第四章 波动方程

4.1 方程的物理背景和定解问题

2.	符号说明: $\mathrm{curl}\mathbf{B}=\nabla \times \mathbf{B}=\left|\begin{array}{ccc}i& j& k\\ {\partial }_1& {\partial }_2& {\partial }_3\\ B_1& B_2& B_3\end{array}\right|=({\partial }_2{B}_3-{\partial }_3{B}_2,{\partial }_3{B}_1-{\partial }_1{B}_3,{\partial }_1{B}_2-{\partial }_2{B}_1)$, $\mathrm{div}\mathbf{u}=\nabla \cdot \mathbf{u}={\partial }_1{u}_1+{\partial }_2{u}_2+{\partial }_3{u}_3$. (1) 有公式 $\nabla \mathrm{div}={\mathrm{curl}}^2+\Delta$, 这是因为 $\nabla (\nabla \cdot \mathbf{a})=\nabla \times (\nabla \times \mathbf{a})+(\nabla \cdot \nabla )\mathbf{a}$ 对 $\mathbf{a}\in {\mathbb{R}}^3$. 因此 ${\partial }_t^2\mathbf{E}=-{\mathrm{curl}}^2\mathbf{E}=\Delta \mathbf{E}$, 第二个等号是因为 $\mathrm{div}\mathbf{E}=0$. 同理有 ${\partial }_t^2\mathbf{B}=\Delta \mathbf{B}$. (2) 作用 $\mathrm{div}$ 得 $\mathrm{div}{\partial }_t^2\mathbf{u}-\mu \mathrm{div}\Delta \mathbf{u}-(\lambda +\mu )\mathrm{div}\nabla (\mathrm{div}\mathbf{u})=0$, 化为 ${\partial }_t^2\mathrm{div}\mathbf{u}-\mu \Delta \mathrm{div}\mathbf{u}-(\lambda +\mu )\Delta (\mathrm{div}\mathbf{u})=0$, 即 ${\partial }_t^{2}v-(\lambda +2\mu )\Delta v=0$. 作用 $\mathrm{curl}$ 得 $\mathrm{curl}{\partial }_t^2\mathbf{u}-\mu \mathrm{curl}\Delta \mathbf{u}-(\lambda +\mu )\mathrm{curl}\nabla (\mathrm{div}\mathbf{u})=0$, 化为 ${\partial }_t^2\mathrm{curl}\mathbf{u}-\mu \Delta \mathrm{curl}\mathbf{u}=0$ (因 $\mathrm{curl}\nabla =\nabla \times \nabla =0$) , 即 ${\partial }_t^{2}w-\mu \Delta w=0$.
3.	要使 ${\partial }_T^2-{\partial }_X^2={\partial }_t^2-3{\partial }_{tx}-4{\partial }_x^2=({\partial }_t-\frac{3}{2}{\partial }_x{)}^2-(\frac{5}{2}{\partial }_x{)}^2$, 只需 $\left(\begin{array}{c}{\partial }_T\\ {\partial }_X\end{array}\right)=\left(\begin{array}{cc}1& -\frac{3}{2}\\ 0& \frac{5}{2}\end{array}\right)\left(\begin{array}{c}{\partial }_t\\ {\partial }_x\end{array}\right)$, 即 $\left(\begin{array}{c}{\partial }_t\\ {\partial }_x\end{array}\right)=\left(\begin{array}{cc}\frac{\partial T}{\partial t}& \frac{\partial X}{\partial t}\\ \frac{\partial T}{\partial x}& \frac{\partial X}{\partial x}\end{array}\right)\left(\begin{array}{c}{\partial }_T\\ {\partial }_X\end{array}\right)=\left(\begin{array}{cc}1& \frac{3}{5}\\ 0& \frac{2}{5}\end{array}\right)\left(\begin{array}{c}{\partial }_T\\ {\partial }_X\end{array}\right)$, 因此取 $\{\begin{array}{rl}T& =t\\ X& =\frac{3}{5}t+\frac{2}{5}x\end{array}.$
5.	即 $-c{\Phi }^{\prime }+6\Phi {\Phi }^{\prime }+{\Phi }^{\prime \prime \prime }=0$, 积分得 $-c\Phi +3{\Phi }^2+{\Phi }^{\prime \prime }$ 为常数, 令 $x\to +\infty$ 得此常数为 $0$. 那么 $-c\Phi {\Phi }^{\prime }+3{\Phi }^2{\Phi }^{\prime }+{\Phi }^{\prime }{\Phi }^{\prime \prime }=0$, 再积分得 $-\frac{c}{2}{\Phi }^2+{\Phi }^3+\frac{1}{2}({\Phi }^{\prime }{)}^2$ 为常数, 且实际上是 $0$, 于是 $\Phi =\pm \frac{{\Phi }^{\prime }}{\sqrt{c-2\Phi }}$. 令 $v=\sqrt{c-2\Phi }$, 则 $\frac{c-v^2}{2}=\pm v^{\prime }$, 从而 $\left(\ln \right|\frac{\sqrt{c}+v}{\sqrt{c}-v}|{)}^{\prime }=\frac{2\sqrt{c}{v}^{\prime }}{c-v^2}=\pm \sqrt{c}$. 答案是 $u(t,x)=\frac{c}{2}{\mathrm{sech}}^2\frac{\sqrt{c}(x-ct)+d}{2},d\in \mathbb{R}$, 其中 $\mathrm{sech}x=\frac{2}{e^x+e^{-x}}$.

