用户: Fyx1123581347/偏微分方程/作业

1A

A1) $$\begin{array}{rl}{L}_t(e^{it\Phi (x)}a(x))& =\frac{1}{1+t|{\Phi }^{\prime }(x){|}^2}(e^{it\Phi (x)}a(x)-i{\Phi }^{\prime }(x)(e^{it\Phi (x)}{a}^{\prime }(x)+it{\Phi }^{\prime }(x)e^{it\Phi (x)}a(x)))\\ & =a(x)e^{it\Phi (x)}-e^{it\Phi (x)}\frac{i{\Phi }^{\prime }(x)}{1+t|{\Phi }^{\prime }(x){|}^2}{a}^{\prime }(x),\end{array}$$因此$$I(t)=I_1(t)+{\int }_{\mathbb{R}}{L}_t(e^{it\Phi (x)}a(x)).$$由分部积分,$${\int }_{\mathbb{R}}{L}_t(e^{it\Phi (x)}a(x))=I_2(t).$$A2) 因 $t>0$, 有$$\begin{array}{r}{\left|1+i{\Phi }^{\prime \prime }(x)-2i\frac{t|{\Phi }^{\prime }(x){|}^2{\Phi }^{\prime \prime }(x)}{1+t|{\Phi }^{\prime }(x){|}^2}\right|}^2\le 1+|{\Phi }^{\prime \prime }(x){|}^2\le \frac{1+c_0^2}{c_0^2}{\Phi }^{\prime \prime }(x{)}^2.\end{array}$$取 $a(x)\ne 0$ 的连通分支 $I_k$, 其为 $\mathbb{R}$ 上开区间, 故至多可数个. 在每一连通分支 $I_k$ 上都有$$\begin{array}{rl}{I}_2(t)\text{的积分项}& \le \frac{\sqrt{1+c_0^2}}{c_0}{\int }_{I_k}\frac{{\Phi }^{\prime \prime }(x)}{1+t{\Phi }^{\prime }(x{)}^2}|a(x)|dx\\ & \le \frac{\sqrt{1+c_0^2}}{c_0}{\int }_{I_k}\frac{{\Phi }^{\prime \prime }(x)}{1+t{\Phi }^{\prime }(x{)}^2}{\int }_{I_k}|a^{\prime }(x)|\\ & \le \frac{\pi \sqrt{1+c_0^2}}{c_0}{\int }_{I_k}|a^{\prime }(x)|.\end{array}$$其中注意 $\frac{{\Phi }^{\prime \prime }(x)}{1+t{\Phi }^{\prime }(x{)}^2}$ 的原函数为 $\frac{1}{\sqrt{t}}\arctan (\sqrt{t}{\Phi }^{\prime }(x))$. 对所有区间求和即得 $I_2(t)\le C_1\frac{1}{\sqrt{t}}\text{ensuremath}{\Vert a^{\prime }(x)\Vert }_{L^1}$.
A3) 由均值不等式 $I_1\le \frac{1}{2\sqrt{t}}\text{ensuremath}{\Vert a^{\prime }(x)\Vert }_{L^1}$. 结合 A2) 即得.
A4) 对 $x$ 分量做 Fourier 变换得$${\partial }_t\hat{u}(t,\xi )-i{\xi }^3\hat{u}(t,\xi )=0.$$因此 ${\hat{u}}_t(\xi )=e^{i{\xi }^3}{\hat{u}}_0\in L^1({\mathbb{R}}_{\xi })$. 注意 $C$ 仅依赖于 $c_0$, 甚至不依赖于 $t,\Phi ,a$.

设紧支光滑函数 $\chi$ 在 $[-2,-\frac{1}{2}]\cup [\frac{1}{2},2]$ 上恒为 $1$, 则$$\hat{u}(t,\xi )=\chi e^{it{\xi }^3}{\hat{u}}_0(\xi ),$$$$u_t(x)={\mathcal{F}}^{-1}(\chi \cdot e^{it{\xi }^3})\ast u_0.$$从而$$\text{ensuremath}{\Vert u(t)\Vert }_{L^{\infty }}\le \text{ensuremath}{\Vert {\mathcal{F}}^{-1}(\chi \cdot e^{it{\xi }^3})\Vert }_{L^{\infty }}\cdot \text{ensuremath}{\Vert u_0\Vert }_{L^1}.$$而$$\begin{array}{r}{\mathcal{F}}^{-1}(\chi \cdot e^{it{\xi }^3})={\int }_{\mathbb{R}}{e}^{it{\xi }^3+i\xi \cdot x}\chi (\xi )d\xi .\end{array}$$取 $\Phi (\xi )={\xi }^3+\frac{\xi x}{t}$, 使用 A3) 的结论即得所需估计.
A5) 同上的论证表明$$\text{ensuremath}{\Vert u(t)\Vert }_{L^{\infty }}\le \frac{C}{\sqrt{t}}\text{ensuremath}{\Vert u_0\Vert }_{L^1}.$$而 $\text{ensuremath}{\Vert {\hat{u}}_t\Vert }_{L^2}=\text{ensuremath}{\Vert {\hat{u}}_0\Vert }_{L^2}$, 故 $\text{ensuremath}{\Vert u\Vert }_{L^2}=\text{ensuremath}{\Vert u_0\Vert }_{L^2}$. 根据 $L^p$ 插值, $$\text{ensuremath}{\Vert u(t)\Vert }_{L^p}\le \frac{C^{\frac{p-2}{p}}}{t^{\frac{p-2}{2p}}}{\left|u_0\right|}_{L^{p^{\prime }}}.$$

