# 求解实例习题参考解答
---
1、 求解弦振动方程的下列混合问题:
(1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_\{tt\}-u\_\{xx\}=0,&0 < x < 2, t > 0,\\\\ u|\_\{t=0\}=\sin \pi x, u\_t|\_\{t=0\}=0,&0\leq x\leq 2,\\\\ u|\_\{x=0\}=u|\_\{x=2\}=0,&t\geq 0;\end\{array\}\right.$
[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) / 由书第 45 页知可设
\begin\{aligned\} u(x,t)=\sum\_\{k=1\}^\infty \left(a\_k\cos\frac\{k\pi t\}\{2\}+b\_k\sin\frac\{k\pi t\}\{2\}\right)\sin \frac\{k\pi x\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将初值条件代入知
\begin\{aligned\} \sum\_\{k=1\}^\infty a\_k\sin\frac\{k\pi x\}\{2\}=\sin \pi x, \sum\_\{k=1\}^\infty b\_k\cos\frac\{k\pi x\}\{2\}=0\Rightarrow a\_k=\left\\{\begin\{array\}\{llllllllllll\}1,&k=2,\\\\ 0,&k\neq 2,\end\{array\}\right. b\_k=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle u(x,t)=\cos \pi t\sin \pi x$. 跟锦数学微信公众号. [在线资料](
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(2)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_\{tt\}-u\_\{xx\}=0,&0 < x < l, t > 0,\\\\ u|\_\{t=0\}=x^2-2lx, u\_t|\_\{t=0\}=0,&0\leq x\leq l,\\\\ u\_x|\_\{x=0\}=u|\_\{x=l\}=0,&t\geq 0;\end\{array\}\right.$
[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) / 由习题 3-1 题 1 (2) 知可设
\begin\{aligned\} u(x,t)=\sum\_\{k=0\}^\infty \left(a\_k\cos\frac\{\left(k+\frac\{1\}\{2\}\right)\pi t\}\{l\} +b\_k\sin\frac\{\left(k+\frac\{1\}\{2\}\right)\pi t\}\{l\}\right)\cos\frac\{\left(k+\frac\{1\}\{2\}\right)\pi x\}\{l\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将初值条件代入知
\begin\{aligned\} \sum\_\{k=0\}^\infty a\_k\frac\{\left(k+\frac\{1\}\{2\}\right)\pi x\}\{l\}=x^2-2lx, \sum\_\{k=0\}^\infty b\_k\frac\{\left(k+\frac\{1\}\{2\}\right)\pi \}\{l\}\cos\frac\{\left(k+\frac\{1\}\{2\}\right)\pi x\}\{l\}=0.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle (I)\_1$ 两边乘以 $\displaystyle \cos\frac\{\left(n+\frac\{1\}\{2\}\right)\pi x\}\{l\}$, 在关于 $\displaystyle x\in [0,l]$ 积分知
\begin\{aligned\} &2a\_n=\int\_0^l (x^2-2lx)\cos \frac\{\left(k+\frac\{1\}\{2\}\right)\pi x\}\{l\}\mathrm\{ d\} x =0\\\\ \Rightarrow& a\_k=\frac\{l^3[4(2k+1)-(-1)^k(2k+1)^2\pi^2]\}\{(2k+1)^3\pi^3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再者 $\displaystyle (I)\_2\Rightarrow b\_k=0$. 于是
\begin\{aligned\} u(x,t)=&\sum\_\{k=0\}^\infty \frac\{l^3[4(2k+1)-(-1)^k(2k+1)^2\pi^2]\}\{(2k+1)^3\pi^3\}\\\\ &\cdot \cos\frac\{\left(k+\frac\{1\}\{2\}\right)\pi t\}\{l\}\cos \frac\{\left(k+\frac\{1\}\{2\}\right)\pi x\}\{l\}. \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)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_\{tt\}-u\_\{xx\}=0,&0 < x < \pi, t > 0,\\\\ u|\_\{t=0\}=x^3, u\_t|\_\{t=0\}=0,&0\leq x\leq \pi,\\\\ u|\_\{x=0\}=u\_x|\_\{x=\pi\}=0,&t\geq 0;\end\{array\}\right.$
[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) / 由习题 3-1 题 1 (1) 知可设
\begin\{aligned\} u(x,t)=\sum\_\{k=0\}^\infty \left\[a\_k\cos\left(k+\frac\{1\}\{2\}\right)t+b\_k\sin \left(k+\frac\{1\}\{2\}\right)t\right\]\sin\left(k+\frac\{1\}\{2\}\right)x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将初值条件代入知
\begin\{aligned\} \sum\_\{k=0\}^\infty a\_k\sin\left(k+\frac\{1\}\{2\}\right)x=x^3, b\_k=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\} \frac\{\pi\}\{2\}a\_k=\int\_0^\pi x^3\sin \left(k+\frac\{1\}\{2\}\right)x\mathrm\{ d\} x \Rightarrow a\_k=\frac\{24(-1)^k[(2k+1)^2\pi^2-8]\}\{(2k+1)^4\pi\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最终,
\begin\{aligned\} u(x,t)=\sum\_\{k=0\}^\infty \frac\{24(-1)^k[(2k+1)^2\pi^2-8]\}\{(2k+1)^4\pi\}\cos\left(k+\frac\{1\}\{2\}\right)t\sin \left(k+\frac\{1\}\{2\}\right)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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(4)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_\{tt\}-a^2u\_\{xx\}=0,&0 < x < l, t > 0,\\\\ u|\_\{t=0\}=x^2(x-l)^2, u\_t|\_\{t=0\}=0,&0\leq x\leq l,\\\\ u\_x|\_\{x=0\}=u\_x|\_\{x=l\}=0,&t\geq 0.\end\{array\}\right.$
[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) / 先求原 pde 具有变量分离形式的解 $\displaystyle u(x,t)=X(x)T(t)$, 则
\begin\{aligned\} &X(x)T''(t)=a^2X''(x)T(t) \Rightarrow \frac\{X''(x)\}\{X(x)\}=\frac\{1\}\{a^2\}\frac\{T''(t)\}\{T(t)\}=-\lambda. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由
\begin\{aligned\} u\_x|\_\{x=0\}=u\_x|\_\{x=l\}=0\Rightarrow X'(0)=X'(l)=0 \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\} \lambda=\lambda\_k=\left(\frac\{k\pi\}\{l\}\right)^2, k\geq 0, X(x)=\cos\frac\{k\pi\}\{l\}x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再求出
\begin\{aligned\} T(t)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{a\_0\}\{2\},&k=0,\\\\ a\_k\cos \frac\{k\pi a\}\{l\}t+b\_k \sin \frac\{k\pi a\}\{l\}t, &k\geq 1.\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\}
于是可设原 pde 的解为
\begin\{aligned\} u(x,t)=\frac\{a\_0\}\{2\}+\sum\_\{k=1\}^\infty \left\[ a\_k\cos \frac\{k\pi a\}\{l\}t +b\_k\sin \frac\{k\pi a\}\{l\}t \right\]\cos\frac\{k\pi\}\{l\}x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
利用初值条件确定出 $\displaystyle a\_k,b\_k$:
\begin\{aligned\} &x^2(x-l)^2 =u(x,0)=\frac\{a\_0\}\{2\}+\sum\_\{k=1\}^\infty a\_k\cos \frac\{k\pi\}\{l\}x,\\\\ &0=u\_t(x,0) =\sum\_\{k=1\}^\infty b\_k\frac\{k\pi a\}\{l\}\cos\frac\{k\pi\}\{l\}x \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
蕴含
\begin\{aligned\} a\_0=&\frac\{2\}\{l\}\int\_0^l x^2(x-l)^2\mathrm\{ d\} x=\frac\{l^4\}\{15\},\\\\ k\geq 1\Rightarrow a\_k=&\frac\{2\}\{l\}\int\_0^l x^2(x-l)^2\cos\frac\{k\pi\}\{l\}x\mathrm\{ d\} x =-\frac\{24l^4(1+\cos k\pi)\}\{k^4\pi^4\}\\\\ =&\left\\{\begin\{array\}\{llllllllllll\}-\frac\{3l^4\}\{m^4\pi^4\},&k=2m,\\\\ 0,&k=2m-1,\end\{array\}\right.\\\\ b\_k=&0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进而
\begin\{aligned\} u(x,t)=\frac\{l^4\}\{30\}-\frac\{3l^4\}\{\pi^4\} \sum\_\{m=1\}^\infty \frac\{1\}\{m^4\}\cos\frac\{2m\pi a\}\{l\}t\cos\frac\{2m\pi\}\{l\}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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---
2、 求解热传导方程的下列混合问题:
(1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_t-u\_\{xx\}=0,0 < x < 1, t > 0,\\\\ u|\_\{t=0\}=\left\\{\begin\{array\}\{llllllllllll\}x,&0\leq x\leq \frac\{1\}\{2\},\\\\ 1-x,&\frac\{1\}\{2\} < x\leq 1,\end\{array\}\right.\\\\ u|\_\{x=0\}=u|\_\{x=1\}=0, t\geq 0;\end\{array\}\right.$
[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) / 我们先求一维热传导方程的第一类初边值问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t-a^2u\_\{xx\}=0,0 < x < l, t > 0,\\\\ u|\_\{t=0\}=\phi(x),0 < x < l,\\\\ u|\_\{x=0\}=0, u|\_\{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\}
的解. 为此, 先求变量分离的解 $\displaystyle u=X(x)T(t)$, 而
\begin\{aligned\} X(x)T'(t)-a^2 X''(x)T(t)=0 \Rightarrow \frac\{T'(t)\}\{a^2T(t)\}=\frac\{X''(x)\}\{X(x)\}=-\lambda. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
联合边界条件知
\begin\{aligned\} X''(x)+\lambda X(x)=0, X(0)=X(l)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当且仅当 (参考书第 44-45 页) $\displaystyle \sqrt\{\lambda\}=\frac\{k\pi\}\{l\}, k\geq 1$ 时, 上述特征值问题有非平凡解. 此时,
\begin\{aligned\} \lambda=\lambda\_k=\left(\frac\{k\pi\}\{l\}\right)^2, k\geq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对应的特征函数为 $\displaystyle X\_k(x)=\sin \frac\{k\pi x\}\{l\}$. 解出
\begin\{aligned\} T'(t)=-\lambda\_k a^2T(t)\Rightarrow T(t)=T\_k(t)=a\_k\mathrm\{e\}^\{-\lambda\_k a^2t\} =a\_k\mathrm\{e\}^\{-\left(\frac\{k\pi a\}\{l\}\right)^2t\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由叠加原理知可设原 pde 的解为
\begin\{aligned\} u(x,t)=\sum\_\{k=1\}^\infty a\_k\mathrm\{e\}^\{-\left(\frac\{k\pi a\}\{l\}\right)^2t\}\sin \frac\{k\pi x\}\{l\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将初值条件代入得
\begin\{aligned\} \phi(x)=u(x,0)=\sum\_\{k=1\}^\infty a\_k\sin\frac\{k\pi x\}\{l\} \Rightarrow a\_k=\frac\{2\}\{l\}\int\_0^l \phi(x)\sin\frac\{k\pi x\}\{l\}\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\}
回到题目. 取 $\displaystyle l=1, \phi(x)=\left\\{\begin\{array\}\{llllllllllll\}x,&0\leq x\leq \frac\{1\}\{2\},\\\\ 1-x,&\frac\{1\}\{2\} < x\leq 1.\end\{array\}\right.$ 则由
\begin\{aligned\} a\_k=2\int\_0^1 \phi(x)\mathrm\{ d\} x=\left\\{\begin\{array\}\{llllllllllll\}0,&k=4m,4m+2,\\\\ \frac\{4\}\{(4m+1)^2\pi^2\},&k=4m+1,\\\\ -\frac\{4\}\{(4m+3)^2\pi^2\},&k=4m+3\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\}
知
\begin\{aligned\} u(x,t)=&\frac\{4\}\{\pi^2\}\sum\_\{m=0\}^\infty \frac\{1\}\{(4m+1)^2\} \mathrm\{e\}^\{-[(4m+1)\pi a]^2t\}\sin\left\[(4m+1)\pi x\right\]\\\\ &-\frac\{4\}\{\pi^2\}\sum\_\{m=0\}^\infty \frac\{1\}\{(4m+3)^2\} \mathrm\{e\}^\{-[(4m+3)\pi a]^2t\}\sin[(4m+3)\pi 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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(2)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_t-a^2u\_\{xx\}=0,&0 < x < 1, t > 0,\\\\ u|\_\{t=0\}=5\cos\frac\{3\pi\}\{2\}x,&0\leq x\leq l,\\\\ u\_x|\_\{x=0\}=u|\_\{x=1\}=0,&t\geq 0;\end\{array\}\right.$
[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) / 先求变量分离的解 $\displaystyle u(x,t)=X(x)T(t)$, 则
\begin\{aligned\} \frac\{X''\}\{X\}=\frac\{T'\}\{a^2T\}=-\lambda, X'(0)=0, X(1)=0.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由习题 3-1 题 1 (2) 知
\begin\{aligned\} \lambda=\lambda\_k=\left\[\left(k+\frac\{1\}\{2\}\right)\pi\right\]^2, \quad X(x)=\cos\left(k+\frac\{1\}\{2\}\right)\pi x, k\geq 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 $\displaystyle (I)$ 得 $\displaystyle T(t)=A \mathrm\{e\}^\{-\left\[\left(k+\frac\{1\}\{2\}\right)\pi a\right\]^2t\}$. 由叠加原理知可设原 pde 的解为
\begin\{aligned\} u(x,t)=\sum\_\{k=0\}^\infty a\_k\mathrm\{e\}^\{-\left\[\left(k+\frac\{1\}\{2\}\right)\pi a\right\]^2t\} \cos \left(k+\frac\{1\}\{2\}\right)\pi x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
代入初值条件知
\begin\{aligned\} &5\cos\frac\{3\pi x\}\{2\}=u(0,t)=\sum\_\{k=0\}^\infty a\_k\cos \left(k+\frac\{1\}\{2\}\right)\pi x \Rightarrow a\_k=\left\\{\begin\{array\}\{llllllllllll\}5,&k=1\\\\ 0,&k\neq 2\end\{array\}\right.\\\\ \Rightarrow&u(x,t)=5\mathrm\{e\}^\{-\left(\frac\{3\pi a\}\{2\}\right)^2t\}\cos \frac\{3\pi x\}\{2\}. \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)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_t-u\_\{xx\}=0,&0 < x < l, t > 0,\\\\ u|\_\{t=0\}=3+\cos \pi x+4\cos 2\pi x,&0\leq x\leq l,\\\\ u\_x|\_\{x=0\}=u\_x|\_\{x=l\}=0,&t\geq 0;\end\{array\}\right.$
