第二章 热方程

2.2 分离变量法和初边值问题解的存在性

5.

令 $v(t,x)=u(t,x)-\cos x$, 满足齐次方程$$\{\begin{array}{rl}{\partial }_{t}v-{\partial }_x^{2}v& =0,\\ v(0,x)& =\cos 2x-\cos x,\\ {\partial }_{t}v(t,0)& ={\partial }_{t}v(t,2\pi )=0.\end{array}$$解为$$v(t,x)=\sum _{k=0}^{\infty }{a}_k{e}^{-{\left(\frac{k}{2}\right)}^{2}t}\cos \frac{kx}{2},$$其中$$a_k=\frac{2}{2\pi }{\int }_0^{2\pi }(\cos 2x-\cos x)\cos \frac{kx}{2}dx={\delta }_{2,\frac{k}{2}}-{\delta }_{1,\frac{k}{2}}=\{\begin{array}{ll}-1,& k=2\\ 1,& k=4\\ 0,& \text{其它}\end{array}.$$所以$$u(t,x)=(1-e^{-t})\cos x+e^{-4t}\cos 2x.$$

6.

令 $\phi (x)$ 满足方程$$\{\begin{array}{rl}{\phi }^{\prime \prime }(x)-\phi (x)& =0,\\ \phi (0)& =0,\\ {\phi }^{\prime }(L)+\phi (L)& =\alpha ,\end{array}$$解得 $\phi (x)=\frac{\alpha }{2}\left(e^{x-L}-e^{-x-L}\right)$. 令 $v(t,x)=e^t(u(t,x)-\phi (x))$, 则 $v$ 满足方程$$\{\begin{array}{rl}{\partial }_{t}v-{\partial }_x^{2}v& =0,\\ v(0,x)& =-\phi (x),\\ v(t,0)& =0,\\ {\partial }_{x}v(t,L)& +v(t,L)=0.\end{array}$$用分离变量法来求解. 寻找非零特解 $v(t,x)=T(t)X(x)$, 代入方程得到 $\frac{T^{\prime }(t)}{T(t)}=\frac{X^{\prime \prime }(x)}{X(x)}=-\lambda$, 再考虑边界条件得到$$\{\begin{array}{rl}-X^{\prime \prime }(x)& =\lambda X(x),\\ X(0)& =0,\\ X^{\prime }(L)& +X(L)=0.\end{array}$$由 Sturm-Liouville 定理, $\{\sin \sqrt{{\lambda }_k}x{\}}_{k=1}^{\infty }$ 构成 $L^2(0,L)$ 的完备正交基, 其中 ${\lambda }_k$ 是方程 $\sqrt{\lambda }\cos \sqrt{\lambda }L+\sin \sqrt{\lambda }L=0$ 的第 $k$ 个正根. 从而得到$$v(t,x)=\sum _{k=1}^{\infty }{a}_k{e}^{-{\lambda }_{k}t}\sin \sqrt{{\lambda }_k}x,$$其中$$a_k=\frac{{\int }_0^L-\phi (x)\sin \sqrt{{\lambda }_k}xdx}{{\int }_0^L{\sin }^2\sqrt{{\lambda }_k}xdx}=\frac{4\alpha e^{-L}}{{\lambda }_k+1}\frac{{\lambda }_k\sinh L\cos \sqrt{{\lambda }_k}L-\sqrt{{\lambda }_k}\cosh L\sin \sqrt{{\lambda }_k}L}{2\sqrt{{\lambda }_k}L-\sin 2\sqrt{{\lambda }_k}L}.$$最后,$$u(t,x)=\phi (x)+v(t,x)e^{-t}.$$

