混合问题
与波动方程的区别在于此时 $T$ 对应的 ODE 是一阶的, 故解是指数函数而不是三角函数.
用分离变量法求解下列混合问题. $$ \text{(2)}\quad
\begin{cases}
u_t = a^2 u_{xx}, & 0 < x < \pi,\ t > 0, \
u\big|{t=0} = \sin x, & 0 \leqslant x \leqslant \pi, \
u_x\big| = 0, & t > 0.
\end{cases}} = u_x\big|_{x=\pi $$ @@ADMONITION_START@@type=note&open=false&title=%E7%AD%94%E6%A1%88@@
$$
u(x,t)=\frac{2}{\pi}+\frac{4}{\pi}\sum\limits_{m=1}^\infty \frac{1}{1-4m^2}e^{-a^2(2m)^2t}\cos(2mx)
$$
注意此时是以 $\{\cos nx\}$ 作为正交基, 所以会有 $n=0$ 的项.
练习 3.3.1
题目
$$
\text{(3)}\quad \begin{cases} u_t = a^2 u_{xx}, & 0 < x < l,\ t > 0, \ u\big|{t=0} = x^2 (l - x), & 0 \leqslant x \leqslant l, \ u_x\big| = 0, & t > 0. \end{cases}} = u\big|_{x=l
$$
$$
u_n(x,t)=\sum\limits_{n=1}^\infty\varphi_n e^{-\left(a(2n+1)\pi/(2l)\right)^2t}\cos\frac{(2n+1)\pi}{2l}x,\quad \varphi_n=\frac{(-1)^n 64(2n+1)\pi l^3-192l^3}{(2n+1)^4\pi^4}
$$
答案
$$
\text{(3)'}\quad \begin{cases} u_t = a^2 u_{xx}, & 0 < x < l,\ t > 0, \ u\big|{t=0} = x^2 (l - x), & 0 \leqslant x \leqslant l, \ u\big| = 0, & t > 0. \end{cases}} = u\big|_{x=l
$$
$$
u(x,t)=\sum\limits_{n=1}^\infty A_ne^{-\left(\frac{an\pi}{l}\right)^2t}\sin\frac{n\pi}{l}x
$$
答案
$$
\text{(4)}\quad \begin{cases} u_t = a^2 u_{xx}, & 0 < x < l,\ t > 0, \ u\big|{t=0} = 0, & 0 \leqslant x \leqslant l, \ u\big| = At, & t > 0. \end{cases}} = 0,\quad u\big|_{x=l
$$
解得
$$
v(x,t)=\sum\limits_{n=1}^\infty \frac{2Al^2(-1)^n}{a^2\pi^3n^3}\left(1-e^{-\left(an\pi/l\right)^2 t}\right)\sin\frac{n\pi}{l}x
$$ 故 $u(x,t)=\frac{Atx}{l}+v(x,t)$
答案
$$
\text{(6)}\quad \begin{cases} u_t - a^2 u_{xx} = 0, & 0 < x < l,\ t > 0, \ u\big|{t=0} = 0, & 0 \leqslant x \leqslant l, \ u_x\big| = q, & t > 0. \end{cases}} = 0,\quad u_x\big|_{x=l
$$
解得
$$
v(x,t)=\frac{a^2q}{l}t-\frac{ql}{6}+\sum\limits_{n=1}^\infty -\frac{2ql(-1)^n}{n^2\pi^2}e^{-\left(an\pi/l\right)^2t}\cos\frac{n\pi}{l}x
$$
故 $u(x,t)=v(x,t)+\frac{x^2q}{2l}$.
答案