[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) / 由一维热传导方程的第二类初边值问题的解知
\begin\{aligned\} u(x,t)=\frac\{A\_0\}\{2\}+\sum\_\{k=1\}^\infty A\_k \mathrm\{e\}^\{-(k\pi a)^2t\}\cos k\pi x, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} &\frac\{A\_0\}\{2\}+\sum\_\{k=1\}^\infty A\_k\cos k\pi x=3+\cos \pi x+4\cos 2\pi x\\\\ \Rightarrow&A\_0=6, A\_1=1, A\_2=4, A\_k=0\left(k\neq 0,1,2\right). \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\} u(x,t)=3+ \mathrm\{e\}^\{-(\pi a)^2t\}\cos \pi x +4\mathrm\{e\}^\{-(2\pi a)^2t\}\cos 2\pi 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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---
3、 求解下列初边值问题:
(1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-u\_\{xx\}+2u\_t+u=0, 0 < x < \pi, t > 0,\\\\ u|\_\{t=0\}=\pi x-x^2, u\_t|\_\{t=0\}=0, 0\leq x\leq \pi,\\\\ u|\_\{x=0\}=u|\_\{x=\pi\}=0, t\geq 0;\end\{array\}\right.$
[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 $\displaystyle v=\mathrm\{e\}^t u$, 则
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}v\_\{tt\}-v\_\{xx\}=0, 0 < x < \pi, t > 0,\\\\ v|\_\{t=0\}=\pi x-x^2, v\_t|\_\{t=0\}=\pi x-x^2, 0\leq x\leq \pi,\\\\ v|\_\{x=0\}=v|\_\{x=\pi\}=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\}
故按一维弦振动方程的混合问题的求解公式知
\begin\{aligned\} v(x,t)=\sum\_\{k=1\}^\infty (a\_k\cos kt+b\_k\sin kt)\sin kx, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle a\_k,b\_k$ 由下式决定:
\begin\{aligned\} \pi x-x^2=\sum\_\{k=1\}^\infty a\_k \sin kx =\sum\_\{k=1\}^\infty kb\_k\sin kx. \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\} a\_k=&\frac\{2\}\{\pi\}\int\_0^\pi (\pi-x^2)\sin kx\mathrm\{ d\} x =\frac\{4[1-(-1)^k]\}\{k^3\pi\}\\\\ =&\left\\{\begin\{array\}\{llllllllllll\}0,&k=2m,\\\\ \frac\{8\}\{(2m-1)^3\pi\},&k=2m-1.\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\}
进而
\begin\{aligned\} u(x,t)=\frac\{8\mathrm\{e\}^\{-t\}\}\{\pi\} \sum\_\{m=1\}^\infty \frac\{(2m-1)\cos(2m-1)t+\sin(2m-1)t\}\{(2m-1)^4\} \sin(2m-1)x. \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)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-u\_\{xx\}+2u\_t+u=0, 0 < x < \pi, t > 0,\\\\ u|\_\{t=0\}=0, u\_t|\_\{t=0\}=x^2(x-\pi), 0\leq x\leq \pi,\\\\ u\_x|\_\{x=0\}=u|\_\{x=\pi\}=0, t\geq 0.\end\{array\}\right.$
[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) / 令 $\displaystyle v=\mathrm\{e\}^t u$, 则
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} v\_\{tt\}-v\_\{xx\}=0, 0 < x < \pi, t > 0,\\\\ v|\_\{t=0\}=0, v\_t|\_\{t=0\}=x^2(x-\pi), 0\leq x\leq \pi,\\\\ v\_x|\_\{x=0\}=0, v\_x|\_\{x=\pi\}=0.\end\{array\}\right.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
先求 $\displaystyle (I)$ 具有变量分离形式的解 $\displaystyle v(x,t)=X(x)T(t)$, 则
\begin\{aligned\} &X(x)T''(t)=a^2X''(x)T(t) \Rightarrow \frac\{X''(x)\}\{X(x)\}=\frac\{1\}\{a^2\}\frac\{T''(t)\}\{T(t)\}=-\lambda. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由
\begin\{aligned\} v\_x|\_\{x=0\}=v|\_\{x=\pi\}=0\Rightarrow X'(0)=X(\pi)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及习题 3-1 题 1 (2) 知
\begin\{aligned\} \lambda=\lambda\_k=\left(k+\frac\{1\}\{2\}\right)^2, k\geq 0, X(x)=\cos \left(k+\frac\{1\}\{2\}\right)x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再求出
\begin\{aligned\} T(t)=a\_k\cos \left(k+\frac\{1\}\{2\}\right)t+b\_k \sin \left(k+\frac\{1\}\{2\}\right)t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是可设 $\displaystyle (I)$ 的解为
\begin\{aligned\} v(x,t)=\sum\_\{k=0\}^\infty \left\[ a\_k\cos \left(k+\frac\{1\}\{2\}\right)t +b\_k\sin \left(k+\frac\{1\}\{2\}\right)t \right\]\cos \left(k+\frac\{1\}\{2\}\right) x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
利用初值条件确定出 $\displaystyle a\_k,b\_k$:
\begin\{aligned\} &0=v(x,0)=\sum\_\{k=0\}^\infty a\_k\cos kx,\\\\ &x^2(x-\pi)=v\_t(x,0) =\sum\_\{k=0\}^\infty \left(k+\frac\{1\}\{2\}\right)b\_k\cos \left(k+\frac\{1\}\{2\}\right)x \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
蕴含 $\displaystyle a\_k=0$,
\begin\{aligned\} &\left(k+\frac\{1\}\{2\}\right)b\_k=-\frac\{8\left\[-24+8(2k+1)\pi(-1)^k\right\]\}\{(2k+1)^4\pi\}\\\\ \Rightarrow& b\_k=\left\\{\begin\{array\}\{llllllllllll\}-\frac\{128\left\[(4m+1)\pi-3\right\]\}\{(4m+1)^5\pi\}, &k=2m,\\\\ \frac\{128\left\[(4m+3)\pi+3\right\]\}\{(4m+3)^5\pi\},&k=2m+1\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\}
进而
\begin\{aligned\} v(x,t)=\sum\_\{m=0\}^\infty \left\[\begin\{array\}\{c\} -\frac\{128\left\[(4m+1)\pi-3\right\]\}\{(4m+1)^5\pi\} \cos\left(2m+\frac\{1\}\{2\}\right)t \cos\left(2m+\frac\{1\}\{2\}\right)x\\\\ +\frac\{128\left\[(4m+3)\pi+3\right\]\}\{(4m+3)^5\pi\} \sin\left(2m+\frac\{3\}\{2\}\right)t \cos\left(2m+\frac\{3\}\{2\}\right)x\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\}
于是 $\displaystyle u(x,t)$