4.2 波动方程的 Cauchy 问题

4.	令 $\tilde{t}=at$, 则有$$\{\begin{array}{rl}& ({\partial }_{\tilde{t}}^2-\Delta )u=0,\\ & u(0,x)=\phi (x),{\partial }_{\tilde{t}}u(0,x)=\frac{1}{a}\psi (x).\end{array}$$由此计算得到, 对一般的 $a$, $n=3$ 时表达式为$$\begin{array}{r}u(t,x)={\partial }_t({-\int }_{\partial B_{at}(x)}t\phi (y)d{S}_y)+{-\int }_{\partial B_{at}(x)}t\psi (y)d{S}_y.\end{array}$$$n=2$ 时表达式为$$\begin{array}{r}u(t,x)={\partial }_t({-\int }_{D_{at}(x)}t\phi (y)\frac{t}{2\sqrt{t^2-|\frac{y-x}{a}{|}^2}}dy)+{-\int }_{D_{at}(x)}t\psi (y)\frac{t}{2\sqrt{t^2-|\frac{y-x}{a}{|}^2}}dy.\end{array}$$ 注. $n=3$ 时表达式的第一项$$\begin{array}{rl}{\partial }_t({-\int }_{\partial B_{at}(x)}t\phi (y)d{S}_y)& ={\partial }_t({-\int }_{\partial B_a(0)}t\phi (x+tz)d{S}_z)\\ & ={-\int }_{\partial B_a(0)}\phi (x+tz)+t\nabla \phi (x+tz)\cdot zd{S}_z\\ & ={-\int }_{\partial B_{at}(x)}\phi (y)+\nabla \phi (y)\cdot (y-x)d{S}_y,\end{array}$$于是 $n=3$ 时的解也写成$$\begin{array}{r}u(t,x)={-\int }_{\partial B_{at}(x)}\phi (y)+\nabla \phi (y)\cdot (y-x)+t\psi (y)d{S}_y.\end{array}$$相应地, $n=2$ 时的解也写成$$\begin{array}{r}u(t,x)={-\int }_{D_{at}(x)}(\phi (y)+\nabla \phi (y)\cdot (y-x)+t\psi (y))\frac{t}{2\sqrt{t^2-|\frac{y-x}{a}{|}^2}}dy.\end{array}$$
5.	记 $C(x,t;a)$ 是以 $(x,t)$ 为顶点、$D_{at}(x)\times 0$ 为底面的圆锥.$$\begin{array}{rl}u(t,x)=& {\partial }_t({-\int }_{D_{at}(x)}t\phi (y)\frac{t}{2\sqrt{t^2-|\frac{y-x}{a}{|}^2}}dy)+{-\int }_{D_{at}(x)}t\psi (y)\frac{t}{2\sqrt{t^2-|\frac{y-x}{a}{|}^2}}dy\\ & +{\int }_0^t{-\int }_{D_{a(t-\tau )}(x)}(t-\tau {)}^2\frac{f(\tau ,y)}{2\sqrt{(t-\tau {)}^2-|\frac{y-x}{a}{|}^2}}dyd\tau \\ =& {-\int }_{D_{at}(x)}(\phi (y)+\nabla \phi (y)\cdot (y-x)+t\psi (y))\frac{t}{2\sqrt{t^2-|\frac{y-x}{a}{|}^2}}dy\\ & +\frac{1}{2\pi a^2}{\int }_{C(x,t;a)}\frac{f(\tau ,y)}{\sqrt{(t-\tau {)}^2-|\frac{y-x}{a}{|}^2}}dyd\tau .\end{array}$$
6.	由于 $d{S}_{y}d(a\tau )=dy$, $$\begin{array}{rl}u(t,x)& ={\partial }_t({-\int }_{\partial B_{at}(x)}t\phi (y)d{S}_y)+{-\int }_{\partial B_{at}(x)}t\psi (y)d{S}_y+{\int }_0^t{-\int }_{\partial B_{a(t-\tau )}(x)}(t-\tau )f(\tau ,y)d{S}_{y}d\tau \\ & ={-\int }_{\partial B_{at}(x)}\phi (y)+\nabla \phi (y)\cdot (y-x)+t\psi (y)d{S}_y+\frac{1}{4\pi a^2}{\int }_{B_{at}(x)}\frac{f(t-|\frac{y-x}{a}|,y)}{|y-x|}dy.\end{array}$$
7.	令 $v=e^{kt}u$, 则 ${\partial }_t^{2}v={\partial }_t(e^{kt}({\partial }_t+k)u)=e^{kt}({\partial }_t+k{)}^{2}u$, 代入后 ${\partial }_t^{2}v=e^{kt}{\partial }_x^{2}u={\partial }_x^{2}v$.
8.	写出 ${\partial }_{t}u,{\partial }_{x}u$ 的表达式代入即可.
9.	令 $v=e^{-\frac{t+x}{2}}u$, 设 $v(t,x)=F(x+t)+G(x-t)$.
