c\} \left\Vert f\_t\right\Vert \_\{C(\bar\{Q\}\_T)\} +\left\Vert f\right\Vert \_\{C(\bar\{Q\}\_T)\}\\\\ +\left\Vert \varphi''\right\Vert \_\{C([0,l])\} +\left\Vert \mu\_1'\right\Vert \_\{C([0,T])\} +\left\Vert \mu\_2'\right\Vert \_\{C([0,T])\}\end\{array\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 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)/ --- 5、 设 $\displaystyle Q\_T=\left\\{(x,t); 0 < x < \pi, 0 < t\leq T\right\\}$, 如果 $\displaystyle u\in C^\{2,1\}(Q\_T)\cap C^\{1,0\}(\bar\{Q\}\_T)$ 满足定解问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}u\_t-2u\_\{xx\}+u\_x^3+u=-1-\sin x,&(x,t)\in Q\_T,\\\\ u|\_\{t=0\}=\cos x-1,&0\leq x\leq\pi,\\\\ u|\_\{x=0\}=u|\_\{x=\pi\}=-t^4,&0\leq t\leq T.\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\} 证明: 在 $\displaystyle \bar\{Q\}\_T$ 上 $\displaystyle u(x,t)\leq 0$. [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 $\displaystyle \exists\ (x\_0,t\_0)\in \overline\{Q\}\_Ts,\mathrm\{ s.t.\} u(x\_0,t\_0) > 0$, 则由 $\displaystyle u|\_\{\partial\_p Q\_T\}\leq 0$ 知 $\displaystyle 0 < u(x\_0,t\_0)=\max\_\{\overline\{Q\}\_T\}u$ 在 $\displaystyle Q\_T$ 内某 $\displaystyle (x\_\star,t\_\star)$ 处达到. 由极值条件知 \begin\{aligned\} u(x\_\star,t\_\star) > 0, u\_t(x\_\star,t\_\star)\geq 0, u\_\{xx\}(x\_\star,t\_\star)\leq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 代入题中第一个等式知 $\displaystyle 0 < \mbox\{左端\}=\mbox\{右端\}\leq 0$. 这是一个矛盾. 故有结论. 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)/ --- 6、 证明 Cauchy 问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}u\_t-a(x,t)u\_\{xx\}+b(x,t)u\_x+c(x,t)u=f(x,t), x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=\varphi(x), x\in\mathbb\{R\}\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\} 的有界解是唯一的, 其中 $\displaystyle a(x,t)\geq a\_0 > 0, c(x,t)\geq 0, a(x,t), b(x,t), c(x,t)$ 有界. [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)、 设 $\displaystyle Q^L\_T=(-L,L)\times (0,T]$, 则 $\displaystyle u$ 在 $\displaystyle \bar\{Q\}^L\_T$ 上的非负最大值 (如果存在的话) 一定在 $\displaystyle \partial\_p Q^L\_T$ 上达到, 即 \begin\{aligned\} \max\_\{\bar\{Q\}^L\_T\}u\leq \max\_\{\partial\_p Q^L\_T\}u^+.\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 $\displaystyle \forall\ \varepsilon > 0$, 令 $\displaystyle v=u-\varepsilon t$, 则 \begin\{aligned\} v\_t-a(x,t)v\_\{xx\}+b(x,t)v\_x+c(x,t)v =f(x,t)-\varepsilon tc(x,t) < 0.\qquad(II) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 往用反证法证明 $\displaystyle v(x,t)$ 在 $\displaystyle \bar\{Q\}^L\_T$ 上的非负最大值如果存在的话, 一定在 $\displaystyle \partial\_p Q^L\_T$ 上达到. 若不然, \begin\{aligned\} \exists\ (x\_0,t\_0)\in Q^L\_T,\mathrm\{ s.t.