2B

B1) 由对称性不妨设 $x\ge 0$,$$g_{\lambda }(x)=\frac{1}{2\pi }{\int }_{\mathbb{R}}\frac{1}{\lambda -i{\xi }^2}{e}^{ix\xi }d\xi .$$取上半圆弧做留数定理, 极点为 $\sqrt{\lambda }{e}^{i\frac{3\pi }{4}}$, 是单极点. 计算得$$g_{\lambda }(x)=\frac{1+i}{2\sqrt{2\lambda }}{e}^{-\sqrt{\frac{\lambda }{2}}(1+i)|x|}.$$B2) 存在性: 取 $u=g_{\lambda }\ast f$, 其中 $g_{\lambda }\in L^1$, 故 $u\in L^p$, 且 $u^{\prime \prime }=g_{\lambda }\ast f^{\prime \prime }\in L^p$. 做 Fourier 变换知 $(A+\lambda )u=f$.

唯一性: 设 $u\in D(A)$ 使得 $Au+\lambda u=0$. 根据 Riesz–Thorin 插值定理, 对 $1\le p\le 2$, Fourier 变换将 $L^p$ 送到 $L^{p^{\prime }}$, 从而是几乎处处定义的函数. 但$$(\lambda -i{\xi }^2)\hat{u}=0,$$因而 $\hat{u}=0$, 进而 $u=0$.
B3)$$\begin{array}{r}\text{ensuremath}{\Vert g_{\lambda }(x)\Vert }_{L^1}={\int }_{\mathbb{R}}\frac{1}{2\sqrt{\lambda }}{e}^{-\sqrt{\frac{\lambda }{2}}|x|}=\frac{\sqrt{2}}{\lambda }.\end{array}$$算子 $(A+\lambda {)}^{-1}$ 即为与 $g_{\lambda }$ 做卷积.
$L^1$:$$\text{ensuremath}{\Vert g_{\lambda }\ast f\Vert }_{L^1}\le \text{ensuremath}{\Vert g\Vert }_{L^1}\cdot \text{ensuremath}{\Vert f\Vert }_{L^1},$$取 ${\chi }_{ϵ}=\frac{1}{ϵ}\chi (x/ϵ)$, 则 $g_{\lambda }\ast {\chi }_{ϵ}\stackrel{L^1}{\to }g$, 故算子范数为 $\frac{\sqrt{2}}{\lambda }$.
$L^2$: 显然 $g_{\lambda }\in L^2$. 作 Fourier 变换, 算子范数相当于乘积算子 $\frac{1}{\lambda -i{\xi }^2}:L^2\to L^2$ 的算子范数, 为 $\frac{1}{\lambda }$.
B4) 对 $1\le p\le 2$, 设 $\theta \in [0,1]$ 使得$$\frac{1}{p}=\frac{1-\theta }{2}+\frac{\theta }{1},$$根据 Riesz–Thorin 定理,$$\text{ensuremath}{\Vert (A+\lambda {)}^{-1}\Vert }_{L^p}\le \text{ensuremath}{{\Vert (A+\lambda {)}^{-1}\Vert }_{L^2}}^{1-\theta }\text{ensuremath}{{\Vert (A+\lambda {)}^{-1}\Vert }_{L^1}}^{\theta }=\frac{2^{\theta /2}}{\lambda }.$$对 $p<2$, 有 $2^{\theta /2}>1$, 不能使用 Hille–Yosida.
B5) $$\text{ensuremath}{{\Vert f_a\Vert }_{L^p}}^p={\int }_{\mathbb{R}}{e}^{-pa{x}^2}dx=\sqrt{\frac{\pi }{ap}}.$$令 $u(t,x)=S(t)f_a$, 则对 $t\ge 0,x\in \mathbb{R}$,$${\partial }_{t}u+i{\partial }_x^{2}u=0.$$做 Fourier 变换有$${\partial }_t\hat{u}-i{\xi }^2\hat{u}=0,$$其中 $\hat{u}\in L^{p^{\prime }}$. 从而 $\hat{u}=e^{it{\xi }^2}\widehat{f_a}(\xi )$,$$\begin{array}{rl}u(t,x)& ={\mathcal{F}}^{-1}(e^{it{\xi }^2})\ast f_a(x)\\ & =\frac{1}{2\sqrt{\pi t}}{e}^{-\frac{i{x}^2}{4t}+\frac{i\pi }{4}}\ast e^{-a{x}^2}\\ & =\frac{1}{2\sqrt{\pi t}}{\int }_{\mathbb{R}}{e}^{-\frac{i{y}^2}{4t}+\frac{i\pi }{4}}\cdot e^{-a(x-y{)}^2}dy\\ & =\frac{1}{\sqrt{1-4ati}}{e}^{-a\frac{1+4ait}{1+16a^2{t}^2}{x}^2}.\end{array}$$因而$$\begin{array}{rl}\text{ensuremath}{{\Vert u\Vert }_{L^p}}^p& =(1+16a^2{t}^2{)}^{-\frac{p}{4}}{\int }_{\mathbb{R}}{e}^{-\frac{pa}{1+16a^2{t}^2}{x}^2}dx\\ & =\sqrt{\frac{\pi }{pa}}(1+16a^2{t}^2{)}^{\frac{1}{2}-\frac{p}{4}}.\end{array}$$B6) $$\begin{array}{r}\underset{a\to \infty }{\lim }\frac{\text{ensuremath}{\Vert S(t)f_a\Vert }_{L^p}}{\text{ensuremath}{\Vert f_a\Vert }_{L^p}}=\underset{a\to \infty }{\lim }(1+16a^2{t}^2{)}^{\frac{2-p}{4p}}\to \infty .\end{array}$$故不能生成半群.