10.	设 $v(t,x)=F(x+at)+G(x-at)$. 可以不妨设 $F(0)=G(0)=0$. 若 $c\ne 0$, 由 $F(x)+G(-x)=c(F^{\prime }(x)+G^{\prime }(-x))$ 得$$(e^{\frac{x}{c}}G(-x){)}^{\prime }=e^{\frac{x}{c}}(-\frac{1}{c}F(x)+F^{\prime }(x)).$$
12.	(1)$$\begin{array}{rl}{\partial }_{r}h& =\frac{1}{{\omega }_{n-1}}{\int }_{S^{n-1}}\nabla \psi (x+r\xi )\cdot \xi d{S}_{\xi }\\ & =\frac{1}{{\omega }_{n-1}{r}^{n-1}}{\int }_{\partial B_r(x)}\frac{\partial \psi }{\partial \mathbf{n}}dS\\ & =\frac{1}{{\omega }_{n-1}{r}^{n-1}}{\int }_{B_r(x)}\Delta \psi dy\\ & =\frac{1}{r^{n-1}}{\int }_0^r\Delta h(x,\rho ){\rho }^{n-1}d\rho ,\end{array}$$因此$${\partial }_r^{2}h+\frac{n-1}{r}{\partial }_{r}h=\frac{1}{r^{n-1}}{\partial }_r(r^{n-1}{\partial }_{r}h)=\Delta h.$$ (2) 记 $x+r\xi =(x_1+r\mu ,\tilde{x}+r\tilde{\xi })$. $$\begin{array}{rl}h(x,r)& =\frac{1}{{\omega }_{n-1}}{\int }_{S^{n-1}}\psi (x+r\xi )d{S}_{\xi }\\ & =\frac{1}{{\omega }_{n-1}}{\int }_{-1}^1{\int }_{\sqrt{1-{\mu }^2}{S}^{n-2}}\varphi (x_1+r\mu )d{S}_{\tilde{\xi }}\frac{d\mu }{\sqrt{1-{\mu }^2}}\\ & =\frac{{\omega }_{n-2}}{{\omega }_{n-1}}{\int }_{-1}^1\varphi (x_1+r\mu ){\sqrt{1-{\mu }^2}}^{n-3}d\mu .\end{array}$$ (3) 把 $x$ 看作参数, $v(x,\cdot )$ 是关于 $r$ 的函数, 得到$$\begin{array}{rl}{\partial }_r^{2}w+\frac{n-1}{r}{\partial }_{r}w& \stackrel{(2)}{=}\frac{{\omega }_{n-2}}{{\omega }_{n-1}}{\int }_{-1}^1{\partial }_t^{2}v(r\mu ,x)(1-{\mu }^2{)}^{\frac{n-3}{2}}d\mu \\ & =\frac{{\omega }_{n-2}}{{\omega }_{n-1}}{\int }_{-1}^1\Delta v(r\mu ,x)(1-{\mu }^2{)}^{\frac{n-3}{2}}d\mu \\ & =\Delta w.\end{array}$$ 在 (2) 中取 $\varphi \equiv 1$ 得到$$w(x,0)=\frac{{\omega }_{n-2}}{{\omega }_{n-1}}{\int }_{-1}^{1}g(x)(1-{\mu }^2{)}^{\frac{n-3}{2}}d\mu \stackrel{\varphi \equiv 1}{=}g(x).$$ (4) 作变量代换 $r=t=\sqrt{s},r\mu =\sqrt{\sigma }$, 代入得到$${\omega }_{n-2}{\int }_0^{s}v(\sqrt{\sigma },x)(s-\sigma {)}^{\frac{n-3}{2}}\frac{d\sigma }{\sqrt{\sigma }}=s^{\frac{n-2}{2}}{\int }_{S^{n-1}}g(x+\sqrt{s}\xi )d{S}_{\xi }.$$两边对 $s$ 求 $\frac{n-1}{2}$ 阶导, 由于 ${\partial }_s={\partial }_{t^2}=\frac{1}{2t}{\partial }_t$, 得到$${\omega }_{n-2}v(t,x)\left(\frac{n-3}{2}\right)!=t\left(\frac{1}{2t}{\partial }_t{)}^{\frac{n-1}{2}}\right(t^{n-2}{\int }_{S^{n-1}}g(x+t\xi )d{S}_{\xi }).$$验算一下系数, $$\begin{array}{r}{\omega }_{n-2}\left(\frac{n-3}{2}\right)!=\frac{2{\pi }^{\frac{n-1}{2}}}{\Gamma (\frac{n-1}{2})}\Gamma \left(\frac{n-1}{2}\right)=\frac{\Gamma (\frac{n}{2}){\omega }_{n-1}}{\sqrt{\pi }}.\end{array}$$ (5) 记 $\bar{x}=(x,0)\in {\mathbb{R}}^{n+1}$, 令 $\bar{u}(t,\bar{x})=u(t,x)$, $\bar{g}(\bar{x})=g(x)$. 