交大课程MA430-偏微分方程续论之总结

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1.Prelimary

1.1 Variational method

For a partial differential equation(for instance, the Poisson equation):

$\left\{ \begin{array}{rl} -\Delta u(x) & = f(x) \quad\mbox{ if $x\in B\subseteq\mathbb{R}$} \\ u(x)|_{\partial B} & = 0 \end{array} \right.$
We may define the operator that values the "Energy" of the function

Definition 1.Energy Operator $J(u)$ is defined as

$J(u)=\int_{B}\left(\frac{1}{2}|\nabla u|^2+f(x)u(x)\right)dx$
We may be interested in the minimum of the operator,
$\bar{u}=\underset{u\in\mathcal{H}_{0}^{1}(B)}{\argmin}J(u)$

By this definition, we may have the following theorem:

Theorem 1.(Existence of Weak Solution) $\bar{u}$ exists, and is the weak solution of the Poisson equation.

Proof: : existence of certain $\bar{u}$

First, note that

$J(u)\ge \frac{1}{2}\int_{B}|\nabla u|^2-\epsilon\int_{B}u^2-\frac{1}{4\epsilon}\int_{B}f^2$
This is done by AM-GM. By Poincar Inequality, we know that
$J(u)\ge \left(\frac{1}{2}-\epsilon C\right)\|u\|^2-\frac{1}{4\epsilon}\int_{B}f^2$
which states that $J(u)$ has a finite infimum.
Second, we may show that if $J(u_n) \searrow m$ ,where $m$ is the infimum. Then $u_n$ has a subsequence that converge to $\bar{u}$ . That is, the uniform bound on $J(u_n)$ results in the uniform bound of $\|u\|^2$ . This is obtained by Cauchy-Schwarz Inequality and Poincaré inequality,

$J(u)\ge \frac{1}{2}\| u \|^2-\| f\|_{L^2}\| u \|_{L^2}\ge \frac{1}{2}\left(\| u \| -C \| f \|_{L^2}\right)^2-\frac{1}{2}C^2 \| f\|_{L^2}$
That is to say, $\| u_n \|\le C\|f\|_{L^2}$
By weak compactness, we know that there is a weak converging sequence $u_{n_k}\to \bar{u}$ , $\|\bar{u}\| \le \underline{\lim}\|u_n\|$
The convexity functional $J(u)$ guarantee that

$m=\lim_{n}J(u_n)=\lim_{n\to\infty}\frac{1}{2}\|u_n\|^2+\lim\int f u_n\ge \frac{1}{2}\|\bar{u}\|^2+\int f\bar{u}=J(\bar{u})$

: We derive that $\bar{u}$ is the weak solution .

Let $h(0)=J(\bar{u})\le J(\bar{u}+tv)=h(t)$ fot $t\in\mathbb{R}$ . Where $v\in \mathcal{H}_{0}^{1}(B)$ . This means $h‘(0)=0$ .
Since

$h(t)=\frac{1}{2}\int |\nabla \bar{u}|^2+t\int \nabla \bar{u}\nabla v+\frac{1}{2}t^2\int|\nabla v|^2+\int f\bar{u} +t\int fv$
The derivative at 0 is given by

$0=\int \nabla \bar{u}\nabla v+\int fv$
Also, by Green‘s formula

$\int_{B}\nabla \bar{u}\nabla v=\int_{\partial B}v\frac{\partial \bar{u}}{\partial n}-\int_B v\Delta \bar{u}=-\int v\Delta u$
Combining these two equations we get

$\int_B (-\Delta \bar{u}+f)vdx\equiv 0 \textrm{ holds for all }v\in\mathcal{H}_{0}^{1}(B)$
And done.

$\square$

This proof also gives a sufficient and necessary condition for $\bar{u}$ to be the weak solution of the Poisson equation, that is

$\int \nabla \bar{u}\nabla v+\int fv\equiv 0\quad \forall v\in\mathcal{H}_{0}^{1}(B)$

The same method also provides us the truth that $J‘(\bar{u})=0$ . But the derivative of an operator remains to be defined, which will be mentioned in the generalized function section below.