\begin\{aligned\} =&-\mathrm\{e\}^\{-t\}\sum\_\{m=0\}^\infty \frac\{128\left\[(4m+1)\pi-3\right\]\}\{(4m+1)^5\pi\} \cos\left(2m+\frac\{1\}\{2\}\right)t \cos\left(2m+\frac\{1\}\{2\}\right)x\\\\ &+\mathrm\{e\}^\{-t\}\sum\_\{m=0\}^\infty \frac\{128\left\[(4m+3)\pi+3\right\]\}\{(4m+3)^5\pi\} \sin\left(2m+\frac\{3\}\{2\}\right)t \cos\left(2m+\frac\{3\}\{2\}\right)x. \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、 求解下列初边值问题:
(1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-(u\_\{xx\}+u\_\{yy\})=0, 0 < x,y < \pi, t > 0,\\\\ u|\_\{t=0\}=3\sin x\cdot \sin 2y, u\_t|\_\{t=0\}=5\sin 3x\cdot \sin 4y, 0\leq x,y\leq \pi,\\\\ u|\_\{x=0,\pi\}=u|\_\{y=0,\pi\}=0, t\geq 0; \end\{array\}\right.$
[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) / 取 $\displaystyle p=q=\pi$, 则由
\begin\{aligned\} a\_\{mn\}=&\frac\{4\}\{\pi^2\}\int\_0^\pi\int\_0^\pi 3\sin x\sin 2y \sin mx\sin ny\mathrm\{ d\} y\mathrm\{ d\} x =\left\\{\begin\{array\}\{llllllllllll\}3,&m=1,n=2,\\\\ 0,&\mbox\{其它\},\end\{array\}\right.\\\\ b\_\{mn\}=&\frac\{4\}\{\sqrt\{m^2+n^2\}\pi^2\} \int\_0^\pi\int\_0^\pi 5\sin 3x\sin 4y\sin mx \sin ny\mathrm\{ d\} y\mathrm\{ d\} x\\\\ =&\left\\{\begin\{array\}\{llllllllllll\}\frac\{5\}\{\sqrt\{m^2+n^2\}\} =1,&m=3, n=4,\\\\ 0,&\mbox\{其它\}\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\}
知
\begin\{aligned\} u(x,y,t)=3\cos\left(\sqrt\{5\}t\right)\sin x\sin 2y +\sin(5t)\sin 3x\sin 4y. \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)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-a^2(u\_\{xx\}+u\_\{yy\})=0, 0 < x < p, 0 < y < q, t > 0,\\\\ u|\_\{t=0\}=Axy(x-p)(y-q), u\_t|\_\{t=0\}=0, 0\leq x\leq p, 0\leq y\leq q,\\\\ u|\_\{x=0,p\}=u|\_\{y=0,q\}=0, t\geq 0. \end\{array\}\right.$
[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 $\displaystyle k\_\{mn\}=\sqrt\{\frac\{m^2\}\{p^2\}+\frac\{n^2\}\{q^2\}\}\pi$, 则
\begin\{aligned\} u(x,y,t)&=\sum\_\{m,n=1\}^\infty \left\[a\_\{mn\}\cos (ak\_\{mn\}t) +b\_\{mn\}\sin (ak\_\{mn\}t)\right\] \sin\frac\{m\pi x\}\{p\}\sin \frac\{n\pi y\}\{q\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} a\_\{mn\}=&\frac\{4\}\{pq\}\int\_0^p \int\_0^q Axy(x-p)(y-q)\sin \frac\{m\pi x\}\{p\} \sin \frac\{n\pi y\}\{q\}\mathrm\{ d\} y\mathrm\{ d\} x\\\\ =&A\cdot\frac\{2\}\{p\}\int\_0^p Ax(x-p) \sin\frac\{m\pi x\}\{p\}\mathrm\{ d\} x \cdot\frac\{2\}\{q\}\int\_0^q Ay(y-q)\sin\frac\{n\pi y\}\{q\}\mathrm\{ d\} y\\\\ =&\left\\{\begin\{array\}\{llllllllllll\} \frac\{64Ap^2q^2\}\{(2k-1)^3(2l-1)^3\pi^3\},&m=2k-1,n=2l-1,\\\\ 0,&\mbox\{其它\},\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\}
\begin\{aligned\} b\_\{mn\}=&\frac\{4\}\{k\_\{mn\}pq\}\int\_0^p\int\_0^q 0 \sin\frac\{m\pi x\}\{p\}\sin \frac\{n\pi x\}\{q\}\mathrm\{ d\} y\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\}
故
\begin\{aligned\} u(x,y,t)=&\frac\{64Ap^2q^2\}\{\pi^6\}\sum\_\{k,l=1\}^\infty \cos\left\\{a\sqrt\{\left\[\left(\frac\{2k-1\}\{p\}\right)^2 +\left(\frac\{2l-1\}\{q\}\right)^2\right\]\}\pi t\right\\}\\\\ &\sin\frac\{(2k-1)\pi x\}\{p\}\sin\frac\{(2l-1)\pi y\}\{q\}. \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、 求解混合问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}+a^2u\_\{xxxx\}=0, 0 < x < l, t > 0,\\\\ u|\_\{t=0\}=x(x-l), u\_t|\_\{t=0\}=0, 0\leq x\leq l,\\\\ u|\_\{x=0\}=u|\_\{x=l\}=u\_\{xx\}|\_\{x=0\}=u\_\{xx\}|\_\{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\}
[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) / 先求变量分离的解 $\displaystyle u(x,t)=X(x)T(t)$, 代入方程得
\begin\{aligned\} X(x)T''(t)+a^2X''''(t)T(t)=0 \Rightarrow \frac\{T''(t)\}\{a^2T(t)\}=-\frac\{X''''(x)\}\{X(x)\}=-\lambda. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 当 $\displaystyle \lambda < 0$ 时, 特征方程为 $\displaystyle \mu^4=\lambda=-\lambda \mathrm\{e\}^\{\mathrm\{ i\}\pi\}$, 解为
\begin\{aligned\} \sqrt[4]\{-\lambda\}\mathrm\{e\}^\{\pm\mathrm\{ i\}\frac\{\pi\}\{4\}\} =\sqrt[4]\{-\lambda\}\left(\frac\{\sqrt\{2\}\}\{2\}\pm \mathrm\{ i\}\frac\{\sqrt\{2\}\}\{2\}\right), \sqrt[4]\{-\lambda\}\mathrm\{e\}^\{\pm \mathrm\{ i\}\frac\{3\pi\}\{4\}\} =\sqrt[4]\{-\lambda\}\left(-\frac\{\sqrt\{2\}\}\{2\}\pm \mathrm\{ i\}\frac\{\sqrt\{2\}\}\{2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而可设
\begin\{aligned\} X(x)=&c\_1\mathrm\{e\}^\{\frac\{\sqrt\{2\}\}\{2\}\sqrt[4]\{-\lambda\}x\} \cos\frac\{\sqrt\{2\}\}\{2\}\sqrt[4]\{-\lambda\}x +c\_2\mathrm\{e\}^\{\frac\{\sqrt\{2\}\}\{2\}\sqrt[4]\{-\lambda\}x\} \sin\frac\{\sqrt\{2\}\}\{2\}\sqrt[4]\{-\lambda\}x\\\\ &+c\_3\mathrm\{e\}^\{-\frac\{\sqrt\{2\}\}\{2\}\sqrt[4]\{-\lambda\}x\} \cos\frac\{\sqrt\{2\}\}\{2\}\sqrt[4]\{-\lambda\}x +c\_4\mathrm\{e\}^\{-\frac\{\sqrt\{2\}\}\{2\}\sqrt[4]\{-\lambda\}x\} \sin\frac\{\sqrt\{2\}\}\{2\}\sqrt[4]\{-\lambda\}x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将边值条件 $\displaystyle X(0)=X(l)=X''(0)=X''(l)=0$ 代入, 经过计算知 $\displaystyle c\_1=c\_2=c\_3=c\_4=0$. 此时, 特征值问题没有非平凡解.
(2)、 当 $\displaystyle \lambda=0$ 时, $\displaystyle X(x)=c\_0+c\_1x+c\_2x^2+c\_3x^3$. 将边值条件 $\displaystyle X(0)=X(l)=X''(0)=X''(l)=0$ 代入, 经过计算知 $\displaystyle c\_1=c\_2=c\_3=c\_4=0$. 此时, 特征值问题没有非平凡解.