由于 $\partial B_t(\bar{x})$ 在 $(\xi ,\pm \sqrt{t^2-|x-\xi {|}^2})$ 处的球面测度与该点到 “圆盘”$B_t(x)$ 上的投影的测度之比为$$\frac{d{S}_{\xi }}{d\xi }=\frac{1}{\sqrt{1-|\frac{\xi -x}{t}{|}^2}},$$升维并利用 $n+1$ 维时解的表达式, 得到$$\begin{array}{rl}u(t,x)& =\bar{u}(t,\bar{x})\\ & =\frac{\sqrt{\pi }t}{\Gamma (\frac{n+1}{2})}\left(\frac{1}{2t}{\partial }_t{)}^{\frac{n}{2}}\right(t^{n-1}\frac{1}{{\omega }_n}{\int }_{S^n}\bar{g}(\bar{x}+t\xi )d{S}_{\xi })\\ & =\frac{\sqrt{\pi }t}{\Gamma (\frac{n+1}{2})}\left(\frac{1}{2t}{\partial }_t{)}^{\frac{n}{2}}\right(t^{n-1}\frac{2}{{\omega }_n}{\int }_{B^n}\frac{g(x+t\xi )}{\sqrt{1-|\xi {|}^2}}d\xi ).\end{array}$$ 注. 本题源自 [LZ] 第二章 1.2–1.4 (pp.19–26). (2) 的中间一个等号用的积分公式之证明见 [Gr] Appendix D.2. (4) 的表达式的系数能化简为$$\frac{\sqrt{\pi }}{\Gamma (\frac{n}{2})}(\frac{1}{2}{)}^{\frac{n-1}{2}}\frac{1}{{\omega }_{n-1}}=\frac{\Gamma (\frac{1}{2})}{\frac{n-2}{2}\cdots \frac{3}{2}\frac{1}{2}\Gamma (\frac{1}{2})}(\frac{1}{2}{)}^{\frac{n-1}{2}}\frac{1}{{\omega }_{n-1}}=\frac{1}{(n-2)!!{\omega }_{n-1}},$$故有$$\begin{array}{r}u(t,x)=\frac{t}{(n-2)!!}\left(\frac{1}{t}{\partial }_t{)}^{\frac{n-1}{2}}\right(t^{n-1}{-\int }_{\partial B_t(x)}g(\xi )d{S}_{\xi }).\end{array}$$(5) 的表达式的系数能化简为$$\frac{\sqrt{\pi }}{\Gamma (\frac{n+1}{2})}(\frac{1}{2}{)}^{\frac{n}{2}}\frac{2}{{\omega }_n}=\frac{\sqrt{\pi }}{\Gamma (\frac{n+1}{2})}\frac{(\frac{n-2}{2})!}{(n-2)!!}\frac{1}{{\omega }_n}=\frac{\sqrt{\pi }}{\Gamma (\frac{n+1}{2})}\frac{\Gamma (\frac{n}{2})}{n!!/n}\frac{1}{{\omega }_n}=\frac{2{\pi }^{\frac{n+1}{2}}/\Gamma (\frac{n+1}{2})}{(2{\pi }^{\frac{n}{2}}/\Gamma (\frac{n}{2}))/n}\frac{1}{n!!{\omega }_n}=\frac{1}{n!!v_n},$$其中 $v_n={\omega }_{n-1}/n$ 是 $B^n$ 的体积, 故有$$\begin{array}{r}u(t,x)=\frac{t}{n!!}\left(\frac{1}{t}{\partial }_t{)}^{\frac{n}{2}}\right(t^{n-1}{-\int }_{B_t(x)}\frac{g(\xi )}{\sqrt{1-|\frac{\xi -x}{t}{|}^2}}d\xi ).\end{array}$$

4.3 波的传播和解关于时间的衰减性

1.	注意 (2),(3) 中 $t$ 方向也是特征方向.
3.	特征方向 $l$ 满足方程$$\sum _{|\alpha |=m}{c}_{\alpha }(x)l^{\alpha }=0.$$ 特征曲面为 $\{x\mid \phi (x)=0\}$, 其中 $\phi \in C^1$ 满足$$\sum _{|\alpha |=m}{c}_{\alpha }(x)(\nabla \phi {)}^{\alpha }=0.$$ 其中对 $l=(l_1,\cdots ,l_n)$, $l^{\alpha }=l_1^{{\alpha }_1}\cdots l_n^{{\alpha }_n}$, $(\nabla \phi {)}^{\alpha }$ 的含义相同.
4.	假设 $\phi ,\psi$ 有一定的衰减性, 则对固定的 $x$, 当 $t\to +\infty$ 时 $u(t,x)\to \frac{1}{2}{\int }_{\mathbb{R}}\psi$.
5.	由于$$\begin{array}{rl}& {\partial }_t({-\int }_{D_t(x)}t\phi (y)\frac{t}{2\sqrt{t^2-|y-x{|}^2}}dy)\\ =& {\partial }_t\left({-\int }_{D_1(0)}\frac{t\phi (x+tz)}{2\sqrt{1-|z{|}^2}}dz\right)\\ =& {-\int }_{D_1(0)}\frac{\phi (x+tz)+t\nabla \phi (x+tz)\cdot z}{2\sqrt{1-|z{|}^2}}dz\\ =& {-\int }_{D_t(x)}\frac{t\phi (y)+t\nabla \phi (y)\cdot (y-x)}{2\sqrt{t^2-|y-x{|}^2}}dy,\end{array}$$得到$$\begin{array}{r}u(t,x)={-\int }_{D_t(x)}\frac{t\phi (y)+t\nabla \phi (y)\cdot (y-x)+t^2\psi (y)}{2\sqrt{t^2-|y-x{|}^2}}dy.\end{array}$$不过这里我们不用关心具体的表达式, 需要的仅是当 $t\ge 1$ 时$$\begin{array}{r}|u(t,x)|\le {-\int }_{D_t(x)}\frac{t^2\Phi (y)}{2\sqrt{t^2-|y-x{|}^2}}dy=\frac{1}{2\pi }{\int }_{D_t(0)}\frac{\Phi (x+y)}{\sqrt{t^2-|y{|}^2}}dy,\end{array}$$其中 $\Phi$ 是一个与 $\phi ,\psi$ 有关的非负光滑紧支函数. 