1.2 Tools From ODE

Consider the ordinary differential equation:

$\begin{equation*} \begin{cases} u‘(t)=f(u(t),t)\\ u(0)=u_0 \end{cases} \end{equation*}$
where

$f$ is continuous on

$(u_0-1,u_0+1)\times(-\infty,+\infty)$ , and Lipschitz in

$u$ . We want to prove the existence and uniqueness of the function. The easy proof is given by the recursion.

$T(\varphi)(t)=u_0+\int_{0}^{t}f(\varphi(s),s)ds$
It‘s easy to check that $T(\varphi)(0)=0$ , and $\exists t$ such that $|T(\varphi)(t)-u_0|\le \frac{1}{2}$ ( this can be done by Lipschitz condition).
We may prove the contraction of the operator.

$\begin{align*}\|T(\varphi_1)(t)-T(\varphi_2)(t)\|&\le\int_{0}^{t}\left|f(\varphi_1(s),s)-f(\varphi_2(s),s)\right|ds\\ &\le M\int_{0}^{t}\|\varphi_1-\varphi_2\|_{L^{\infty}}ds\le M\delta\|\varphi_1-\varphi_2\|_{L^{\infty}}\\ &\le \frac{1}{2}\|\varphi_1-\varphi_2\|_{L^{\infty}} \end{align*}$
This holds for $M\delta\le \frac{1}{2}$ , $\delta=\min\{\frac{1}{2M},1\}$ . This means $T$ is a contraction map(Since $M$ does not change, so $\delta$ does not change).
()In this section, we prove that $T$ has a unique fixed point.

$\forall \varphi^{(0)}\in X$ , we denote $\varphi^{(n)}=\overbrace{T\circ T\circ \cdots \circ T}^{n\textrm{ times}}(\varphi^{(0)})$ .

So for all $n\in\mathbb{N}$ ,

$\|\varphi^{(n)}(t)-\varphi^{(0)}(t)\|_{L^{\infty}}\le\sum_{k=1}^{n}\|\varphi^{(k)}(t)-\varphi^{(k-1)}(t)\|_{L^{\infty}}\le 2\|\varphi^{(1)}(t)-\varphi^{(0)}(t)\|\le 2P$
where $P$ is a number greater than left hand side.

So $\forall \epsilon>0,\exists N\in\mathbb{N},\frac{1}{2^{N-1}}\le\epsilon$ , $\forall m>n>N$ , we have

$\|\varphi^{(m)}(t)-\varphi^{(n)}(t)\|_{L^{\infty}}\le \frac{1}{2^{n}}\|\varphi^{(m-n)}(t)-\varphi^{(0)}(t)\|_{L^{\infty}}\le \epsilon$
By the completeness of $L^{\infty}$ we know that $\varphi^{(n)}$ is a Cauchy sequence and converge to a point $\varphi$ . And since $\lim_{n\to\infty}\|\varphi^{(n)}(t)-\varphi^{(n-1)}(t)\|_{L^{\infty}}=0$ we know that $T(\varphi)=\varphi$ .

To show the uniqueness, if $T(\psi)=\psi$ , we have

$0\le\|\psi-\phi\|_{L^{\infty}}=\|T(\varphi)-T(\psi)\|_{L^{\infty}}\le\frac{1}{2}\|\psi-\phi\|_{L^{\infty}}$
which means $\phi\equiv \psi$ . and the proof is done.