\} v(x\_0,t\_0)=\max\_\{\bar\{Q\}^L\_T\}v(x,t)\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 由极值条件知在 $\displaystyle (x\_0,t\_0)$ 处, \begin\{aligned\} v\_t\geq 0, v\_x=0, v\_\{xx\}\leq 0\Rightarrow v\_t-av\_\{xx\}+bv\_x+cv\geq c(x\_0,t\_0)M\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 $\displaystyle (II)$ 矛盾. 这就证明了 $\displaystyle (I)$. (2)、 设 \begin\{aligned\} &M=\sup\_\{H\_T\}|u|, A=\sup\_\{H\_T\}|a|, B=\sup\_\{H\_T\}|b|, F=\sup\_\{H\_T\}|f|, B=\sup\_\{\mathbb\{R\}\}|\varphi|,\\\\ &\pm u(x,t)=v(x,t)+Ft+B+\frac\{M\}\{L^2\}(x^2+CLt), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 题中 pde 线性算子为 $\displaystyle \mathcal\{L\}$, 则 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}\mathcal\{L\} v=\pm f-F-\frac\{M\}\{L^2\}(CL-2a+2bx) -c\left\[Ft+B+\frac\{M\}\{L^2\}(x^2+CLt)\right\],\\\\ v(x,0)=\varphi(x)-B-\frac\{M\}\{L^2\}x^2\leq 0,\\\\ v(\pm L,t)=u(\pm L,t)-Ft-B-M-\frac\{M\}\{L\}Ct\leq 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\} 取 $\displaystyle C > 0$ 使得 $\displaystyle C > \frac\{2a\_0+2BL\}\{L\}$, 则 $\displaystyle \mathcal\{L\} v\leq 0$. 由第 1 步知 \begin\{aligned\} \max\_\{\bar\{Q\}^L\_T\}v\leq \max\_\{\partial\_p Q^L\_T\} v\leq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 对 $\displaystyle \forall\ (x,t)\in H\_T$, 对充分的 $\displaystyle L$, $\displaystyle (x,t)\in Q^L\_T$, 而由上式知 \begin\{aligned\} |u(x,t)|\leq FT+B+\frac\{M\}\{L^2\}(x^2+CLt). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 令 $\displaystyle L\to\infty$ 即得 $\displaystyle |u(x,t)|\leq FT+B$. (3)、 回到题目. 设 $\displaystyle \bar\{u\}$ 也是原定解问题的解, 令 $\displaystyle v=u-\bar\{u\}$, 则 $\displaystyle \mathcal\{L\} v=0, v|\_\{t=0\}=0$. 由第 2 步的估计知 $\displaystyle v\equiv 0$. 这就证明了唯一性. 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)/ --- 7、 设 $\displaystyle u(x,t)$ 在 $\displaystyle H\_T=\left\\{x\in \mathbb\{R\}, 0 < t\leq T\right\\}$ 上属于 $\displaystyle C^\{2,1\}(H\_T)\cap C(\bar\{H\}\_T)$ 且 $\displaystyle u(x,t)\geq -m\ (m > 0)$, 又满足方程 \begin\{aligned\} \mathcal\{L\} u=u\_t-a(x,t)u\_\{xx\}+b(x,t)u\_x+c(x,t)u\geq 0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中系数满足 \begin\{aligned\} 0 < a(x,t)\leq M(x^2+1), |b(x,t)|\leq M(|x|+1), |c(x,t)|\leq M, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} $\displaystyle M$ 为常数. 证明: 如果 $\displaystyle u|\_\{t=0\}\geq 0$, 则在 $\displaystyle H\_T$ 上 $\displaystyle u(x,t)\geq 0$. [恕我愚昧, 不知道怎么做哦.] [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) / 恕我愚昧, 不知道怎么做哦. 