下面对上式给出两种估计. 方法 I. 设 $\mathrm{supp}\Phi \subset D_R(0)$, 则$$\begin{array}{r}|u(t,x)|\le \frac{1}{2\pi }{\int }_{D_t(0)\cap D_R(-x)}\frac{dy}{\sqrt{t^2-|y{|}^2}}\Vert \Phi {\Vert }_{L^{\infty }}.\end{array}$$由于 $\frac{1}{\sqrt{t^2-|y{|}^2}}$ 是在靠近 $D_t(0)$ 的边缘处很大, 那么若 $D_R(-x)$ 在 $D_t(0)$ 的中央, 上式就不会太大. 具体写出来就是: 第一种情况, $|x|\le \frac{t}{2}$. 这时 $|y|\le |x|+R\le \frac{t}{2}+R$, 于是$$\begin{array}{r}|u(t,x)|\le \frac{|D_R(-x)|}{2\pi }\frac{1}{\sqrt{t^2-(\frac{t}{2}+R{)}^2}}\Vert \Phi {\Vert }_{L^{\infty }}\sim \frac{R^2}{\sqrt{3}t}\Vert \Phi {\Vert }_{L^{\infty }}.\end{array}$$接下来要考虑 $D_R(-x)$ 同 $D_t(0)$ 的边缘区域有重叠的情况, 这时 $|x|$ 相对 $t$ 不太小, 衰减性来自小圆 $D_R(-x)$ 相对于大圆 $D_t(0)$ 所占的圆心角 $\theta$ 仅有大约 $\frac{R}{|x|}$. 也就是说: 第二种情况, $|x|>\frac{t}{2}$. 对 $y=r\xi (r\in \mathbb{R},\xi \in S^1)$, $dy=rdrd{S}_{\xi }$. 这时$$|u(t,x)|\le \frac{\theta }{2\pi }{\int }_{[|x|-R,|x|+R]\cap [0,t]}\frac{rdr}{\sqrt{t^2-r^2}}\Vert \Phi {\Vert }_{L^{\infty }}.$$由于$$\frac{\theta }{2\pi }\le \frac{\sin \frac{\theta }{2}}{2}=\frac{R}{2|x|}$$以及$${\int }_{[|x|-R,|x|+R]\cap [0,t]}\frac{rdr}{\sqrt{t^2-r^2}}\le {\int }_{t-2R}^t\frac{rdr}{\sqrt{t^2-r^2}}=-\sqrt{t^2-r^2}{|}_{t-2R}^t\le 2\sqrt{Rt},$$相乘得$$|u(t,x)|\le \frac{R\sqrt{Rt}}{|x|}\Vert \Phi {\Vert }_{L^{\infty }}\le \frac{2R\sqrt{R}}{\sqrt{t}}\Vert \Phi {\Vert }_{L^{\infty }}.$$方法 II. 写成径向积分$$\begin{array}{r}|u(t,x)|\le \frac{1}{2\pi }{\int }_0^{t}r{\int }_{S^1}\Phi (x+r\xi )d{S}_{\xi }\frac{dr}{\sqrt{t^2-r^2}}.\end{array}$$分成两段, 对 $r$ 不太接近 $t$ 的部分可以直接估计: $$\begin{array}{rl}& {\int }_0^{t-\epsilon }r{\int }_{S^1}\Phi (x+r\xi )d{S}_{\xi }\frac{dr}{\sqrt{t^2-r^2}}\\ \le & \frac{1}{\sqrt{t^2-(t-\epsilon {)}^2}}{\int }_0^{t-\epsilon }r{\int }_{S^1}\Phi (x+r\xi )d{S}_{\xi }dr\\ \le & \frac{1}{\sqrt{2\epsilon t-{\epsilon }^2}}\Vert \Phi {\Vert }_{L^1}.\end{array}$$而当 $r$ 接近于 $t$ 时, 需要更好地利用 $\Phi$ 的衰减性: 注意到由于 $\Phi$ 是紧支的, 有$$\begin{array}{rl}& r{\int }_{S^1}\Phi (x+r\xi )d{S}_{\xi }\\ =& -r{\int }_{S^1}{\int }_r^{\infty }\frac{\partial \Phi }{\partial s}(x+s\xi )dsd{S}_{\xi }\\ \le & {\int }_r^{\infty }s{\int }_{S^1}|\nabla \Phi (x+s\xi )|d{S}_{\xi }ds\\ \le & \Vert \nabla \Phi {\Vert }_{L^1},\end{array}$$因此$$\begin{array}{rl}& {\int }_{t-\epsilon }^{t}r{\int }_{S^1}\Phi (x+r\xi )d{S}_{\xi }\frac{dr}{\sqrt{t^2-r^2}}\\ \le & {\int }_{t-\epsilon }^t\frac{dr}{\sqrt{t^2-r^2}}\Vert \nabla \Phi {\Vert }_{L^1}\\ \le & \frac{1}{\sqrt{t}}{\int }_{t-\epsilon }^t\frac{dr}{\sqrt{t-r}}\Vert \nabla \Phi {\Vert }_{L^1}\\ =& \frac{2\sqrt{\epsilon }}{\sqrt{t}}\Vert \nabla \Phi {\Vert }_{L^1}.\end{array}$$最后 $\epsilon$ 取为随便一个常数即可, 两式相加就有 $|u(t,x)|\le \frac{C}{\sqrt{t}}\Vert \Phi {\Vert }_{W^{1,1}}$ 对 $x\in {\mathbb{R}}^2$ 一致地成立. 注. (i) 第二个估计的优点在于, 这一上界不直接与支集半径 $R$ 有关. (ii) $n$ 维的结论见 [LZ] 第四章定理 6.1 (p.85), 推论 6.1 (pp.90–1).