To show the tools from the ODE, we consider the Heat equation

$\left\{ \begin{array}{rl} u_t & = \Delta u \quad\mbox{ in $\mathbb{R}$} \\ u(x,0) & = u_0(x) \end{array} \right.$
By the analysis in PDE course, we derive that

$u(x,t)=e^{\Delta t}u_0=\int_{\mathbb{R}^n}H(x-y,t)u_0(y)dy$
where

$H(x,t)$ denotes the heat kernel, with representation

$H(x,t)=\frac{1}{(4\pi t)^{d/2}}e^{-|x-y|^2/4t}$
However, when handling this kind of problem

$\left\{ \begin{array}{rl} u_t & = \Delta u +f(u,x,t) \\ u(x,0) & = u_0(x) \end{array} \right.$
The addition of non-linear term makes it hard to analysis, so we use the similar method in ODE, that is omitting the space term and get

$u‘=Au+f(u,t)\Rightarrow u(t)=e^{At}u(0)+\int_{0}^{t}e^{A(t-\tau)}f(u_0(\tau),\tau)d\tau$
where

$A$ denotes an operator, and the first term can be analyzed by the heat equation before,

$\begin{align*}u(x,t)&=e^{\Delta t}u_0+\int_{0}^{t}e^{\Delta(t-\tau)}f(u(x,\tau),x,\tau)d\tau \\&=\int_{\mathbb{R}^n}H(x-y,t)u_0(y)dy+\int_{0}^{t}\int_{\mathbb{R}^n}H(x-y,t-\tau)f(u(y,\tau),y,\tau)dyd\tau \end{align*}$

2.Generalized function

2.1 The generalized function

The generalized functions, sometimes called the distributions, are functions generalized to make some functions that with physical means(such as locally integrable) differentiable. The derivatives are generalized so as to meet the use of much physical applications.

The generalization should be done with some principles in hand, such as the Fundamental Theorem of Calculus should also holds in the generalized function space, the functions that is originally differentiable will also be differentiable, and the derivatives should be the same.

To generalize, we use interaction with functions. Instead of observing the function itself, we use some "test functions" to interact with the original one to see the property. The fundamental interaction is the inner product in $L^2$ space, that is

$\langle f,g\rangle=\int_{\mathbb{R}^n}f(x)g(x)dx$
where

$f$ is locally integrable and

$g$ is infinitely differentiable on a compact support set

$C_{0}^{\infty}(\mathbb{R}^n)=D$

We define the generalized function as follows: $f:C_{0}^{\infty}(\mathbb{R})\to \mathbb{R}(\mathbb{C})$ that satisfies

Linearity, $f(a g_1+bg_2)=af(g_1)+bf(g_2)$ .
Continuity, $f(g_n)\to f(g)$ if $g_n\to g$ , here $g_n\to g$ is defined as (a) $D^{\alpha}g_n\to D^{\alpha}g$ uniformly and (b) $\exists K\subset\subset \mathbb{R}^n$ such that $g_n(x)=0$ if $x\not\in K$

We denote $D‘$ as the space of generalized function.

()Show that locally integrable functions are generalized functions. That is $f\in L_{loc}^{1}(\mathbb{R}^n)\Rightarrow f\in D‘$

Proof:

We have $f\in L_{loc}^{1}(\mathbb{R}^n)$ and $\varphi\in C_{0}^{\infty}(\mathbb{R}^n)$ , we define operator

$T_{f}(\varphi):=\int_{\mathbb{R}^n}f(x)\varphi(x)dx$
This is clearly linear.

The only thing we needs to show is that $\varphi_n\to\varphi$ implies $T_{f}(\varphi_n)\to T_{f}(\varphi)$

$\begin{align*} |T_f(\varphi)|&=\left|\int_{\mathbb{R}^n}f(x)\varphi(x)dx\right|\le\int_{\mathbb{R}^n}|f(x)||\varphi(x)|dx\\ &=\int_{K}|f(x)||\varphi(x)|dx \le\|\varphi\|_{L^{\infty}}\int_{K}|f(x)|dx \end{align*}$
where

$K$ is the support of

$\varphi$ . This holds for all

$\varphi$ , so we substitute

$\varphi_n-\varphi$ into it and gets

$|T_f(\varphi_n)-T_{f}(\varphi)|\le\|\varphi_n-\varphi\|_{L^{\infty}}\int_{K}|f(x)|dx$
By definition we know that it converges to zero, that is

$T_f \in D‘$

$\square$

() Define $D^{\alpha}$ in $D‘$ so that it is consistent with the definition in classical function spaces.