2.3 Fourier 变换和 Cauchy 问题解的存在性

5.	(1) 把$$u(t,x)=\frac{1}{(4\pi t{)}^{\frac{n}{2}}}{\int }_{\mathbb{R}}\phi (y)e^{-\frac{|x-y{|}^2}{4t}}dy,t>0,x\in \mathbb{R}$$延拓为$$u(t,z)=\frac{1}{(4\pi t{)}^{\frac{n}{2}}}{\int }_{\mathbb{R}}\phi (y)e^{-\frac{|z-y{|}^2}{4t}}dy,t>0,z\in \mathbb{C},$$这是关于 $z$ 的全纯函数, 限制回 $\mathbb{R}$ 仍是解析函数.
6.	(1) $\alpha =1$. (2) $f$ 满足 $f+\xi f^{\prime }+2f^{\prime \prime }=0$. $H(t,x)=\frac{1}{\sqrt{4\pi t}}{e}^{-{\left(\frac{x}{\sqrt{t}}\right)}^2/4}$ 启示我们可以设 $f(\xi )=\frac{1}{\sqrt{4\pi }}{e}^{-\frac{{\xi }^2}{4}}g(\xi )$. 代入化简得到 $-\frac{\xi }{2}{g}^{\prime }+g^{\prime \prime }=0$, 于是 $g(\xi )=C{\int }_0^{\xi }{e}^{\frac{s^2}{4}}ds+C^{\prime }$. 由 $\underset{\xi \to \infty }{\lim }{f}^{\prime }(\xi )=0$ 得 $C=0$, 由 ${\int }_{\mathbb{R}}f=1$ 得 $C^{\prime }=1$, 因此 $f(\xi )=\frac{1}{\sqrt{4\pi }}{e}^{-\frac{{\xi }^2}{4}}$.
7.	可参考 [Evans] 2.3.1 Theorem 1, 2 (P47-51), 或 [Han] Theorem 5.2.3, 5.2.11 (P160-2,173-4).
8.	方法 I. 由 (2.23) 与命题 2.3.4, $$\begin{array}{rl}(H(t,\cdot )\ast H(s,\cdot ))(x-z)& =(F^{-1}[e^{-|\cdot {|}^{2}t}]\ast F^{-1}[e^{-|\cdot {|}^{2}s}])(x-z)\\ & =F^{-1}[e^{-|\cdot {|}^2(t+s)}](x-z)=H(t+s,x-z).\end{array}$$ 方法 II. 记左式积分为 $I$, 可展开为$$\begin{array}{rl}I& ={\int }_{{\mathbb{R}}^n}H(t,x-y)H(s,y-z)dy\\ & ={\int }_{{\mathbb{R}}^n}\frac{\exp \left(-\frac{(x-y{)}^2}{4t}-\frac{(y-z{)}^2}{4s}\right)}{(4\pi t{)}^{\frac{n}{2}}(4\pi s{)}^{\frac{n}{2}}}dy\\ & =D{\int }_{{\mathbb{R}}^n}\exp (-(A{y}^2-B\cdot y+C))dy,\end{array}$$其中定义$$D=\frac{1}{(4\pi t{)}^{\frac{n}{2}}(4\pi s{)}^{\frac{n}{2}}},A=\frac{1}{4t}+\frac{1}{4s},B=\frac{x}{2t}+\frac{z}{2s},C=\frac{x^2}{4t}+\frac{z^2}{4s}.$$考虑配凑平方有$$\begin{array}{rl}I& =D\exp \left(\frac{B^2}{4A}-C\right){\int }_{{\mathbb{R}}^n}^{}\exp \left(-A{\left(y-\frac{B}{2A}\right)}^2\right)dy\\ & =D\exp \left(\frac{B^2}{4A}-C\right){\left({\int }_{\mathbb{R}}^{}\exp \left(-A{y}^2\right)dy\right)}^n\\ & =D{\left(\frac{\sqrt{\pi }}{\sqrt{A}}\right)}^n\exp \left(\frac{B^2}{4A}-C\right).\end{array}$$分别计算$$\begin{array}{rl}D{\left(\frac{\pi }{A}\right)}^{\frac{n}{2}}& =\frac{1}{(4\pi t{)}^{\frac{n}{2}}(4\pi s{)}^{\frac{n}{2}}}{\left(\frac{4\pi }{\frac{s+t}{st}}\right)}^{\frac{n}{2}}=\frac{1}{(4\pi (s+t){)}^{\frac{n}{2}}},\\ \frac{B^2}{4A}-C& =\frac{{\left(\frac{x}{2t}+\frac{z}{2s}\right)}^2}{\frac{1}{t}+\frac{1}{s}}-\left(\frac{x^2}{4t}+\frac{z^2}{4s}\right)=-\frac{(x-z{)}^2}{4(s+t)},\end{array}$$可见计算结果正是 $H(s+t,x-z)$.
9.	$$\begin{array}{rl}{\int }_0^{\infty }H(t,x-y)dt& ={\int }_0^{\infty }\frac{1}{(4\pi t{)}^{\frac{n}{2}}}{e}^{-\frac{|x-y{|}^2}{4t}}dt\\ & ={\int }_0^{\infty }\frac{1}{(\pi \frac{|x-y{|}^2}{s}{)}^{\frac{n}{2}}}{e}^{-s}d\frac{|x-y{|}^2}{4s}\\ & =\frac{\Gamma (\frac{n}{2}-1)}{4{\pi }^{\frac{n}{2}}|x-y{|}^{n-2}}\\ & =\frac{\Gamma (\frac{n}{2})}{(n-2)2{\pi }^{\frac{n}{2}}|x-y{|}^{n-2}}\\ & =\frac{1}{(n-2){\omega }_{n-1}|x-y{|}^{n-2}}.\end{array}$$
11.	把 $\hat{u}(\xi )={\int }_{\mathbb{R}}u(x)e^{-i\xi x}dx,\xi \in \mathbb{R}$ 延拓为全纯函数 $\hat{u}(z)={\int }_{\mathbb{R}}u(x)e^{-izx}dx,z\in \mathbb{C}$, 由全纯函数的零点离散及 $\hat{u}(\xi )$ 具有紧支集得 $\hat{u}\equiv 0$, 于是 $u\equiv 0$.