4.4 分离变量法和初边值问题解的存在性

2.	令 $v=e^{\frac{t}{2}}u$, 则$$\{\begin{array}{rl}& {\partial }_t^{2}v-{\partial }_x^{2}v=\frac{v}{4},\\ & v(0,x)=\phi (x),{\partial }_{t}v(0,x)=\psi (x)+\frac{1}{2}\phi (x),\\ & v(t,0)=v(t,L)=0.\end{array}$$考虑非零特解 $v(t,x)=T(t)X(x)$, 代入方程得到 $\frac{T^{\prime \prime }(t)}{T(t)}=\frac{X^{\prime \prime }(x)}{X(x)}+\frac{1}{4}=-\lambda$, 由边界条件$$X(0)=X(L)=0$$得到必须 $\lambda =\frac{k^2{\pi }^2}{L^2}-\frac{1}{4},k\in {\mathbb{Z}}^{+}$, 此时 $X_k(x)=C_k\sin \frac{k\pi x}{L}$, $$T_k(t)=\{\begin{array}{ll}{A}_k\cos \left(\sqrt{\frac{k^2{\pi }^2}{L^2}-\frac{1}{4}}t\right)+B_k\sin \left(\sqrt{\frac{k^2{\pi }^2}{L^2}-\frac{1}{4}}t\right),& \lambda =\frac{k^2{\pi }^2}{L^2}-\frac{1}{4}>0,\text{ 即 }k>\frac{L}{2\pi },\\ A_k\exp \left(\sqrt{-\frac{k^2{\pi }^2}{L^2}+\frac{1}{4}}t\right)+B_k\exp (-\sqrt{-\frac{k^2{\pi }^2}{L^2}+\frac{1}{4}}t),& \lambda =\frac{k^2{\pi }^2}{L^2}-\frac{1}{4}<0,\text{ 即 }k<\frac{L}{2\pi },\\ At+B,& \lambda =\frac{k^2{\pi }^2}{L^2}-\frac{1}{4}=0,\text{ 即 }k=\frac{L}{2\pi }.\end{array}$$
3.	从相容性条件$$\{\begin{array}{rl}& 0=u(0,\pi )=\alpha {\pi }^4+\beta {\pi }^3,\\ & 0={\partial }_{t}u(0,\pi )=-\gamma ,\\ & 0={\partial }_t^{2}u(0,\pi )={\partial }_x^{2}u(0,\pi )=12\alpha {\pi }^2+6\beta \pi ,\end{array}$$解得 $\alpha =\beta =\gamma =0$. 求解方程得到$$u(t,x)=\cos t\sin x.$$
4.	(1) 由于$$h^{\prime }(t)={\int }_0^1(2{\partial }_{t}u{\partial }_t^{2}u+8{\partial }_{x}u{\partial }_{xt}u)dx=(8{\partial }_{t}u{\partial }_{x}u){|}_{x=0}^1=0,$$得到$$h\left(\frac{1}{3}\right)=h(0)={\int }_0^1(30x(1-x){)}^2+4(12\pi {\sin }^2\pi x\cos \pi x{)}^{2}dx=30+36{\pi }^2.$$ (2) 由解的表达式$$u(t,x)=\sum _{k=1}^{\infty }(A_k\cos 2k\pi t+B_k\sin 2k\pi t)\sin k\pi x$$得到$$u(2,x)=u(0,x)=4{\sin }^3\pi x.$$
5.	设 $v(t,x)=u(t,x)+{\lambda }_1(x)g_1(t)+{\lambda }_2(x)g_2(t)$. 要让$$\{\begin{array}{rl}(-{\partial }_x+\alpha )v(t,0)=0& \\ ({\partial }_x+\beta )v(t,L)=0& \end{array},$$可以令 (1) $v(t,x)=u(t,x)-\frac{(x-L{)}^2}{2L+\alpha L^2}{g}_1(t)-\frac{x^2}{2L+\beta L^2}{g}_2(t)$. (2) $v(t,x)=u(t,x)-\frac{(x-L{)}^2}{2L}{g}_1(t)-\frac{x^2}{2L}{g}_2(t)$. (3) $v(t,x)=u(t,x)-\frac{1}{\alpha }{g}_1(t)-\frac{x^2}{2L}{g}_2(t)$.

4.5 能量方法和解的唯一性与稳定性

   由于习题课可能讲不到这一节, 这里给出所有的解答.

1.	同 4.5.2 节的推导, 得到$$\frac{dE(t)}{dt}-{\int }_{B_{t_0-t}(x_0)}{\partial }_{t}ufdx=-\frac{dF(t)}{dt},$$于是$$E(t)+F(t)=E(0)+{\int }_0^t{\int }_{B_{t_0-s}(x_0)}{\partial }_{t}ufdxds.$$定义$$G(t)={\int }_0^t{\int }_{B_{t_0-s}(x_0)}(({\partial }_{t}u{)}^2+|\nabla u{|}^2)(s,x)dxds,$$则$$\begin{array}{rl}\frac{dG(t)}{dt}& ={\int }_{B_{t_0-t}(x_0)}({\partial }_{t}u{)}^2+|\nabla u{|}^{2}dx\\ & =2E(t)\\ & \le 2(E(0)+{\int }_0^t{\int }_{B_{t_0-s}(x_0)}{\partial }_{t}ufdxds)\\ & \le 2E(0)+G(t)+{\int }_0^t{\int }_{B_{t_0-s}(x_0)}{f}^{2}dxds.\end{array}$$由 Gronwall 不等式,$$\begin{array}{rl}G(t)& \le C(T)(2E(0)+{\int }_0^t{\int }_{B_{t_0-s}(x_0)}{f}^{2}dxds)\\ & \le C(T)(2\Vert \psi {\Vert }_{L^2}^2+2\Vert \nabla \phi {\Vert }_{L^2}^2+{\int }_0^T{\int }_{B_{t_0-s}(x_0)}{f}^{2}dxds).\end{array}$$
2.	$$\frac{dE(t)}{dt}=-{\int }_{\Omega }({\partial }_{t}u{)}^{2}dx\le 0.$$
3.	我们有$$\begin{array}{rl}{\int }_{\Omega }{u}^2(t,x)dx& ={\int }_{\Omega }{u}^2(0,x)dx+{\int }_{\Omega }{\int }_0^t{\partial }_t(u^2)dtdx\\ & \le {\int }_{\Omega }{u}^2(0,x)dx+{\int }_{\Omega }{\int }_0^t{u}^2+({\partial }_{t}u{)}^{2}dtdx\\ & \le \Vert \phi {\Vert }_{L^2}^2+C{\int }_0^{t}E(t)dt.\end{array}$$然后用定理 4.5.1 即可. 上面最后一个不等号用了 Poincaré 不等式, 即: $\forall s\in (0,t)$, $\partial \Omega$ 上 $u(s,\cdot )=0$ 推出$$\Vert u(s,\cdot ){\Vert }_{L^2(\Omega )}\le C\Vert \nabla u(s,\cdot ){\Vert }_{L^2(\Omega )}.$$不用了 Poincaré 不等式的方法是, 同下一题转而考虑 $F(t)$ 而不是 $E(t)$.