Proof:By

$D^{\alpha}f(\varphi)=\int_{\mathbb{R}^n}D^{\alpha}f(x)\varphi(x)dx=(-1)^{|\alpha|}\int_{\mathbb{R}^n}f(x)D^{\alpha}\varphi(x)dx$
this is consistent with all

$\varphi\in C_{0}^{\infty}(\mathbb{R}^n)$ by integration by parts. Thus we define

$D^{\alpha}f(\varphi):=(-1)^{|\alpha|}\int_{\mathbb{R}^n}f(x)D^{\alpha}\varphi(x)dx$

()Prove that $D^{\alpha} f\in D‘$ if $f\in D‘$

Proof:Clearly,

$D^{\alpha} f(a\varphi_1+b\varphi_2)=(-1)^{|\alpha|}\int_{\mathbb{R}^n}f(x)D^{\alpha}(a\varphi_1+b\varphi_2)dx=aD^{\alpha}f(\varphi_1)+bD^{\alpha}f(\varphi_2)$
the linearity is obvious.

Also by

$|D^{\alpha}f(\varphi-\varphi_n)|=\left|\int_{\mathbb{R}^n}f(x)D^{\alpha}(\varphi-\varphi_n)\right|$
and

$D^{\alpha}\varphi\to D^{\alpha}\varphi_n$ uniformly, that is

$\sup_{x\in K}|\varphi-\varphi_n|\to 0$

This means

$D^{\alpha}f(\varphi-\varphi_n)|\le\sup_{x\in K}|\varphi-\varphi_n|\left|\int_{K}fdx\right|\to0$
this shows that

$D^{\alpha}f\in D‘\square$

Here we will show two examples of special generalized function.

()Let $h(x)$ be that

$h(x)=\left\{ \begin{array}{rl} 0 \quad (x\le 0)\\ 1 \quad (x>0) \end{array} \right.$ Show that

$Dh(x)=\delta$ .

Proof: $\forall\varphi\in C_{0}^{\infty}(\mathbb{R}^n)$ , we have

$Dh(\varphi(x))=-\int_{\mathbb{R}}h(x)\varphi(x)dx=-\int_{0}^{+\infty}\varphi‘(x)=\varphi(0)$
This means

$h‘(x)\equiv \delta(x)$ and done.

$\square$

(Homework 5(a))Let

$\Gamma(x)=\frac{1}{(n-2)|S^{n-1}||x|^{n-2}}.\quad(n\ge 3,n\in \mathbb{Z})$

$|S|^{n-1}$ is the surface area of unit sphere with dimension $n$ . Show that

$-\Delta \Gamma(x)=\delta_0$

Proof:Let $B_R$ be the sphere with radius $R$ centered at the origin, we get

$\int_{B_R}\Delta \Gamma\varphi dx=\int_{B_R\backslash B_{\varepsilon}}\Delta \Gamma\varphi dx+\int_{B_{\varepsilon}} \Delta\Gamma\varphi dx$
When

$x\not = 0$ , we have

$\Delta\Gamma(x)=\sum_{i=1}^{n}\frac{\partial^2}{\partial x_i^2}\Gamma(x)=-C\cdot\sum_{i=1}^{n}\frac{(n-2)(\sum_{k\not = i}x_k^2-(n-1)x_i^2)}{(\sum_{i=1}^{n}x_{i}^2)^{n-2}}\equiv0$
This means the integration does not depends on the radius:

$\int_{B_{R}}\Delta\Gamma\varphi dx=\int_{B_{\varepsilon}}\Delta\Gamma\varphi dx$
By Green‘s formula

$\begin{align*}\int_{B_{\varepsilon}}\Delta\Gamma\varphi dx &=-\int_{B_{\varepsilon}}\nabla\Gamma\cdot\nabla\varphi dx+\int_{\partial B_{\varepsilon}}\nabla\Gamma\cdot \vec{n}\varphi d\sigma \\ &=\int_{B_{\varepsilon}}\Gamma\Delta\varphi dx+\int_{\partial B_{\varepsilon}}(\nabla\Gamma\cdot \vec{n}\varphi-\Gamma\nabla\varphi\cdot \vec{n})d\sigma \end{align*}$
Since