(3)、 当 $\displaystyle \lambda > 0$ 时, $\displaystyle \mu^4=\lambda$ 的解为
\begin\{aligned\} \sqrt[4]\{\lambda\}, -\sqrt[4]\{\lambda\}, \mathrm\{e\}^\{\pm \frac\{\pi\}\{2\}\mathrm\{ i\}\}\sqrt[4]\{\lambda\} =\pm \sqrt[4]\{\lambda\}\mathrm\{ i\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故可设
\begin\{aligned\} X(x)=C\_1\mathrm\{e\}^\{\sqrt[4]\{\lambda\}x\} +C\_2\mathrm\{e\}^\{-\sqrt[4]\{\lambda\}x\} +C\_3\cos \left(\sqrt[4]\{\lambda\}x\right) +C\_4\sin \left(\sqrt[4]\{\lambda\}x\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
边值条件 $\displaystyle X(0)=X(l)=X''(0)=X''(l)=0$ 代入,
\begin\{aligned\} C\_1+C\_2+C\_3=&0,\\\\ C\_1\mathrm\{e\}^\{\sqrt[4]\{\lambda\}l\} +C\_2\mathrm\{e\}^\{-\sqrt[4]\{\lambda\}l\} +C\_3\cos \left(\sqrt[4]\{\lambda\}l\right)+C\_4\sin\left(\sqrt[4]\{\lambda\}l\right)=&0,\\\\ C\_1+C\_2-C\_3=&0,\\\\ C\_1\mathrm\{e\}^\{\sqrt[4]\{\lambda\}l\} +C\_2\mathrm\{e\}^\{-\sqrt[4]\{\lambda\}l\} -C\_3\cos\left(\sqrt[4]\{\lambda\}l\right) -C\_4\sin\left(\sqrt[4]\{\lambda\}l\right)=&0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle C\_3=0, C\_2=-C\_1$,
\begin\{aligned\} 2C\_1\sinh \left(\sqrt[4]\{\lambda\}l\right)+C\_4\sin \left(\sqrt[4]\{\lambda\}l\right)=0, 2C\_1\sinh\left(\sqrt[4]\{\lambda\}l\right)-C\_4\sin \left(\sqrt[4]\{\lambda\}l\right)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
为使得特征值问题有非平凡解, 必须让上述关于 $\displaystyle C\_1,C\_4$ 的线性方程组的系数矩阵行列式
\begin\{aligned\} &-4\sin\left(\sqrt[4]\{\lambda\}l\right)\sinh\left(\sqrt[4]\{\lambda\}l\right)=0\\\\ \Rightarrow& \sqrt[4]\{\lambda\}=k\pi, k\geq 1 \Rightarrow \lambda=\lambda\_k=\left(\frac\{k\pi\}\{l\}\right)^4, k\geq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此时, 特征函数为 $\displaystyle X\_k(x)=\sin\frac\{k\pi\}\{l\}x$. 解出
\begin\{aligned\} &T''(t)+a^2\lambda\_kT(t)=0\\\\ \Rightarrow& T(t)=a\_k\cos\left\[a\left(\frac\{k\pi\}\{l\}\right)^2t\right\] +b\_k\sin\left\[a\left(\frac\{k\pi\}\{l\}\right)^2t\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由叠加原理知可设原 pde 的解为
\begin\{aligned\} u(x,t)=\sum\_\{k=1\}^\infty \left\\{a\_k\cos\left\[a\left(\frac\{k\pi\}\{l\}\right)^2t\right\] +b\_k\sin\left\[a\left(\frac\{k\pi\}\{l\}\right)^2t\right\]\right\\}\sin\frac\{k\pi\}\{l\}x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将初值条件代入得
\begin\{aligned\} x(x-l)=&u(x,0)=\sum\_\{k=1\}^\infty a\_k\sin\frac\{k\pi\}\{l\}x,\\\\ 0=&u\_t(x,0)=\sum\_\{k=1\}^\infty b\_k a\left(\frac\{k\pi\}\{l\}\right)^2\sin\frac\{k\pi\}\{l\}x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle b\_k=0$,
\begin\{aligned\} a\_k=&\frac\{2\}\{l\}\int\_0^l x(x-l)\sin\frac\{k\pi\}\{l\}x\mathrm\{ d\} x =\left\\{\begin\{array\}\{llllllllllll\}0,&k=m,\\\\ -\frac\{8l^3\}\{(2m-1)^3\pi^4\},&k=2m-1.\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\}
故
\begin\{aligned\} u(x,t)=-\frac\{8l^3\}\{\pi^4\}\sum\_\{m=1\}^\infty \frac\{1\}\{(2m-1)^3\} \cos\left\\{\left\[\frac\{(2m-1)\pi\}\{l\}\right\]^2at\right\\} \sin\frac\{(2m-1)\pi x\}\{l\}. \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、 研究圆柱形区域或两侧绝热的圆板上的热传导问题. 设侧边的温度保持为常数, 此时热传导问题可以归结为求解下述的定解问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t-(u\_\{xx\}+u\_\{yy\})=0,\\\\ u(x,y,0)=\varphi(x,y),\\\\ u|\_\{x^2+y^2=R^2\}=0, R > 0\mbox\{为常数\}.\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\}
[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) / 先求变量分离形式的解 $\displaystyle u(x,y,t)=T(t)V(x,y)$, 则
\begin\{aligned\} T'(t)+\lambda T(t)=0, V\_\{xx\}+V\_\{yy\}+\lambda V=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 $\displaystyle \lambda$ 是常数. 于是
\begin\{aligned\} T(t)=A\mathrm\{e\}^\{-\lambda t\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
考虑特征值问题
\begin\{aligned\} V\_\{xx\}+V\_\{yy\}+\lambda V=0, V|\_\{x^2+y^2=R^2\}=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们作极坐标变换, 而
\begin\{aligned\} V\_\{rr\}+\frac\{1\}\{r\}V\_r+\frac\{1\}\{r^2\}V\_\{\theta\theta\}+\lambda V=0, V(1,\theta)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再设 $\displaystyle V(r,\theta)=R(r)\varTheta(\theta)$, 则
\begin\{aligned\} \varTheta''(\theta)+\mu \varTheta(\theta)=0, r^2R''(r)+rR'(r)+(\lambda r^2-\mu)R(r)=0,\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \mu$ 与 $\displaystyle r,\theta$ 无关. 由 $\displaystyle V$ 是函数 (而是单值的) 知 $\displaystyle \varTheta(\theta)$ 具有周期 $\displaystyle 2\pi$, 而 $\displaystyle \mu=n^2\ (n\geq 0)$, 对应的特征函数为
\begin\{aligned\} \varTheta\_0(\theta)=\frac\{a\_0\}\{2\}, \varTheta\_n(\theta)=a\_n\cos n\theta+b\_n\sin n\theta\left(n\geq 1\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
现在考查 $\displaystyle (I)$. $\displaystyle R(r)$ 在 $\displaystyle r=0$ 处应有界, 由 $\displaystyle V|\_\{x^2+y^2=R^2\}=0$ 知 $\displaystyle R(r)|\_\{r=R\}=0$. 令 $\displaystyle \rho=r\sqrt\{\lambda\}$, 并将 $\displaystyle \mu=n^2$ 代入得 $\displaystyle n$ 阶 Bessel 方程
\begin\{aligned\} \rho^2 R''(\rho)+\rho R'(\rho)+(\rho^2-n^2)R(\rho)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
已经要求 $\displaystyle R(\rho)$ 在 $\displaystyle r=0$ 处有界, 而这种解除去一个常数因子外是唯一确定的, 它是第一类 $\displaystyle n$ 阶 Bessel 函数 $\displaystyle J\_n(\rho)$, 其有可数个正根 $\displaystyle \mu^\{(n)\}\_k, k\geq 1$. 于是
\begin\{aligned\} R(J\_n(\mu^\{(n)\}\_k))=J\_n(\mu^\{(n)\}\_k)=0\Rightarrow R\sqrt\{\lambda\}=\mu^\{(n)\}\_k \Leftrightarrow \sqrt\{\lambda\}=\frac\{\mu^\{(n)\}\_k\}\{R\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而原 pde 的解可写成
\begin\{aligned\} &u(x,y,t)=u(r,\theta,t)\\\\ =&\sum\_\{n=0\}^\infty \sum\_\{m=1\}^\infty \mathrm\{e\}^\{-\left(\frac\{\mu^\{(n)\}\_k\}\{R\}\right)^2t\} \left(a^\{(m)\}\_n\cos n\theta +b^\{(m)\}\_n\sin n\theta\right)J\_n\left(\frac\{\mu^\{(n)\}\_k\}\{R\}r\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 $\displaystyle a^\{(m)\}\_n, b^\{(m)\}\_n$ 由初值条件 $\displaystyle u(x,y,0)=\varphi(x,y)$ 决定. 由 Bessel 函数的性质知