4.	设 $u_1$, $u_2$ 都是解, 令 $v=u_1-u_2$, 则$$\{\begin{array}{rl}& {\partial }_t^{2}v-\Delta v-{\partial }_{t}v=f(u_1)-f(u_2),\\ & v(0,x)=0,{\partial }_{t}v(0,x)=0,\\ & \frac{\partial v}{\partial \mathbf{n}}(t,x)=0.\end{array}$$由 $f$ 是 Lipschitz 的, 有$$|f(u_1)-f(u_2)|\le \alpha |u_1-u_2|=\alpha |v|,$$则$$\begin{array}{rl}\frac{dE(t)}{dt}& ={\int }_{\Omega }{\partial }_{t}v({\partial }_{t}v+f(u_1)-f(u_2))dx\\ & \le (1+\frac{\alpha }{2}){\int }_{\Omega }({\partial }_{t}v{)}^2+v^{2}dx.\end{array}$$为了获得合适的微分不等式, 令$$F(t)={\int }_{\Omega }{v}^2+({\partial }_{t}v{)}^2+|\nabla v{|}^{2}dx,$$则$$\begin{array}{rl}\frac{dF(t)}{dt}& ={\int }_{\Omega }2v{\partial }_{t}vdx+\frac{dE(t)}{dt}\\ & \le (2+\frac{\alpha }{2}){\int }_{\Omega }({\partial }_{t}v{)}^2+v^{2}dx\\ & \le (2+\frac{\alpha }{2})F(t).\end{array}$$由 Gronwall 不等式,$$\begin{array}{rl}F(t)& \le C(t,\alpha )F(0)=0,\end{array}$$因此 $v\equiv 0$.
5.	(1)$$\frac{dE(t)}{dt}={\int }_{\Omega }{\partial }_{t}u(-u^3)dx=-\frac{d}{dt}\frac{1}{4}{\int }_{\Omega }{u}^{4}dx,$$即 $E(t)+{\frac{1}{4}\int }_{\Omega }{u}^{4}dx$ 是常数. (2) 此时 $E(0)+{\frac{1}{4}\int }_{\Omega }{u}^{4}dx{|}_{t=0}=0$.
6.	同 4.5.2 的推导, 得到$$\frac{dE(t)}{dt}=-{\int }_{B_{t_0-t}(x_0)}({\partial }_{t}u{)}^{2n+2}dx-\frac{dF(t)}{dt}\le -\frac{dF(t)}{dt}.$$对 $t_0=s$, 当 $|x_0|$ 很大时 $B_s(x_0)$ 与 $\mathrm{supp}\phi \cup \mathrm{supp}\psi$ 不相交, 则 $E(0)=0$, 从而 $E\equiv 0$, 于是 $u(s,x_0)=0$.
7.	设 $u_1$, $u_2$ 都是解, 令 $v=u_1-u_2$, 则$$\{\begin{array}{rl}& {\partial }_t^{2}v-\Delta v+|{\partial }_t{u}_1{|}^{p-1}{\partial }_t{u}_1-|{\partial }_t{u}_2{|}^{p-1}{\partial }_t{u}_2=0,\\ & v(0,x)=0,{\partial }_{t}v(0,x)=0.\end{array}$$注意到由于 $|x{|}^{p-1}x$ 递增, 有$${\partial }_{t}v(|{\partial }_t{u}_1{|}^{p-1}{\partial }_t{u}_1-|{\partial }_t{u}_2{|}^{p-1}{\partial }_t{u}_2)\ge 0,$$因此$${\partial }_{t}v({\partial }_t^{2}v-\Delta v)\le 0.$$那么, 同 4.5.2 节的推导得到$$\frac{dE(t)}{dt}\le -\frac{dF(t)}{dt}.$$
8.	(1) 由 (2) 可得. (2) 以下每次出现的 $C$ 代表不同的常数. 记$$Lu=-\sum _{i,j=1}^n{\partial }_{x_i}(a_{ij}{\partial }_{x_j}u)+\sum _{k=1}^n{b}_k{\partial }_{x_k}u+cu.$$与波动方程类似, 来估计$${\int }_0^t{\int }_{\Omega }{\partial }_{t}u({\partial }_t^{2}u+Lu)dxdt.$$首先, 由 $u$ 在 $\partial \Omega$ 恒为 $0$ 得到 ${\partial }_{t}u$ 在 $\partial \Omega$ 也为 $0$, 于是$$\begin{array}{rl}& -{\int }_{\Omega }{\partial }_{t}u\sum _{i,j=1}^n{\partial }_{x_i}(a_{ij}{\partial }_{x_j}u)dx\\ =& {\int }_{\Omega }\sum _{i,j=1}^n{\partial }_{x_{i}t}u{a}_{ij}{\partial }_{x_j}udx\\ =& \frac{1}{2}({\int }_{\Omega }\sum _{i,j=1}^n{\partial }_{x_{i}t}u{a}_{ij}{\partial }_{x_j}udx+{\int }_{\Omega }\sum _{i,j=1}^n{\partial }_{x_{j}t}u{a}_{ij}{\partial }_{x_i}udx)\\ =& \frac{1}{2}{\int }_{\Omega }{\partial }_t\left(\sum _{i,j=1}^n{a}_{ij}{\partial }_{x_i}u{\partial }_{x_j}u\right)dx-\frac{1}{2}{\int }_{\Omega }\sum _{i,j=1}^n{\partial }_t{a}_{ij}{\partial }_{x_i}u{\partial }_{x_j}udx,\end{array}$$中间的等号用到了 $a_{ij}=a_{ji}$. 