$\Delta\varphi$ and

$\nabla\varphi\cdot \vec{n}$ are both bounded in

$B_{\varepsilon}$ . Also, by AM-GM

$\int_{B_{\varepsilon}}\frac{1}{|r|^{n-2}}dx=\int_{B_{\varepsilon}}\frac{1}{\left|\sqrt{\sum x_i^2}\right|^{n-2}}dx\le\int_{B_{\varepsilon}}\frac{1}{|\prod x_i|^{\frac{n-2}{n}}}dx\le \prod\left(\int_{-\epsilon}^{\epsilon}\frac{dx_i}{|x_i|^{\frac{n-2}{n}}}\right)\to 0 \mbox{ as $\varepsilon\to 0$}$
We obtain

$\int_{B_{\varepsilon}}\Gamma dx$ converges to zero while

$\varepsilon\to0$ .

So the only term that does not vanish is $\int_{\partial B_{\varepsilon}}\nabla\Gamma\cdot \vec{n}\varphi d\sigma$ .

Obviously

$\nabla\Gamma\cdot \vec{n}=\frac{\partial}{\partial r}\Gamma=-\frac{1}{|S^{n-1}||x|^{n-1}}.$
Which gives the following equality:

$\lim_{\varepsilon\to 0}\int_{\partial B_\epsilon}-\frac{\varphi(x)}{|S^{n-1}||x|^{n-1}}dx=\lim_{\varepsilon\to 0}\int_{\partial B_\epsilon}-\frac{\varphi(x)}{|S^{n-1}||\varepsilon|^{n-1}}dx$
By continuity of

$\varphi(x)$ it is clear that this limit equals

$-\varphi(0)$ . That is

$\int_{\mathbb{R}^{n}}-\Delta\Gamma(x)\varphi(x)dx=\varphi(0)$

2.2 Spaces of functions

In this section we define some spaces that will aid our research of generalized functions.

Definition 2.A normed linear subspace is a linear space equipped with a norm, where the norm on a linear space $X$ is defined as s map $X\to \mathbb{R}$ such that

$\|x\|\ge 0,\|x\|=0$ iff $x=0$ .

$\|\lambda x\|=|\lambda|\|x\|$ .

$\|x+y\|\le\|x\|+\|y\|$ .

A Banach space is a complete NLS.

The completion is given by "every Cauchy sequence converges" in the metric space.

Example 1. Let

$L^p(\Omega)=\left\{f:\Omega\to\mathbb{R}\mid f\textrm{ is measurable, that is } \int_{\Omega}|f(x)|^pdx<+\infty\right\}\quad 1\le p\le \infty$
with the norm $\|f\|_{L^p}:=(\int_{\Omega}|f(x)|^pdx)^{1/p}$

When $p=\infty$ , the norm is defined as
$\|f\|_{L^\infty}=\esssup_{x\in\Omega}|f(x)|$

And now we define the H?lder semi-norm:

Definition 3.We define $f\in C_{b}^{0,\alpha}(\Omega),(0<\alpha\le 1)$ if

$[f]_{C^{\alpha}(\Omega)}=\sup_{x\not=y,x,y\in\Omega}\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}<+\infty$

()Prove that $\alpha>1$ , $[f]_{C^{\alpha}(\Omega)}<\infty$ implies $f\equiv C$ .

Proof:Since $\forall x,y\in\Omega$ ,

$\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}<c$
And this constant is fixed and finite.So we take the definition of derivative:
for

$x$ and

$x+\epsilon$ , where

$\epsilon\to0$ , we obtain:

$|f‘(x)|=\lim_{\epsilon\to 0}\frac{|f(x)-f(x+\epsilon)|}{|\epsilon|}<\lim_{\epsilon\to 0}c\epsilon^{\alpha-1}=0$
This means for any point

$x$ , the derivative at

$x$ exists and equals zero.This directly implies

$f\equiv C$ for some

$C$ .