\begin\{aligned\} \int\_0^1 xJ\_n\left(\mu^\{(n)\}\_mx\right)J\_n\left(\mu^\{(n)\}\_kx\right)=\delta\_\{mk\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且 $\displaystyle \left\\{\sqrt\{x\}J\_n\left(\mu^\{(n)\}\_mx\right)\right\\}\_\{m=1\}^\infty$ 构成 $\displaystyle L^2(\sqrt\{x\}\mathrm\{ d\} x)$ 的规范正交基. 再由
\begin\{aligned\} 1,\sin n\theta, \cos n\theta\left(n\geq 1\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
构成 $\displaystyle L^2[0,2\pi]$ 的规范正交基知
\begin\{aligned\} J\_n\left(\frac\{\mu^\{(n)\}\_k\}\{R\}r\right)\cos n\theta\left(n\geq 0\right), J\_n\left(\frac\{\mu^\{(n)\}\_k\}\{R\}r\right)\sin n\theta\left(n\geq 1\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
构成 $\displaystyle L^2\left(\sqrt\{r\}\mathrm\{ d\} r, \mathrm\{ d\} \theta\right)$ 的规范正交基. 于是 $\displaystyle \varphi(r,\theta)$ 可以展开成级数形式
\begin\{aligned\} \varphi(r,\theta)=\sum\_\{n=0\}^\infty \sum\_\{m=1\}^\infty \left( \varphi^\{(m)\}\_n\cos n\theta+\psi^\{(m)\}\_n\sin n\theta\right)J\_n\left( \frac\{\mu^\{(n)\}\_k\}\{R\}r\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而
\begin\{aligned\} &u(x,y,t)=u(r,\theta,t)\\\\ =&\sum\_\{n=0\}^\infty \sum\_\{m=1\}^\infty \mathrm\{e\}^\{-\left(\frac\{\mu^\{(n)\}\_k\}\{R\}\right)^2t\} \left(\varphi^\{(m)\}\_n\cos n\theta +\psi^\{(m)\}\_n\sin n\theta\right)J\_n\left(\frac\{\mu^\{(n)\}\_k\}\{R\}r\right), \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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---
7、 求矩形区域 $\displaystyle D: 0 < x < a, 0 < y < b$ 上的 Laplace 方程满足边界条件:
\begin\{aligned\} u(x,0)=u(x,b)=u(0,y)=0, u(a,y)=\left\\{\begin\{array\}\{llllllllllll\}y,&0\leq y < \frac\{b\}\{2\},\\\\ b-y,&\frac\{b\}\{2\}\leq y\leq b\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) / 先求变量分离的解 $\displaystyle u(x,y)=X(x)Y(y)$. 而
\begin\{aligned\} &X''(x)Y(y)+Y''(y)X(x)=0 \Rightarrow \frac\{X''(x)\}\{X(x)\}=\frac\{Y''(y)\}\{Y(y)\}=\lambda\\\\ \Rightarrow& X''(x)-\lambda X(x)=0, Y''(y)+\lambda Y(y)=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 $\displaystyle Y(0)=Y(b)=0$, 而当且仅当
\begin\{aligned\} \sqrt\{\lambda\}=\frac\{k\pi\}\{b\}\left(k\geq 1\right) \Leftrightarrow \lambda=\lambda\_k=\left(\frac\{k\pi\}\{b\}\right)^2\left(k\geq 1\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
时, 特征值问题有非平凡解, 且特征函数为 $\displaystyle Y\_k(x)=\sin\frac\{k\pi y\}\{b\}$. 求解出
\begin\{aligned\} X''(x)-\left(\frac\{k\pi\}\{b\}\right)^2X(x)=0 \Rightarrow X(x)=a\_k\mathrm\{e\}^\{\frac\{k\pi x\}\{b\}\}+b\_k\mathrm\{e\}^\{-\frac\{k\pi x\}\{b\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是原 pde 的解为
\begin\{aligned\} u(x,y,t)=\sum\_\{k=1\}^\infty \left(a\_k\mathrm\{e\}^\{\frac\{k\pi x\}\{b\}\}+b\_k\mathrm\{e\}^\{-\frac\{k\pi x\}\{b\}\}\right)\sin\frac\{k\pi y\}\{b\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而 $\displaystyle a\_k,b\_k$ 由下式决定:
\begin\{aligned\} &\sum\_\{k=1\}^\infty (a\_k+b\_k)\sin\frac\{k\pi y\}\{b\}=0,\\\\ &\sum\_\{k=1\}^\infty \left(a\_k \mathrm\{e\}^\{\frac\{k\pi a\}\{b\}\} +b\_k \mathrm\{e\}^\{-\frac\{k\pi a\}\{b\}\}\right)\sin\frac\{k\pi y\}\{b\} =\varphi(y)=\left\\{\begin\{array\}\{llllllllllll\}y,&0\leq y < \frac\{b\}\{2\},\\\\ b-y,&\frac\{b\}\{2\}\leq y\leq b.\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 $\displaystyle \varphi(y)=\sum\_\{k=1\}^\infty B\_k\sin\frac\{k\pi y\}\{b\}$,
\begin\{aligned\} B\_k=\frac\{2\}\{b\}\int\_0^b \varphi(y) \sin\frac\{k\pi y\}\{b\}\mathrm\{ d\} y =\left\\{\begin\{array\}\{llllllllllll\}0,&k=2m,\\\\ -\frac\{4(-1)^mb\}\{(2m-1)^2\pi^2\},&k=2m-1\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\}
知
\begin\{aligned\} a\_k \mathrm\{e\}^\{\frac\{k\pi a\}\{b\}\} +b\_k \mathrm\{e\}^\{-\frac\{k\pi a\}\{b\}\}=B\_k, a\_k+b\_k=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
解出 $\displaystyle a\_k,b\_k$, 并代入原 pde 的解中得
\begin\{aligned\} u(x,y,t)=\sum\_\{m=1\}^\infty &\frac\{4(-1)^m\}\{\left\[\mathrm\{e\}^\{\frac\{a(2m-1)\pi\}\{b\}\} -\mathrm\{e\}^\{-\frac\{a(2m-1)\pi\}\{b\}\}\right\] (2m-1)^2\pi^2\}\\\\ &\cdot\left\[\mathrm\{e\}^\{\frac\{(2m-1)\pi x\}\{b\}\}-\mathrm\{e\}^\{-\frac\{(2m-1)\pi x\}\{b\}\}\right\]\sin\frac\{(2m-1)\pi y\}\{b\}. \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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---
8、 在上述矩形区域中求 Laplace 方程 Neumann 问题的解:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}\Delta\_2u=0,&(x,y)\in D,\\\\ u\_y|\_\{y=0\}=g(x), u\_y|\_\{y=b\}=h(x),&0\leq x\leq a,\\\\ u\_x|\_\{x=0\}, u\_x|\_\{x=a\}=0, &0\leq y\leq b,\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 $\displaystyle g(x)$ 和 $\displaystyle h(x)$ 满足
\begin\{aligned\} \int\_0^a \left\[g(x)-h(x)\right\]\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\}
[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) / 先求变量分离的解 $\displaystyle u(x,y)=X(x)Y(y)$, 则
\begin\{aligned\} X''(x)Y(y)+Y''(y)X(x)=0\Rightarrow \frac\{X''(x)\}\{X(x)\}=-\frac\{Y''(y)\}\{Y(y)\}=-\lambda. \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 $\displaystyle X'(0)=0, X'(a)=0$, 而当且仅当 [参考书第 49-50 页]
\begin\{aligned\} \sqrt\{\lambda\}=\frac\{k\pi\}\{a\}\left(k\geq 0\right) \Leftrightarrow \lambda=\lambda\_k=\left(\frac\{k\pi\}\{a\}\right)^k\left(k\geq 0\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
时, 特征值问题有非平凡解, 且特征函数为 $\displaystyle X\_k(x)=\cos\frac\{k\pi x\}\{a\}$. 又由 $\displaystyle Y''(y)-\left(\frac\{k\pi\}\{a\}\right)^2Y(y)=0$ 知
