6.1热传导方程定解问题的求解习题参考解答

zhangzujin

# 热传导方程定解问题的求解习题参考解答 --- 1、 求解热传导方程 \begin\{aligned\} u\_t-u\_\{xx\}=0, x\in\mathbb\{R\}, t > 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 的 Cauchy 问题, 已知初始条件为 (1)、 $\displaystyle u|\_\{t=0\}=\sin x$; [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) / \begin\{aligned\} &u(x,t)=\frac\{1\}\{2\sqrt\{\pi t\}\}\int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-\frac\{(x-y)^2\}\{4t\}\} \sin y\mathrm\{ d\} y\\\\ \stackrel\{y-x=\xi\}\{=\}&\frac\{1\}\{2\sqrt\{\pi t\}\} \int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-\frac\{\xi^2\}\{4t\}\} \sin(x+\xi)\mathrm\{ d\} \xi\\\\ =&\sin x\left\[\frac\{1\}\{2\sqrt\{\pi t\}\}\int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-\frac\{\xi^2\}\{4t\}\}\cos\xi\mathrm\{ d\} \xi\right\] +\cos x\left\[\frac\{1\}\{2\sqrt\{\pi t\}\}\int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-\frac\{\xi^2\}\{4t\}\}\sin \xi\mathrm\{ d\} \xi\right\]\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&\sin x\cdot \frac\{1\}\{\sqrt\{\pi t\}\}\int\_0^\infty \mathrm\{e\}^\{-\frac\{\xi^2\}\{4t\}\}\cos\xi\mathrm\{ d\} \xi\\\\ \stackrel\{\frac\{\xi\}\{2\sqrt\{t\}\}=u\}\{=\}& \sin x\cdot \frac\{2\}\{\sqrt\{\pi\}\}\int\_0^\infty \mathrm\{e\}^\{-u^2\} \cos \left(2\sqrt\{t\}u\right)\mathrm\{ d\} u =\mathrm\{e\}^\{-t\}\sin x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 最后一步我们利用了 \begin\{aligned\} I(y)=\int\_0^\infty \mathrm\{e\}^\{-x^2\}\cos (xy)\mathrm\{ d\} x=\frac\{\sqrt\{\pi\}\}\{2\}\mathrm\{e\}^\{-\frac\{y^2\}\{4\}\}. \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, 由 $\displaystyle I(0)=\int\_0^\infty \mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x=\frac\{\sqrt\{\pi\}\}\{2\}$, \begin\{aligned\} I'(y)&=\int\_0^\infty -x\mathrm\{e\}^\{-x^2\} \sin (xy)\mathrm\{ d\} x =\frac\{1\}\{2\}\int\_0^\infty \sin (xy)\mathrm\{ d\} \mathrm\{e\}^\{-x^2\}\\\\ &=-\frac\{1\}\{2\}\int\_0^\infty y\cos(xy)\mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x =-\frac\{y\}\{2\}I(y) \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\{\mathrm\{ d\} I\}\{I\}=-\frac\{y\}\{2\}\mathrm\{ d\} y \Rightarrow \ln I=-\frac\{y^2\}\{4\}+C\_1\\\\ \Rightarrow& I(y)=C\mathrm\{e\}^\{-\frac\{y^2\}\{4\}\} \stackrel\{I(0)=\frac\{\sqrt\{\pi\}\}\{2\}\}\{\Rightarrow\} I(y)=\frac\{\sqrt\{\pi\}\}\{2\}\mathrm\{e\}^\{-\frac\{y^2\}\{4\}\}. \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)、 $\displaystyle u|\_\{t=0\}=x^2+1$; [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) / \begin\{aligned\} u(x,t)&=\frac\{1\}\{2\sqrt\{\pi t\}\}\int\_\mathbb\{R\} \mathrm\{e\}^\{-\frac\{(x-y)^2\}\{4t\}\}(y^2+1)\mathrm\{ d\} y \stackrel\{\frac\{y-x\}\{2\sqrt\{t\}\}=s\}\{=\}\frac\{1\}\{\sqrt\{\pi\}\} \int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-s^2\} \left(x+2\sqrt\{t\}s\right)^2\mathrm\{ d\} s+1\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{1\}\{\sqrt\{\pi\}\}\left\[x^2\int\_\{\mathbb\{R\}\}\mathrm\{e\}^\{-s^2\}\mathrm\{ d\} s +4t\int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-s^2\}s^2\mathrm\{ d\} s\right\]+1\\\\ &\stackrel\{s^2=u\}\{=\}x^2+\frac\{4t\}\{\sqrt\{\pi\}\} \cdot 2 \int\_0^\infty \mathrm\{e\}^\{-u\}u\cdot\frac\{1\}\{2\}u^\{-\frac\{1\}\{2\}\}\mathrm\{ d\} u +1\\\\ &=x^2+\frac\{4t\}\{\sqrt\{\pi\}\}\cdot\varGamma\left(\frac\{3\}\{2\}\right)+1 =x^2+2t+1. \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)、 $\displaystyle u|\_\{t=0\}=x+x^2$. [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) / \begin\{aligned\} u(x,t)&=\frac\{1\}\{2\sqrt\{\pi t\}\}\int\_\mathbb\{R\} \mathrm\{e\}^\{-\frac\{(x-y)^2\}\{4t\}\}\mathrm\{e\}^\{-|y|\}\mathrm\{ d\} y\\\\ &\stackrel\{\frac\{y-x\}\{2\sqrt\{t\}\}=s\}\{=\}\frac\{1\}\{\sqrt\{\pi\}\} \int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-s^2\}\left\[(x+\sqrt\{2\}ts)+(x+2\sqrt\{t\}s)^2\right\]\mathrm\{ d\} s\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{1\}\{\sqrt\{\pi\}\} \int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-s^2\} (x+x^2+4ts^2)\mathrm\{ d\} s\\\\ &\stackrel\{s^2=u\}\{=\}x+x^2 +\frac\{4t\}\{\sqrt\{\pi\}\}\cdot 2\int\_0^\infty \mathrm\{e\}^\{-u\}u\cdot\frac\{1\}\{2\}u^\{-\frac\{1\}\{2\}\}\mathrm\{ d\} u\\\\ &=x+x^2+\frac\{4t\}\{\sqrt\{\pi\}\}\varGamma\left(\frac\{3\}\{2\}\right) =x+x^2+2t. \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 $\displaystyle \varphi(x)\in C(\mathbb\{R\})$ 且具有紧支集 (即在一有界集外恒为零), $\displaystyle u(x,t)$ 为 Poisson 公式 (1.11) 给出的热传导方程 Cauchy 问题的解, 试证 \begin\{aligned\} \lim\_\{t\to\infty\}u(x,t)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 对 $\displaystyle x\in (-\infty,\infty)$ 一致成立. [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 $\displaystyle \left\\{x; \varphi(x)\neq 0\right\\}\subset [-A,A]$, $\displaystyle M=\max\_\{[-A,A]\}|\varphi|$, 则 \begin\{aligned\} \sup\_\{x\in\mathbb\{R\}\} |u(x,t)|&=\sup\_\{x\in\mathbb\{R\}\} \left|\frac\{1\}\{2a\sqrt\{\pi t\}\}\int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-\frac\{(x-y)^2\}\{4a^2t\}\}\varphi(y)\mathrm\{ d\} y\right|\\\\ &\leq \frac\{1\}\{2a\sqrt\{\pi t\}\}\int\_\{\mathbb\{R\}\} |\varphi(y)|\mathrm\{ d\} y \leq \frac\{MA\}\{a\sqrt\{\pi t\}\}\xrightarrow\{t\to\infty\}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、 求解初值问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t-a^2u\_\{xx\}=g(t)u+f(x,t), x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=0, 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\} 其中 $\displaystyle a$ 为常数, $\displaystyle g(t)\in C\left([0,\infty)\right), f(x,t)\in C\left(\mathbb\{R\}\times [0,\infty)\right)$. [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 $\displaystyle v=\mathrm\{e\}^\{-\int\_0^t g(s)\mathrm\{ d\} s\}u$, 则 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} v\_t-a^2v\_\{xx\}=\mathrm\{e\}^\{-\int\_0^t g(s)\mathrm\{ d\} s\}f(x,t), x\in\mathbb\{R\}, t > 0,\\\\ v|\_\{t=0\}=0, 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\} 由齐次化原理知 \begin\{aligned\} &v(x,t)=\int\_0^t \frac\{1\}\{2a\sqrt\{\pi(t-\tau)\}\}\mathrm\{ d\}\tau \int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-\frac\{(x-y)^2\}\{4a^2(t-\tau)\}\} \mathrm\{e\}^\{-\int\_0^\tau g(s)\mathrm\{ d\} s\} f(y,\tau)\mathrm\{ d\} y\\\\ \Rightarrow&u(x,t)=\mathrm\{e\}^\{\int\_0^t g(s)\mathrm\{ d\} s\}\int\_0^t \frac\{1\}\{2a\sqrt\{\pi(t-\tau)\}\}\mathrm\{ d\}\tau \int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-\frac\{(x-y)^2\}\{4a^2(t-\tau)\}\} \mathrm\{e\}^\{-\int\_0^\tau g(s)\mathrm\{ d\} s\} f(y,\tau)\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\} 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、 求解下列 Cauchy 问题: (1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_t-2u\_\{xx\}-4u\_x-u=0, &x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=x,&x\in\mathbb\{R\};\end\{array\}\right.$ [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 $\displaystyle v=\mathrm\{e\}^\{t+x\}u$, 则 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}v\_t-2v\_\{xx\}=0,&x\in\mathbb\{R\}, t > 0,\\\\ v|\_\{t=0\}=x\mathrm\{e\}^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\} 由 Poisson 公式知 \begin\{aligned\} v(x,t)=&\frac\{1\}\{2\sqrt\{2\}\sqrt\{\pi t\}\}\int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-\frac\{(x-y)^2\}\{4\cdot 2t\}\}y\mathrm\{e\}^y\mathrm\{ d\} y =\frac\{1\}\{2\sqrt\{2\}\sqrt\{\pi t\}\}\int\_\{\mathbb\{R\}\} y\mathrm\{e\}^\{\frac\{[y-(x+4t)]^2\}\{8t\}+2t+x\}\mathrm\{ d\} y\\\\ =&\frac\{\mathrm\{e\}^\{2t+x\}\}\{2\sqrt\{2\pi t\}\}\int\_\{\mathbb\{R\}\} [\tau+(x+4t)]\mathrm\{e\}^\{-\frac\{\tau^2\}\{8t\}\}\mathrm\{ d\} \tau \xlongequal\{\tiny\mbox\{对称性\}\} \frac\{\mathrm\{e\}^\{2t+x\}\}\{2\sqrt\{2\pi t\}\}\int\_\{\mathbb\{R\}\} (x+4t)\mathrm\{e\}^\{-\frac\{\tau^2\}\{8t\}\}\mathrm\{ d\} \tau\\\\ =&\mathrm\{e\}^\{2t+x\}(x+4t). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle u=\mathrm\{e\}^t(x+4t)$. 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)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_t-2u\_\{xx\}-4tu=0, &x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=x\mathrm\{e\}^x,&x\in\mathbb\{R\};\end\{array\}\right.$ [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 $\displaystyle v=\mathrm\{e\}^\{-2t^2\}u$, 则 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}v\_t-v\_\{xx\}=0,&x\in\mathbb\{R\}, t > 0,\\\\ v|\_\{t=0\}=x\mathrm\{e\}^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\} 类似第 1 小题的计算过程知 \begin\{aligned\} v(x,t)=\frac\{1\}\{2\sqrt\{\pi t\}\}\int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-\frac\{(x-y)^2\}\{4t\}\}y\mathrm\{e\}^y\mathrm\{ d\} y=\mathrm\{e\}^\{t+x\}(x+2t). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle u=\mathrm\{e\}^\{t+2t^2+x\}(x+2t)$. 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)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_t-u\_\{xx\}=3t^2u, &x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=x^2+1,&x\in\mathbb\{R\}.\end\{array\}\right.$ [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 $\displaystyle v=\mathrm\{e\}^\{-t^3\}u$, 则 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}v\_t-v\_\{xx\}=0,&x\in\mathbb\{R\}, t > 0,\\\\ v|\_\{t=0\}=x^2+1,&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\} 类似第 1 小题的计算过程知 \begin\{aligned\} v(x,t)=\frac\{1\}\{2\sqrt\{\pi t\}\}\int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-\frac\{(x-y)^2\}\{4t\}\}(y^2+1)\mathrm\{ d\} y=1+2t+x^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 故 $\displaystyle u=\mathrm\{e\}^\{t^3\}(1+2t+x^2)$. 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、 证明函数 \begin\{aligned\} V(x,y,t; \xi,\eta,\tau)=\frac\{1\}\{4\pi a^2(t-\tau)\}\mathrm\{e\}^\{-\frac\{(x-\xi)^2+(y-\eta)^2\}\{4a^2(t-\tau)\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 关于变量 $\displaystyle x,y,t$ 满足方程 \begin\{aligned\} V\_t-a^2(V\_\{xx\}+V\_\{yy\})=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 关于变量 $\displaystyle \xi,\eta,\tau$ 满足方程 \begin\{aligned\} V\_\tau+a^2(V\_\{\xi\xi\}+V\_\{\eta\eta\})=0. \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) / 由 \begin\{aligned\} V\_t=&-\frac\{1\}\{4\pi a^2(t-\tau)^2\}\mathrm\{e\}^\{-\frac\{(x-\xi)^2+(y-\eta)^2\}\{4a^2(t-\tau)\}\}\\\\ &+\frac\{1\}\{4\pi a^2(t-\tau)\}\mathrm\{e\}^\{-\frac\{(x-\xi)^2+(y-\eta)^2\}\{4a^2(t-\tau)\}\} \frac\{(x-\xi)^2+(y-\eta)^2\}\{4a^2(t-\tau)^2\},\\\\ V\_x=&\frac\{1\}\{4\pi a^2(t-\tau)\}\mathrm\{e\}^\{-\frac\{(x-\xi)^2+(y-\eta)^2\}\{4a^2(t-\tau)\}\} \frac\{-(x-\xi)\}\{2a^2(t-\tau)\},\\\\ V\_\{xx\}=&\frac\{1\}\{4\pi a^2(t-\tau)\}\mathrm\{e\}^\{-\frac\{(x-\xi)^2+(y-\eta)^2\}\{4a^2(t-\tau)\}\} \frac\{(x-\xi)^2\}\{4a^4(t-\tau)\}\\\\ &+\frac\{1\}\{4\pi a^2(t-\tau)\}\mathrm\{e\}^\{-\frac\{(x-\xi)^2+(y-\eta)^2\}\{4a^2(t-\tau)\}\} \frac\{-1\}\{2a^2(t-\tau)\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} \begin\{aligned\} V\_\{yy\}=&\frac\{1\}\{4\pi a^2(t-\tau)\}\mathrm\{e\}^\{-\frac\{(x-\xi)^2+(y-\eta)^2\}\{4a^2(t-\tau)\}\} \frac\{(y-\eta)^2\}\{4a^4(t-\tau)\}\\\\ &+\frac\{1\}\{4\pi a^2(t-\tau)\}\mathrm\{e\}^\{-\frac\{(x-\xi)^2+(y-\eta)^2\}\{4a^2(t-\tau)\}\} \frac\{-1\}\{2a^2(t-\tau)\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} 知 $\displaystyle V\_t-a^2(V\_\{xx\}+V\_\{yy\})=0$. 