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)/ --- 8、 设 $\displaystyle C^\{2,1\}(Q\_T)\cap C^\{1,0\}(\bar\{Q\}\_T)$ 是问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t-a^2u\_\{xx\}+b(x,t)u\_x+c(x,t)u=f(x,t), (x,t)\in Q\_T,\\\\ u(x,0)=\varphi(x), 0\leq x\leq l,\\\\ u(0,t)=u(l,t)=0, 0\leq t\leq T \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 $\displaystyle a$ 为正常数. 若 \begin\{aligned\} B=\sup |b(x,t)|, C=\sup |c(x,t)| \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 存在, 则 $\displaystyle u$ 满足以下估计: \begin\{aligned\} &\sup\_\{0\leq t\leq T\}\int\_0^l u^2(x,t)\mathrm\{ d\} x +\int\_0^T \int\_0^l u\_x^2(x,t)\mathrm\{ d\} x\mathrm\{ d\} t\\\\ \leq& M\left\[\int\_0^l u^2(x,t)\mathrm\{ d\} x +\int\_0^T \int\_0^l f^2(x,t)\mathrm\{ d\} x\mathrm\{ d\} t\right\], \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle M$ 只依赖于 $\displaystyle T,l,a,B$ 和 $\displaystyle C$. [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 $\displaystyle u$, 得 \begin\{aligned\} &\frac\{1\}\{2\}\int\_0^l u^2\mathrm\{ d\} x +a^2\int\_0^l u\_x^2\mathrm\{ d\} x =-\int\_0^l bu\_xu\mathrm\{ d\} x -\int\_0^l c u^2\mathrm\{ d\} x +\int\_0^l fu\mathrm\{ d\} x\\\\ \leq&\int\_0^l \left|\frac\{Bu\}\{a\}\right|\cdot |au\_x|\mathrm\{ d\} x+ C\int\_0^l u^2\mathrm\{ d\} x +\int\_0^l \frac\{f^2+u^2\}\{2\}\mathrm\{ d\} x\\\\ \leq &\left(\frac\{B^2\}\{2a^2\}+C+\frac\{1\}\{2\}\right)\int\_0^l u^2\mathrm\{ d\} x+\frac\{a^2\}\{2\}\int\_0^l u\_x^2\mathrm\{ d\} x+\frac\{1\}\{2\}\int\_0^l f^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\} 故 \begin\{aligned\} \frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\int\_0^l u^2\mathrm\{ d\} x +a^2\int\_0^l u\_x^2\mathrm\{ d\} x \leq \left(\frac\{B^2\}\{a^2\}+2C+1\right)\int\_0^l u^2\mathrm\{ d\} x+\frac\{1\}\{2\}\int\_0^lf^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\} 设 \begin\{aligned\} E(t)=\int\_0^l u^2\mathrm\{ d\} x+a^2\int\_0^t \int\_0^l u\_x^2\mathrm\{ d\} x\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'(t)\leq \left(\frac\{B^2\}\{a^2\}+2C+1\right)E(t)+\int\_0^l f^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\} 由 Gronwall 不等式知 \begin\{aligned\} E(t)\leq \mathrm\{e\}^\{\left(\frac\{B^2\}\{a^2\}+2C+1\right)t\}\left\[E(0) +\int\_0^t \int\_0^l f^2\mathrm\{ d\} x\mathrm\{ d\} s\right\], \forall\ 0\leq t\leq T. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 令 $\displaystyle K=\min\left\\{1,a^2\right\\}, L=\mathrm\{e\}^\{\frac\{B^2\}\{a^2\}+2C+1\}, M=\frac\{L\}\{K\}$, 则 \begin\{aligned\} &\sup\_\{0\leq t\leq T\}\int\_0^l u^2(x,t)\mathrm\{ d\} x +\int\_0^T \int\_0^l u\_x^2(x,t)\mathrm\{ d\} x\mathrm\{ d\} t\\\\ \leq& M\left\[\int\_0^l u^2(x,t)\mathrm\{ d\} x +\int\_0^T \int\_0^l f^2(x,t)\mathrm\{ d\} x\mathrm\{ d\} t\right\]. \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)/ --- 9、 设 $\displaystyle C^\{2,1\}(Q\_T)\cap C^\{1,0\}(\bar\{Q\}\_T)$ 满足以下定解问题: \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t-a(x,t)u\_\{xx\}+b(x,t)u\_x+c(x,t)u=f(x,t), (x,t)\in Q\_T,\\\\ u(x,0)=\varphi(x), 0\leq x\leq l,\\\\ \left.