2.4 极值原理及其应用

2.	令 $v=e^{-c_{0}t}u$, $v_{\epsilon }=v-\epsilon t,\epsilon >0$. 假设 $\underset{\overline{{\Omega }_T}}{\max }{v}_{\epsilon }>0$, 由于在 ${\Gamma }_T$ 有 $v_{\epsilon }\le 0$, $\underset{\overline{{\Omega }_T}}{\max }{v}_{\epsilon }$ 在 ${\Omega }_T$ 内某点取到. 在该点 ${\partial }_t{v}_{\epsilon }\ge 0$, $\Delta v_{\epsilon }\le 0$, 然而 $({\partial }_t-\Delta )v_{\epsilon }=e^{-c_{0}t}({\partial }_t-\Delta +c)u-(c+c_0)v-\epsilon <0$, 矛盾. 因此 $v_{\epsilon }\le 0$, 令 $\epsilon \to 0$ 得 $v\le 0$, 从而 $u\le 0$.
3.	本题是 [Han] Theorem 5.3.8 (P185-6).
4.	本题是 [Han] Theorem 5.3.9 (P186-8). 要让$$\{\begin{array}{rlrl}({\partial }_t-\Delta +c)(\pm u)& \le ({\partial }_t-\Delta +c)w,& (t,x)& \in (B_R{)}_T,\\ \pm u& \le w,& (t,x)& \in {\Gamma }_T,\end{array}$$ 只需$$\{\begin{array}{rlrl}0& \le ({\partial }_t-\Delta +c)(e^{c_{0}t}{v}_R),& (t,x)& \in (B_R{)}_T,\\ |u|& \le v_R,& (t,x)& \in (0,T]\times \partial B_R,\\ 0& \le v_R(0,x),& x& \in B_R.\end{array}$$ 可以取$$v_R=\frac{\underset{(0,T]\times \partial B_R}{\sup }|u|}{R^2}(|x{|}^2+2nt).$$
8.	把 $u_0(t)=e^{-\frac{1}{t^{\alpha }}},t>0$ 延拓为全纯函数 $u_0(z)=e^{-\frac{1}{z^{\alpha }}},\mathrm{Re}z>0$. 取 $c\in (0,1)$ 很小使得对于 $|z-t|=ct$ 有 $\mathrm{Re}\left(-\frac{1}{z^{\alpha }}\right)<-\frac{1}{2t^{\alpha }}$, 由 Cauchy 积分公式, $$u_0^{(k)}(t)=\frac{k!}{2\pi i}{\int }_{|z-t|=ct}\frac{e^{-\frac{1}{z^{\alpha }}}}{(z-t{)}^{k+1}}dz.$$于是$$|u_0^{(k)}(t)|\le k!\frac{\underset{|z-t|=ct}{\min }{e}^{\mathrm{Re}\left(-\frac{1}{z^{\alpha }}\right)}}{(ct{)}^k}<\frac{k!}{(ct{)}^k}{e}^{-\frac{1}{2t^{\alpha }}}.$$
9.	(1) ${\partial }_{t}u=\Delta u=\Delta (\phi \ast H)=(\Delta \phi )\ast H\ge 0$. (2) 对给定的 $x\in {\mathbb{R}}^n$, $$\Delta u={\partial }_{t}u=\phi \ast ({\partial }_{t}H)=\phi \ast \left(\right(\frac{|x{|}^2}{4t^2}-\frac{n}{2t}\left)H\right)\le C\Vert (\frac{|x{|}^2}{4t^2}-\frac{n}{2t})H{\Vert }_{L^1}\le \frac{C^{\prime }}{t}\to 0$$当 $t\to \infty$, 且这个收敛在 $x$ 的有界集上是一致的. 那么, 对于$$\frac{1}{|\partial B_R(x^0)|}{\int }_{\partial B_R(x^0)}u(t,x)d{S}_x-u(t,x^0)=\frac{1}{w_{n-1}}{\int }_0^R{\int }_{B_r(x^0)}\Delta udxdr$$(见定理 1.2.1 的证明) , 令 $t\to +\infty$ 得到 $v$ 满足平均值性质. (3) Liouville 定理.
10.	把 $t$ 限制在 $(0,T]$ 上. 需构造满足 $\{\begin{array}{c}{\varphi }^{\prime \prime }\le -\mu \varphi \\ \varphi \ge \underset{\Omega }{\sup }|u_0|\\ \varphi \le C\underset{\Omega }{\sup }|u_0|\end{array}$ 的 $\varphi$. 例如$$\varphi (x_n)=(-x_n^2+\underset{\Omega }{\sup }{x}_n^2+1)\underset{\Omega }{\sup }|u_0|,$$只需要 $-2\le -\mu (\underset{\Omega }{\sup }{x}_n^2+1),C=\underset{\Omega }{\sup }{x}_n^2+1$.
11.	见 [John] 7.1(d) (P222-6) .

2.5 能量方法和解的唯一性

太简单了.

参考文献

[Evans]	L. C. Evans. Partial Differential Equations. Graduate Studies in Mathematics 19. Amer. Math. Soc, 2010.
[Han]	Q. Han. A Basic Course in Partial Differential Equations. Graduate Studies in Mathematics 120. Amer. Math. Soc, 2011.
[John]	F. John. Partial Differential Equations. Applied Mathematical Sciences 1. Springer, 1991.