进一步, \begin\{aligned\} V\_\tau=-V\_t, V\_\{xx\}=V\_\{\xi\xi\}, V\_\{yy\}=V\_\{\eta\eta\} \Rightarrow V\_\tau+a^2(V\_\{\xi\xi\}+V\_\{\eta\eta\})=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)/ --- 6、 证明若 $\displaystyle u\_i(x,t)\ (i=1,2,\cdots,n)$ 分别是下述 $\displaystyle n$ 个 Cauchy 问题的解: \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \frac\{\partial u\_i\}\{\partial t\}-a^2\frac\{\partial^2u\_i\}\{\partial x\_i^2\}=0, \\\\ u\_i|\_\{t=0\}=\varphi\_i(x\_i), i=1,2,\cdots,n.\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 $\displaystyle u(x\_1,\cdots,x\_n,t)=u\_1(x\_1,t)u\_2(x\_2,t)\cdots u\_n(x\_n,t)$ 是 Cauchy 问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t-a^2(u\_\{x\_1x\_1\}+u\_\{x\_2x\_2\}+\cdots+u\_\{x\_nx\_n\})=0,\\\\ u|\_\{t=0\}=\varphi\_1(x\_1)\varphi\_2(x\_2)\cdots \varphi\_n(x\_n)\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. [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\} \frac\{\partial u\}\{\partial t\}=&\sum\_\{i=1\}^n \frac\{\partial \}\{\partial t\} u\_i(x\_i,t) \left\[\prod\_\{j=1\atop j\neq i\}^n u\_j(x\_j,t)\right\]\\\\ =&\sum\_\{i=1\}^n a^2\frac\{\partial^2 \}\{\partial x\_i^2\} u\_i(x\_i,t) \left\[\prod\_\{j=1\atop j\neq i\}^n u\_j(x\_j,t)\right\]\\\\ =&a^2\sum\_\{i=1\}^n \frac\{\partial^2 \}\{\partial x\_i^2\}\left\[\prod\_\{j=1\}^n u\_j(x\_j,t)\right\] =a^2\Delta\_nu, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\} $\displaystyle u|\_\{t=0\}=\prod\_\{i=1\}^n u\_i(x\_i,0)=\prod\_\{i=1\}^n \varphi(x\_i)$ 知结论成立. 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 $\displaystyle u(x,t)$ 是 Cauchy 问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}u\_t-5u\_\{xx\}=0,&x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=x^5\mathrm\{e\}^\{x^2+1\}\cos 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\} 的解, 试求 $\displaystyle u(x,t)$ 在点 $\displaystyle (x,t)=(0,5)$ 处的值. [怎么回事? 都不能解到 $\displaystyle t=5$...] [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) / [怎么回事? 都不能解到 $\displaystyle t=5$...] 下面是形式计算, 毕竟广义积分发散! \begin\{aligned\} u(0,5)=\frac\{1\}\{2\sqrt\{5\pi t\}\}\int\_\mathbb\{R\} \mathrm\{e\}^\{-\frac\{y^2\}\{4\cdot 5\cdot 5\}\}\cdot y^5\mathrm\{e\}^\{y^2+1\}\cos y\mathrm\{ d\} y \xlongequal\{\tiny\mbox\{对称性\}\} 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)/ --- 8、 设 $\displaystyle u(x,t)$ 是 Cauchy 问题 \begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}u\_t-5u\_\{xx\}=0,&x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=x^5\mathrm\{e\}^\{x^2+1\}\cos 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\} 的解, 试求 $\displaystyle u\_x(x,t)$ 在点 $\displaystyle (x,t)=(0,5)$ 处的值. [怎么回事? 都不能解到 $\displaystyle t=5$...] [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) / [怎么回事? 都不能解到 $\displaystyle t=5$...] 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)/