\begin\{aligned\} Y\_0(y)=C+\frac\{A\_0\}\{2\}y, Y\_k(y)=a\_k\mathrm\{e\}^\{\frac\{k\pi y\}\{a\}\} +b\_k\mathrm\{e\}^\{-\frac\{k\pi y\}\{a\}\}, k\geq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故原 pde 的解可设为
\begin\{aligned\} u(x,y)=C+\frac\{A\_0\}\{2\}y +\sum\_\{k=1\}^\infty \left(a\_k\mathrm\{e\}^\{\frac\{k\pi y\}\{a\}\} +b\_k \mathrm\{e\}^\{-\frac\{k\pi y\}\{a\}\}\right)\cos\frac\{k\pi x\}\{a\}.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进而
\begin\{aligned\} u\_y(x,y)=\frac\{A\_0\}\{2\}+\sum\_\{k=1\}^\infty \left(a\_k\frac\{k\pi\}\{a\}\mathrm\{e\}^\{\frac\{k\pi y\}\{a\}\}-b\_k\frac\{k\pi\}\{a\}\mathrm\{e\}^\{-\frac\{k\pi y\}\{a\}\}\right)\cos\frac\{k\pi x\}\{a\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle y=0,y=a$, 则
\begin\{aligned\} g(x)&=\frac\{A\_0\}\{2\}+\sum\_\{k=1\}^\infty \left(a\_k\frac\{k\pi\}\{a\}-b\_k\frac\{k\pi\}\{a\}\right)\cos\frac\{k\pi x\}\{a\},\\\\ h(x)&=\frac\{A\_0\}\{2\}+\sum\_\{k=1\}^\infty \left(a\_k\frac\{k\pi\}\{a\}\mathrm\{e\}^\frac\{k\pi b\}\{a\} -b\_k\frac\{k\pi\}\{a\}\mathrm\{e\}^\{-\frac\{k\pi b\}\{a\}\}\right)\cos\frac\{k\pi x\}\{a\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故将 $\displaystyle g(x),h(x)$ 在 $\displaystyle [0,a]$ 上展开成余弦级数
\begin\{aligned\} g(x)=\frac\{A\_0\}\{2\}+\sum\_\{k=1\}^\infty A\_k \cos \frac\{k\pi x\}\{a\}, h(x)=\frac\{A\_0\}\{2\}+\sum\_\{k=1\}^\infty B\_k\cos\frac\{k\pi x\}\{a\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中
\begin\{aligned\} A\_0&=\frac\{2\}\{a\}\int\_0^a g(x)\mathrm\{ d\} x=\frac\{2\}\{a\}\int\_0^a h(x)\mathrm\{ d\} x,\\\\ k\geq 1\Rightarrow A\_k&=\frac\{2\}\{a\} \int\_0^a g(x)\cos\frac\{k\pi x\}\{a\}\mathrm\{ d\} x, \\\\ B\_k&=\frac\{2\}\{a\}\int\_0^a h(x)\cos\frac\{k\pi x\}\{a\}\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\}
则
\begin\{aligned\} A\_k=a\_k\frac\{k\pi\}\{a\}-b\_k\frac\{k\pi\}\{a\}, B\_k=a\_k\frac\{k\pi\}\{a\}\mathrm\{e\}^\frac\{k\pi b\}\{a\} -b\_k\frac\{k\pi\}\{a\}\mathrm\{e\}^\{-\frac\{k\pi b\}\{a\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由此, 即可求出 $\displaystyle a\_k,b\_k$. 代入 $\displaystyle (I)$ 得到原 pde 的解. 注意: 因为是 Neumann 问题, 所以解在相差一个常数 $\displaystyle C$ 的意义下才唯一! 跟锦数学微信公众号. [在线资料](
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---
9、 求解定解问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \Delta\_2u=0, 1 < r < 2, 0 < \theta < \pi,\\\\ u(1,\theta)=\sin \theta, u(2,\theta)=0,\\\\ u(r,0)=u(r,\pi)=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\}
[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) / 先求具有形式 $\displaystyle u(x,y)=u(r,\theta)=R(r)\varTheta(\theta)$ 的解, 则
\begin\{aligned\} 0&=\Delta\_2u=u\_\{rr\}+\frac\{1\}\{r\}u\_r+\frac\{1\}\{r^2\}u\_\{\theta\}\\\\ &=R''(r)\varTheta(\theta) +\frac\{1\}\{r\}R'(r)\varTheta(\theta) +\frac\{1\}\{r^2\}R(r)\varTheta''(t). \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\{r^2R''(r)+rR'(r)\}\{R(r)\}=-\frac\{\varTheta''(\theta)\}\{\varTheta(\theta)\}=\lambda. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} \varTheta''(\theta)+\lambda \varTheta(\theta)=0, \varTheta(0)=0, \varTheta(\pi)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知当且仅当 $\displaystyle \lambda=\lambda\_k=k^2\ (k\geq 1)$ 时, 上述特征值问题才有非平凡解. 此时, 特征函数为 $\displaystyle \varTheta\_k(\theta)=\sin k\theta$. 进一步,
\begin\{aligned\} r^2R''(r)+rR'(r)-k^2R(r)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是一个 Euler 方程, 由 $\displaystyle k\geq 1$ 时, 它的解为
\begin\{aligned\} R\_k(r)=a\_kr^k+b\_kr^\{-k\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故原 pde 的解可设为
\begin\{aligned\} u(x,y)=u(r,\theta)=\sum\_\{k=1\}^\infty (a\_kr^k+b\_kr^\{-k\}) \sin k\theta. \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
\begin\{aligned\} \sin\theta=&u(1,\theta)=\sum\_\{k=1\}^\infty (a\_k+b\_k)\sin k\theta,\\\\ 0=&u(2,\theta)=\sum\_\{k=1\}^\infty (a\_k2^k+b\_k2^\{-k\})\sin k\theta \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\} a\_1+b\_1=1, a\_k+b\_k=0\ (k\geq 2), a\_k2^k+b\_k2^\{-k\}=0\left(k\geq 1\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
解得
\begin\{aligned\} &a\_1=-\frac\{1\}\{3\}, a\_2=\frac\{4\}\{3\}, a\_k=b\_k=0\left(k\geq 2\right)\\\\ \Rightarrow&u(x,y)=u(r,\theta)=\left(-\frac\{1\}\{3\}r+\frac\{4\}\{3r\}\right) \sin\theta. \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](
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---
10、 求解边值问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{xx\}+u\_\{yy\}=0, 0 < x < 1, 0 < y < \infty,\\\\ u|\_\{x=0\}=u|\_\{x=1\}=0, u|\_\{y=0\}=x(1-x),\\\\ \lim\_\{y\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\}
[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) / 先求具有变量分离形式的解 $\displaystyle u(x,y)=X(x)Y(y)$, 则
\begin\{aligned\} X''(x)Y(y)+Y''(y)X(x)=0\Rightarrow \frac\{X''(x)\}\{X(x)\}=-\frac\{Y''(y)\}\{Y(y)\}=-\lambda. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} X''(x)+\lambda X(x)=0, X(0)=0, X(1)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知当且仅当
\begin\{aligned\} \sqrt\{\lambda\}=k\pi\left(k\geq 1\right)\Leftrightarrow \lambda=\lambda\_k=(k\pi)^2\left(k\geq 1\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
时, 上述特征值问题才有非平凡解, 且特征函数为 $\displaystyle X\_k(x)=\sin k\pi x$. 再解出
\begin\{aligned\} Y''(y)=\lambda\_kY(y)\Rightarrow Y\_k(y)=a\_k\mathrm\{e\}^\{k\pi y\}+b\_k\mathrm\{e\}^\{-k\pi y\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是原 pde 的解可设为
\begin\{aligned\} u(x,y)=\sum\_\{k=1\}^\infty \left(a\_k\mathrm\{e\}^\{k\pi y\}+b\_k\mathrm\{e\}^\{-k\pi y\}\right) \sin k\pi x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle \lim\_\{y\to+\infty\}u=0$ 知 $\displaystyle a\_k=0$. 再由