\left\[-\frac\{\partial u\}\{\partial x\}+\alpha u\right\]\right|\_\{x=0\} =\left.\left\[\frac\{\partial u\}\{\partial x\}+\beta u\right\]\right|\_\{x=l\}=0, 0\leq t\leq T, \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 $\displaystyle a(x,t)\geq a\_0 > 0, \alpha\geq 0, \beta\geq 0$, $\displaystyle a(x,t), a\_x(x,t), b(x,t), c(x,t)$ 在 $\displaystyle Q\_T$ 上有界, 证明 \begin\{aligned\} &\sup\_\{0\leq t\leq T\}\int\_0^l u^2(x,t)\mathrm\{ d\} x +\int\_0^T \int\_0^l u\_x^2(x,t)\mathrm\{ d\} x\mathrm\{ d\} t\\\\ \leq& M\left\[\int\_0^l u^2(x,t)\mathrm\{ d\} x +\int\_0^T \int\_0^l f^2(x,t)\mathrm\{ d\} x\mathrm\{ d\} t\right\], \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 其中 $\displaystyle M$ 只依赖于 $\displaystyle T,a\_0$ 以及 $\displaystyle a,b,c$ 的界. [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 $\displaystyle A=\max\_\{\bar\{Q\}\_T\}\left\\{|a|,|a\_x|,|b|,|c|\right\\}$, 则由 \begin\{aligned\} &-\int\_0^l au\_\{xx\}u\mathrm\{ d\} x =-\int\_0^l au\mathrm\{ d\} u\_x \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} -\left\[auu\_x|\_0^l-\int\_0^l (a\_xu+au\_x)u\_x\mathrm\{ d\} x\right\]\\\\ =&-a(l,t)u(l,t)u\_x(l,t) +a(0,t)u(0,t)u\_x(0,t) +\int\_0^l a\_xuu\_x\mathrm\{ d\} x +\int\_0^l au\_x^2\mathrm\{ d\} x\\\\ =&-a(l,t)u(l,t)[-\beta u(l,t)] +a(0,t)u(0,t)[\alpha u(0,t)]\\\\ &+\int\_0^l a\_xuu\_x\mathrm\{ d\} x +\int\_0^l au\_x^2\mathrm\{ d\} x\\\\ \geq&a\_0\int\_0^l u\_x^2\mathrm\{ d\} x -A\int\_0^l |uu\_x|\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\} 知方程两边 $\displaystyle \cdot u$ 得 \begin\{aligned\} &\frac\{1\}\{2\}\frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} t\}\int\_0^l u^2\mathrm\{ d\} x +a\_0\int\_0^l u\_x^2\mathrm\{ d\} x-A\int\_0^l |uu\_x|\mathrm\{ d\} x\\\\ \leq&\int\_0^l (f-bu\_x-cu)u\mathrm\{ d\} x \leq A\int\_0^l u^2\mathrm\{ d\} x +A\int\_0^l |uu\_x|\mathrm\{ d\} x+\int\_0^l fu\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\} 故 \begin\{aligned\} &\frac\{1\}\{2\}\frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} t\}\int\_0^l u^2\mathrm\{ d\} x +a\_0\int\_0^l u\_x^2\mathrm\{ d\} x\\\\ \leq& A \int\_0^l u^2\mathrm\{ d\} x +\int\_0^l \left|\frac\{2A\}\{\sqrt\{a\_0\}\}u\right| \cdot \left|\sqrt\{a\_0\}u\right|\mathrm\{ d\} x +\int\_0^l fu\mathrm\{ d\} x\\\\ \leq&A\int\_0^l u^2\mathrm\{ d\} x +\int\_0^l \left\[\frac\{1\}\{2\}\frac\{4A^2\}\{a\_0\}u^2+\frac\{a\_0\}\{2\}u\_x^2\right\]\mathrm\{ d\} x +\int\_0^l \frac\{f^2+u^2\}\{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\} 于是 \begin\{aligned\} \frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} t\}\int\_0^l u^2\mathrm\{ d\} x +a\_0\int\_0^l u\_x^2\mathrm\{ d\} x \leq \left(2A+\frac\{4A^2\}\{a\_0\}+1\right) \int\_0^l u^2\mathrm\{ d\} x +\int\_0^l f^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\} 设 \begin\{aligned\} E(t)=\int\_0^l u^2(x,t)\mathrm\{ d\} x+a\_0\int\_0^t \int\_0^l u\_x^2(x,s)\mathrm\{ d\} x\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'(t)\leq \left(2A+\frac\{4A^2\}\{a\_0\}+1\right)E(t) +\int\_0^l f^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\} 由 Gronwall 不等式, \begin\{aligned\} E(t)\leq \mathrm\{e\}^\{\left(2A+\frac\{4A^2\}\{a\_0\}+1\right)t\}\left\[E(0)+\int\_0^t \int\_0^l f^2\mathrm\{ d\} x\mathrm\{ d\} s\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 设 $\displaystyle K=\min\left\\{1,a\_0\right\\}$, $\displaystyle L=\mathrm\{e\}^\{\left(2A+\frac\{4A^2\}\{a\_0\}+1\right)T\}, M=\frac\{L\}\{K\}$, 则 \begin\{aligned\} &\sup\_\{0\leq t\leq T\}\int\_0^l u^2(x,t)\mathrm\{ d\} x +\int\_0^T \int\_0^l u\_x^2(x,t)\mathrm\{ d\} x\mathrm\{ d\} t\\\\ \leq& M\left\[\int\_0^l u^2(x,t)\mathrm\{ d\} x +\int\_0^T \int\_0^l f^2(x,t)\mathrm\{ d\} x\mathrm\{ d\} t\right\], \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)/ --- 10、 设 $\displaystyle Q=\left\\{(x,t); 0 < x < l, 0 < t < T\right\\}$, $\displaystyle u\in C^\{2,1\}(Q)\cap C(\bar\{Q\})$ 是方程组 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t+\left\[u(u^2+v^2)\right\]\_x=u\_\{xx\},\\\\ v\_t+\left\[v(u^2+v^2)\right\]\_x=v\_\{xx\} \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\} 的解且满足 $\displaystyle u(0,t)=u(l,t)=v(0,t)=v(l,t)=0$. 证明其能量 \begin\{aligned\} E(t)=\int\_0^l (u^2+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\} 关于时间 $\displaystyle t$ 单调递减. [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)=&2\int\_0^l \left(uu\_t+vv\_t\right)\mathrm\{ d\} x =2\int\_0^l \left\\{\begin\{array\}\{c\} u\left\[u\_\{xx\}-\left(u(u^2+v^2)\right)\_x\right\]\\\\ +v\left\[v\_\{xx\}-\left(v(u^2+v^2)\right)\_x\right\]\end\{array\}\right\\}\mathrm\{ d\} x\\\\ =&-2\int\_0^l (u\_x^2+v\_x^2)\mathrm\{ d\} x -2\int\_0^l (u^2+v^2)(u^2+v^2)\_x\mathrm\{ d\} x\\\\ =&-2\int\_0^l (u\_x^2+v\_x^2)\mathrm\{ d\} x-(u^2+v^2)^2|\_\{x=0\}^\{x=l\}\\\\ =&-2\int\_0^l (u\_x^2+v\_x^2)\mathrm\{ d\} x\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 $\displaystyle E(t)\searrow$. 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)/