由 ${\partial }_t{a}_{ij}$ 的有界性, 上式的第二项$$\begin{array}{r}\left|\frac{1}{2}{\int }_{\Omega }\sum _{i,j=1}^n{\partial }_t{a}_{ij}{\partial }_{x_i}u{\partial }_{x_j}udx\right|\le CE(t).\end{array}$$此外, $$|{\int }_{\Omega }\sum _{k=1}^n{\partial }_{t}u(b_k{\partial }_{x_k}u+cu)dx|\le CE(t).$$又$${\int }_0^t\frac{1}{2}{\int }_{\Omega }{\partial }_t\left(\sum _{i,j=1}^n{a}_{ij}{\partial }_{x_i}u{\partial }_{x_j}u\right)dxdt=\frac{1}{2}{\int }_{\Omega }\sum _{i,j=1}^n{a}_{ij}{\partial }_{x_i}u{\partial }_{x_j}udx{|}_0^t,$$以及$${\int }_0^t{\int }_{\Omega }{\partial }_{t}u{\partial }_t^{2}udxdt=\frac{1}{2}{\int }_{\Omega }({\partial }_{t}u{)}^{2}dx{|}_0^t.$$结合以上四式得到$$\begin{array}{rl}& \frac{1}{2}{\int }_{\Omega }({\partial }_{t}u{)}^2+a_{ij}{\partial }_{x_i}u{\partial }_{x_j}udx{|}_0^t-C{\int }_0^{t}E(t)dt\\ \le & {\int }_0^t{\int }_{\Omega }{\partial }_{t}u({\partial }_t^{2}u+Lu)dxdt\\ =& {\int }_0^t{\int }_{\Omega }{\partial }_{t}ufdxdt\\ \le & \frac{1}{2}{\int }_0^t{\int }_{\Omega }({\partial }_{t}u{)}^{2}dxdt+\frac{1}{2}{\int }_0^t{\int }_{\Omega }|f{|}^{2}dxdt,\end{array}$$即$$\begin{array}{rl}& {\int }_{\Omega }({\partial }_{t}u{)}^2+a_{ij}{\partial }_{x_i}u{\partial }_{x_j}udx{|}_t\\ \le & {\int }_{\Omega }({\partial }_{t}u{)}^2+a_{ij}{\partial }_{x_i}u{\partial }_{x_j}udx{|}_0+{\int }_0^t{\int }_{\Omega }({\partial }_{t}u{)}^{2}dxdt+{\int }_0^t{\int }_{\Omega }|f{|}^{2}dxdt+C{\int }_0^{t}E(t)dt.\end{array}$$由 $(a_{ij}{)}_{i,j}$ 的一致椭圆性和有界性, 从上式进一步得到$$\begin{array}{r}{\int }_{\Omega }({\partial }_{t}u{)}^2+|\nabla u{|}^{2}dx{|}_t\le C(E(0)+{\int }_0^t{\int }_{\Omega }|f{|}^{2}dxdt+{\int }_0^{t}E(t)dt).\end{array}$$还剩下 ${\int }_{\Omega }{u}^{2}dx{|}_t$ 要估计. 我们有$$\begin{array}{rl}{\int }_{\Omega }{u}^2(t,x)dx& ={\int }_{\Omega }{u}^2(0,x)dx+{\int }_{\Omega }{\int }_0^t{\partial }_t(u^2)dtdx\\ & \le {\int }_{\Omega }{u}^2(0,x)dx+{\int }_{\Omega }{\int }_0^t{u}^2+({\partial }_{t}u{)}^{2}dtdx\\ & \le E(0)+{\int }_0^{t}E(t)dt.\end{array}$$以上两式相加得$$\begin{array}{rl}E(t)& \le C(E(0)+{\int }_0^t{\int }_{\Omega }|f{|}^{2}dxdt+{\int }_0^{t}E(t)dt)\\ & \le C(E(0)+{\int }_0^T{\int }_{\Omega }|f{|}^{2}dxdt+{\int }_0^{t}E(t)dt).\end{array}$$由 Gronwall 不等式得$$\begin{array}{r}E(t)\le C(T)(E(0)+{\int }_0^T{\int }_{\Omega }|f{|}^{2}dxdt).\end{array}$$(3) 为了仍有$${\int }_{\partial \Omega }{\partial }_{t}u\sum _{i,j=1}^n{a}_{ij}{\partial }_{x_j}u{\mathbf{n}}_{i}dS=0,$$其中 $\mathbf{n}=({\mathbf{n}}_1,\cdots ,{\mathbf{n}}_n)$, 需要的边界条件为$$\sum _{i,j=1}^n{a}_{ij}{\partial }_{x_j}u{\mathbf{n}}_i=0,(t,x)\in (0,T)\times \partial \Omega .$$

参考文献

[Gr]	L. Grafakos. Classical Fourier Analysis. Graduate Texts in Mathematics 249. Springer, 2014.
[LZ]	李大潜, 周忆. 非线性波动方程. 现代数学丛书. 上海科学技术出版社, 2016. ➢Springer 版下载