$\square$

So we may derive the norm in H?lder space

$\|f\|_{C^{k,\alpha}}=\sum_{|\beta|\le k}\|D^{\beta}f\|_{C^0(\bar{\Omega})}+\sum_{|\beta|= k}[D^{\beta}f]_{C^\alpha(\bar{\Omega})}$

2.3 Weak Derivative

In this section, we will define the weak derivative that is extremely important in Sobolev space.

The weak derivative is somehow similar to the generalized derivative that we have defined before, but a slight difference is made.

Definition 4.For $f\in L_{loc}^{1}(\Omega)$ ,if there exists $v\in L^{1}_{loc}(\Omega)$ such that for all $\varphi\in C_{0}^{\infty}(\Omega)$ we have

$(-1)^{|\alpha|}\int_{\Omega}f(x)D^{\alpha}\varphi(x)dx=\int_{\Omega}v(x)\varphi(x)dx$
we say $v(x)$ is the $\alpha-$ th weak derivative of $f$

Here we gave some examples to show the properties of weak derivative.

$\exists f\in L^{1}(-1,1)$ that $f$ is not weak differentiable
()Let $h(x)$ be

$h(x)=\left\{ \begin{array}{rl} 0 \quad (x\le 0)\\ 1 \quad (x>0) \end{array} \right.$
prove that it is not weak differentiable.

Proof:If weak derivative exists, denoted by $u$ , $\forall \varphi\in C_{0}^{\infty}(-1,1)$ we have
$\int_{u}\varphi dx=\varphi(0).$
Since $u$ is locally integrable, $\forall \varepsilon>0$ , $\exists \delta>0$ such that $M(A)<\delta$ , $\int_{A}|u|dx<\frac{1}{2}$ . Where $M(\cdot)$ denotes the measure of a set.

So we take

$\varphi_{\delta}(x)=\left\{ \begin{array}{rl} 0 \quad (|x|\ge \frac{\delta}{2})\\ e^{-\frac{|x|}{\delta^2-4|x|^2}} \quad (|x|<\frac{\delta}{2}) \end{array} \right.$
Obviously, $\varphi_{\delta}\in C_{0}^{\infty}(-1,1)$ , $|\varphi_{\delta}(x)|\le1$ .
That is
$1=\varphi(0)=\int_{-1}^{1}u\varphi_{\delta}dx\le\int_{-\delta}^{\delta}|u||\varphi_{\delta}|dx\le\int_{-\delta}^{\delta}|u|dx\le\frac{1}{2}$
Which makes a contradiction. $\square$
The two weak derivative of $f$ are equal almost everywhere. That is $D^{\alpha}f=v$ , $D^{\alpha}f=u$ , $u,v\in L_{loc}^{1}(\Omega)\Rightarrow u=v$ a.e.
() $D^{\alpha} f=u, D^{\alpha} f=v$ , show that $u=v$ a.e.
Proof:Since $\int_{\Omega}(u(x)-v(x))f(x)dx\equiv 0$ for all $f\in C_{0}^{\infty}(\Omega)$ , by the next homework we know that it is true.
()
Weak derivative satisfies the testing principle, that is $f\in L_{loc}^{1}(\Omega)$ , if
$\int_{\Omega}f(x)\varphi(x)dx=0$ holds $\forall \varphi\in C_{0}^{\infty}(\Omega)$ , then $f=0$ a.e.
Proof:Easily, there exists sequence $\phi_n \in C^\infty_c(\Omega)$ such that $\phi_n \rightarrow \chi_{B(x,r)}$ locally uniformly.

