用户: Yao/古代偏微分方程

1Laplace equation

$$\{\begin{array}{ll}-\Delta u=f,& x\in \Omega ,\\ u=\phi ,& x\in \partial \Omega .\end{array}$$

Proposition 1.1. $u$ is nonnegative and harmonic in $B_R(x_0)$, then $|Du(x_0)|\le \frac{n}{R}u(x_0)$.

Proof. $|Du(x_0)|=\frac{\partial u}{\partial n}(x_0)={-\int }_{B_R(x_0)}\frac{\partial u}{\partial n}=\frac{n}{R}{-\int }_{\partial B_R(x_0)}u\cdot n=\frac{n}{R}u(x_0)$.

$\square$

Corollary 1.2. A harmonic function in ${\mathbb{R}}^n$ bounded from above or below is constant.

Similarly we have the gradient estimates for general harmonic functions, and the corollary follows from induction.

Proposition 1.3. $u$ is harmonic in $B_R(x_0)$, then $|Du(x_0)|\le \frac{n}{R}\underset{{\bar{B}}_R(x_0)}{\max }|u|$.

Corollary 1.4. $|D^{\alpha }u(x_0)|\le \frac{c_{|\alpha |,n}}{R^{|\alpha |}}\underset{{\bar{B}}_R(x_0)}{\max }|u|$.

Definition 1.5. $$\Gamma (x)=\{\begin{array}{ll}\frac{1}{2\pi }\ln \frac{1}{|x|},& n=2\\ \frac{1}{(n-2){\omega }_{n-1}}|x{|}^{2-n},& n\ge 3.\end{array}.$$

The constants are taken as such to make ${\int }_{\partial B_R(y)}\frac{\partial \Gamma }{\partial n}(x-y)dx=-1$.

Definition 1.6. For fixed $x\in \partial \Omega$, $G(x,y)$ as a function of $y$ with singularity at $x$ solves $$\{\begin{array}{ll}{\Delta }_y(G(x,y)-\Gamma (x-y))=0,& y\in \Omega ,\\ G(x,y)=0,& y\in \partial \Omega .\end{array}$$

Proposition 1.7. $$u(x)=-{\int }_{\Omega }G(x,y)\Delta u(y)dy-{\int }_{\partial \Omega }u(y)\frac{\partial G}{\partial n}(x,y)dy.$$

Example 1.8. $G(x,y)=\Gamma (x-y)-\Gamma (R\frac{x}{|x|}-\frac{|x|}{R}y)$ for $x,y\in B_R(a)$,

$G(x,y)=\Gamma (x-y)-\Gamma (\tilde{x}-y)$ for $x,y\in {\mathbb{R}}_{+}^n=\{x_n>0\}$ where $\tilde{x}=(x_1,\cdots ,x_{n-1},-x_n)$.

Proposition 1.9. (1) $u$ is harmonic in $B_R(0)$, then $u(x)={\int }_{\partial B_R(0)}\frac{R^2-|x{|}^2}{{\omega }_{n}R|x-y{|}^n}u(y)dy$.

(2) $u$ is harmonic in ${\mathbb{R}}_{+}^n$, then $u(x)={\int }_{\partial {\mathbb{R}}_{+}^n}\frac{2x_n}{n{\omega }_n|x-y{|}^n}u(y)dy$.

Lemma 1.10. (Hopf) $B$ is an open ball and $x_0\in \partial B$, $Lu\ge 0$ with $c\le 0$ in $B$. If $u(x_0)>u(x),\forall x\in B$ and $u(x_0)\ge 0$, then for any outward direction $v$ at $x_0$ ($v\cdot n(x_0)>0$) we have $$\underset{t\to 0^{+}}{\mathrm{liminf}}\frac{u(x_0)-u(x_0-tv)}{t}>0.$$

2Heat equation

$$\{\begin{array}{ll}{\partial }_{t}u-\Delta u=0,& (t,x)\in (0,+\infty )\times {\mathbb{R}}^n\\ u(0,x)=\phi (x),& x\in {\mathbb{R}}^n\end{array}$$

Proposition 2.1. A solution is $$u(t,x)=\frac{1}{(4\pi t{)}^{\frac{n}{2}}}{\int }_{{\mathbb{R}}^n}\phi (y)e^{-\frac{|x-y{|}^2}{4t}}dy,$$and is unique provided that $|u(t,x)|\le C{e}^{A|x{|}^2}$. Plus, all solutions are smooth.

Proposition 2.2. Write ${\Omega }_T=(0,T]\times \Omega$ and ${\Gamma }_T=\{0\}\times U\cup [0,T]\times \partial U$. $u$ attains its maximum and minimum on $\overline{{\Omega }_T}$ only at ${\Gamma }_T$ unless $u$ is constant.