\begin\{aligned\} x(1-x)=u(x,0)=\sum\_\{k=1\}^\infty b\_k\sin k\pi 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\} b\_k=2\int\_0^1 x(1-x)\sin k\pi x\mathrm\{ d\} x =\left\\{\begin\{array\}\{llllllllllll\}0,&k=2m,\\\\ \frac\{8\}\{(2m-1)^3\pi^3\},&k=2m-1.\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\}
故
\begin\{aligned\} u(x,y)=\frac\{8\}\{\pi^3\}\sum\_\{m=1\}^\infty \frac\{\mathrm\{e\}^\{-(2m-1)\pi y\}\sin(2m-1)\pi x\}\{(2m-1)^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、 求解半圆盘区域上的边值问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} \Delta\_2u=0, &0 < r < R, 0 < \theta < \pi,\\\\ u\_r(R,\theta)=\theta(\pi-\theta), &0\leq \theta\leq \pi,\\\\ u(r,0)=0, u(r,\pi)=0, &0\leq r\leq 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\}
[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) / 先求具有形式 $\displaystyle u(x,y)=u(r,\theta)=R(r)\varTheta(\theta)$ 的解, 则
\begin\{aligned\} 0&=\Delta\_2u=u\_\{rr\}+\frac\{1\}\{r\}u\_r+\frac\{1\}\{r^2\}u\_\{\theta\}\\\\ &=R''(r)\varTheta(\theta) +\frac\{1\}\{r\}R'(r)\varTheta(\theta) +\frac\{1\}\{r^2\}R(r)\varTheta''(t). \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\{r^2R''(r)+rR'(r)\}\{R(r)\}=-\frac\{\varTheta''(\theta)\}\{\varTheta(\theta)\}=\lambda. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} \varTheta''(\theta)+\lambda \varTheta(\theta)=0, \varTheta(0)=0, \varTheta(\pi)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知当且仅当 $\displaystyle \lambda=\lambda\_k=k^2\ (k\geq 1)$ 时, 上述特征值问题才有非平凡解. 此时, 特征函数为 $\displaystyle \varTheta\_k(\theta)=\sin k\theta$. 进一步,
\begin\{aligned\} r^2R''(r)+rR'(r)-k^2R(r)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
是一个 Euler 方程, 由 $\displaystyle k\geq 1$ 时, 它的解为
\begin\{aligned\} R\_k(r)=a\_kr^k+b\_kr^\{-k\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故原 pde 的解可设为
\begin\{aligned\} u(x,y)=u(r,\theta)=\sum\_\{k=1\}^\infty (a\_kr^k+b\_kr^\{-k\}) \sin k\theta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由于 $\displaystyle u$ 在原点附近是有界的, 而 $\displaystyle b\_k=0$. 再由边界条件
\begin\{aligned\} \theta(\pi-\theta)=&u\_r(R,\theta)=\sum\_\{k=1\}^\infty ka\_kR^\{k-1\}\sin k\theta \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\} ka\_kR^\{k-1\}=\frac\{2\}\{\pi\}\int\_0^\pi \theta(\pi-\theta)\sin k\theta\mathrm\{ d\} \theta =\left\\{\begin\{array\}\{llllllllllll\}0,&k=2m,\\\\ \frac\{8\}\{(2m-1)^3\pi\},&k=2m-1.\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\}
故
\begin\{aligned\} u(x,y)=u(r,\theta)=\frac\{8\}\{\pi\}\sum\_\{m=1\}^\infty \frac\{r^\{2m-1\}\sin (2m-1)x\}\{(2m-1)^4R^\{2m-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)/
---
12、 利用分离变量法求解下述问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}-u\_\{xx\}=0,&0 < x < \pi, t > 0,\\\\ u|\_\{t=0\}=\varphi(x), u\_t|\_\{t=0\}=\psi(x), &0\leq x\leq \pi,\\\\ (-u\_x+u)|\_\{x=0\}=(u\_x+u)|\_\{x=\pi\}=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\}
其中 $\displaystyle \varphi\in C^4[0,\pi], \psi\in C^3[0,\pi]$, 且 $\displaystyle \varphi,\psi$ 都在 $\displaystyle 0,\pi$ 附近为零.
[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) / 先求具有变量分离形式的解 $\displaystyle u(x,t)=X(x)T(t)$, 则
\begin\{aligned\} T'(t)X(x)-X''(x)T(t)=0 \Rightarrow \frac\{T'(t)\}\{T(t)\}=\frac\{X''(x)\}\{X(x)\}=-\lambda. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} X''(x)=\lambda X(x), -X'(0)+X(0)=0, X'(\pi)+X(\pi)=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及第 3.1 节分离变量法的理论基础知上述特征值问题有可数个特征值 $\displaystyle \lambda\_k > 0$. 此时, 设特征函数为 $\displaystyle X(x)=c\_1\cos \sqrt\{\lambda\}x+c\_2\sin \sqrt\{\lambda\}x$ 后, 将边值条件代入, 得
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}1&-\sqrt\{\lambda\}\\\\ \cos\sqrt\{\lambda\}\pi-\sqrt\{\lambda\}\sin \sqrt\{\lambda\}\pi &\sqrt\{\lambda\}\cos \sqrt\{\lambda\} \pi+\sin\sqrt\{\lambda\}\pi\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}c\_1\\\\c\_2\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}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\}
它的系数矩阵行列式为 $\displaystyle 0$, 而
\begin\{aligned\} \frac\{2\sqrt\{\lambda\}\}\{\lambda-1\}=\tan \sqrt\{\lambda\}\pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle \lambda\_k > 0$ 是上述超越方程的第 $\displaystyle k$ 个正根, 则它就是所考虑特征值问题的特征值. 此时,
\begin\{aligned\} X(x)=\sqrt\{\lambda\_k\}c\_2\cos \sqrt\{\lambda\_k\}x+c\_2\sin \sqrt\{\lambda\_k\}x =c\_2\left(\sqrt\{\lambda\_k\}\cos\sqrt\{\lambda\_k\}x+\sin\sqrt\{\lambda\_k\}x\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle X\_k(x)=\frac\{\sqrt\{\lambda\_k\}\cos\sqrt\{\lambda\_k\}x+\sin\sqrt\{\lambda\_k\}x\}\{\left\Vert \sqrt\{\lambda\_k\}\cos\sqrt\{\lambda\_k\}x+\sin\sqrt\{\lambda\_k\}x\right\Vert \_\{L^2\}\}$, 则由第 3.1 解知 $\displaystyle X\_k$ 构成 $\displaystyle L^2[0,\pi]$ 的一个规范正交基. 解出
\begin\{aligned\} T''(t)=-\lambda\_kT(t) \Rightarrow T(t)=a\_k\cos \sqrt\{\lambda\_k\}t+b\_k\sin \sqrt\{\lambda\_kt\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而可设原 pde 的解为
\begin\{aligned\} u(x,t)=\sum\_\{k=1\}^\infty \left(a\_k\cos \sqrt\{\lambda\_k\}t+b\_k\sin \sqrt\{\lambda\_k\}t\right)X\_k(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
利用初值条件即可定出 $\displaystyle a\_k,b\_k$. 事实上,
\begin\{aligned\} \varphi(x)=\sum\_\{k=1\}^\infty a\_k X\_k(x)\Rightarrow a\_k=\int\_0^\pi \varphi(x)X\_k(x)\mathrm\{ d\} x,\\\\ \psi(x)=\sum\_\{K=1\}^\infty b\_k\sqrt\{\lambda\_k\}X\_k(x)\Rightarrow b\_k=\frac\{1\}\{\sqrt\{\lambda\_k\}\}\int\_0^\pi \psi(x)X\_k(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\}
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 07:47:53