For sufficiently large $n$ if all $\phi_n$ are supported inside $K$ then you have the estimate for any $\epsilon >0$

$\left|\int_\Omega (f\phi_n - f\chi_{B(x,r)})dx \right| \leq \int_\Omega |f||\phi_n - \chi_{B(x,r)}|dx \leq \epsilon\|f\|_{L^1(K\cap B(x,r))}$

But as $\int_\Omega f\phi_n dx = 0$ hence $\int_\Omega f\chi_{B(x,r)} dx = 0$ .

Considering lebesgue differentiation theorem,

$f=\lim_{\delta\to 0}\frac{1}{|B(x,\delta)|}\int_\Omega ( f\chi_{B(x,r)})dx=0 \mbox{ a.e.}$ .

3.Sobolev space and its properties

3.1 Basic ideas

When the definition of weak derivatives is finished, we may come to our final goal: the study of Sobolev space.

Definition 5.The Sobolev space is defined as follows:

$W^{k,p}(\Omega)=\left\{f\in L^{p}(\Omega)|\forall \alpha,|\alpha|\le k, D^{\alpha}f \textrm{ exists as weak derivative and }\sum_{|\alpha|\le k}\int_{\Omega}|D^{\alpha}f|^{p}dx<\infty \right\}$
Naturally, the norm of this space is defined as
$\|f\|_{W^{k,p}(\Omega)}=\left(\sum_{|\alpha|\le k}\int_{\Omega}|D^{\alpha}f|^{p}dx\right)^{\frac{1}{p}}\cong \sum_{|\alpha|\le k}\left(\int_{\Omega}|D^{\alpha}f|^{p}dx\right)^{\frac{1}{p}}$

Specially, $p=2$ , $W^{k,p}(\Omega)\equiv H^{k}(\Omega)$ .

To show the goodness of this space ,we may first show its completeness under this norm. That is , it is a Banach space.

Theorem 2.The Sobolev space $W^{k,p}(\Omega)$ is a Banach space.

Proof:The outline of the proof is shown as follows:

Sobolev norm is really a norm.
If $\{f_m\}_{m=1}^{\infty}$ is a Cauchy sequence in $W^{k,p}$ , then $\{D^{\alpha}f_{m}\}_{m=1}^{\infty}$ is a Cauchy sequence in $L^p$ .
$D^{\alpha}f_{m}$ converges to $f_{\alpha}$ in $L^p$ , thus $D^{\alpha}f=f_{\alpha}$ , and $f_m\to f$ .

So we may begin our proof.

For 1, Linearity and positivity is clear. To show the triangle inequality, we may use the Minkowski‘s inequality and done.

For 2,If a Cauchy sequence in $W^{k,p}$ is constructed, name $\{f_m\}_{m=1}^{\infty}$ . $\forall$ index $\alpha$ , $|\alpha|\le k,D^{\alpha}f_n\in L^p$ and

$\|D^{\alpha}f_n-D^{\alpha}f_m\|_{L^p}\le\|f_n-f_m\|_{W^{k,p}}\to 0$
By completeness of

$L^p$ , we have

$f_m\to f$ in

$L^p$ as well as

$D^{\alpha}f_m\to f_{\alpha}$ in

$L^p$ .

For 3,We may first check that $D^{\alpha}f=f_{\alpha}$ , thus $f\in W^{k,p}$ .

Observing that

$D^{\alpha}f(\varphi)=(-1)^{|\alpha|}\int_{\Omega}f(x)D^{\alpha}\varphi(x) dx=\lim_{n\to\infty }(D^{\alpha}f_n)(\varphi(x))=f_{\alpha}(\varphi(x))$
We know that

$D^{\alpha}f=f_\alpha\in L^p$ . Thus

$f\in W^{k,p}$ , and

$D^{\alpha}f=f_{\alpha}$ for all

$|\alpha|\le k$ .

Since $D^{\alpha}f=f_{\alpha}$ , it‘s obvious that $\|f_n-f\|_{W^{k,p}}\to 0$ by the addition of $|D^{\alpha}(f_n-f)|$ all converges to zero. $\square$

(未完待续..)

交大课程MA430-偏微分方程续论之总结

原文：http://www.cnblogs.com/misaka01034/p/MA430.html

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