Proposition 2.3. Write $E(x,t;r)=\{(y,s)\in {\mathbb{R}}^{n+1}\mid s\le t,\frac{1}{(4\pi (t-s){)}^{\frac{n}{2}}}{e}^{-\frac{|x-y{|}^2}{4(t-s)}}\ge \frac{1}{r^n}\}$, then $$u(t,x)=\frac{1}{4r^n}{\int }_{E(x,t;r)}u(y,s)\frac{|x-y{|}^2}{(t-s{)}^2}dyds.$$

Consider the boundary-value problem$$\{\begin{array}{ll}{\partial }_{t}u-\Delta u=f(t,x),& (t,x)\in (0,T)\times \Omega ,\\ u(0,x)=\phi (x),{\partial }_{t}u(0,x)=\psi (x),& x\in \Omega .\end{array}$$Let$$E(t)=\frac{1}{2}{\int }_{\Omega }{u}^2(t,x).$$We have$$E^{\prime }(t)={\int }_{\Omega }u{\partial }_{t}u={\int }_{\Omega }u\Delta u=-{\int }_{\Omega }|\nabla u{|}^2\le 0.$$

3Wave equation

$$\{\begin{array}{ll}{\partial }_t^{2}u-\Delta u=0,& (t,x)\in (0,+\infty )\times {\mathbb{R}}^n,\\ u(0,x)=\phi (x),{\partial }_{t}u(0,x)=\psi (x),& x\in {\mathbb{R}}^n.\end{array}$$

Proposition 3.1. The solution is $$u(t,x)=\frac{1}{2}(\phi (x+t)+\phi (x-t))+\frac{1}{2}{\int }_{x-t}^{x+t}\psi (y)dy$$when $n=1$.

Proposition 3.2. The solution is $$\begin{array}{rl}u(t,x)& ={\partial }_t(t{-\int }_{\partial B_t(x)}\phi )+t{-\int }_{\partial B(t,x)}\psi \\ & ={-\int }_{\partial B_t(x)}\phi (y)+D\phi (y)\cdot (y-x)+t\psi (y)dy.\end{array}$$when $n=3$.

Proposition 3.3. The solution is $$\begin{array}{rl}u(t,x)& ={\partial }_t(t{-\int }_{\partial B_t(x)}\frac{\phi }{2\sqrt{1-|\frac{y-x}{t}{|}^2}}dy)+t{-\int }_{\partial B(t,x)}\frac{\psi }{2\sqrt{1-|\frac{y-x}{t}{|}^2}}dy\\ & ={-\int }_{\partial B_t(x)}\frac{\phi (y)+D\phi (y)\cdot (y-x)+t\psi (y)}{2\sqrt{1-|\frac{y-x}{t}{|}^2}}dy.\end{array}$$when $n=2$.

Consider the initial&boundary-value problem$$\{\begin{array}{ll}{\partial }_t^{2}u-\Delta u=f(t,x),& (t,x)\in (0,T)\times \Omega ,\\ u(0,x)=\phi (x),{\partial }_{t}u(0,x)=\psi (x),& x\in \Omega ,\\ u(t,x)=0,& (t,x)\in (0,T)\times \partial \Omega .\end{array}$$Let$$E(t)=\frac{1}{2}{\int }_{\Omega }({\partial }_{t}u{)}^2+|\nabla u{|}^2.$$We have$$\begin{array}{rl}{E}^{\prime }(t)& ={\int }_{\Omega }{\partial }_{t}u(\Delta u+f)+\nabla u\cdot {\partial }_t\nabla u\\ & ={\int }_{\Omega }\mathrm{div}({\partial }_{t}u\nabla u)+{\partial }_{t}uf\\ & ={\int }_{\Omega }{\partial }_{t}uf\text{by divergence theorem and }{\partial }_{t}u{|}_{(0,T)\times \partial \Omega }=0\\ & \le E(t)+\frac{1}{2}{\int }_{\Omega }{f}^2,\end{array}$$so by Gronwall$$E(t)\le C(\Vert \psi {\Vert }_{L^2}^2+\Vert \nabla \phi {\Vert }_{L^2}^2+\Vert f{\Vert }_{L^2((0,T)\times \Omega )}^2).$$

Consider the Cauchy problem when $n=3$.

For fixed $(t_0,x_0)$, let$$\begin{array}{rl}E(t)& =\frac{1}{2}{\int }_{B_{t_0-t}(x_0)}({\partial }_{t}u{)}^2+|\nabla u{|}^2,\\ F(t)& =\frac{1}{2}{\int }_0^t{\int }_{\partial B_{t_0-s}}|n{\partial }_{t}u-\nabla u{|}^2.\end{array}$$We have$$\begin{array}{rl}{E}^{\prime }(t)& =\frac{1}{2}(-{\int }_{\partial B_{t_0-t}(x_0)}({\partial }_{t}u{)}^2+|\nabla u{|}^2+{\int }_{B_{t_0-t}(x_0)}{\partial }_t(({\partial }_{t}u{)}^2+|\nabla u{|}^2\left)\right)\\ & =\frac{1}{2}(-{\int }_{\partial B_{t_0-t}(x_0)}({\partial }_{t}u{)}^2+|\nabla u{|}^2+2{\int }_{B_{t_0-t}(x_0)}\mathrm{div}({\partial }_{t}u\nabla u))\text{as }{\partial }_t^{2}u=\Delta u\\ & =\frac{1}{2}(-{\int }_{\partial B_{t_0-t}(x_0)}({\partial }_{t}u{)}^2+|\nabla u{|}^2-2{\partial }_{t}u\frac{\partial u}{\partial n})\\ & =-F^{\prime }(t),\end{array}$$so$$E(t)\le E(t)+F(t)=E(0).$$