\newcommand\argmin{\operatorname{arg\,min}}
\newcommand\esssup{\operatorname{ess\,sup}}
\newcommand\supp{\operatorname{supp}}
\def\Xint#1{\mathchoice {\XXint\displaystyle\textstyle{#1}}% {\XXint\textstyle\scriptstyle{#1}}% {\XXint\scriptstyle\scriptscriptstyle{#1}}% {\XXint\scriptscriptstyle\scriptscriptstyle{#1}}% \!\int} \def\XXint#1#2#3{{\setbox0=\hbox{$#1{#2#3}{\int}$ } \vcenter{\hbox{$#2#3$ }}\kern-.6\wd0}} \def\ddashint{\Xint=} \def\dashint{\Xint-}

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.

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: First part: existence of certain \bar{u}

1. 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 Poincaré 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.

2. 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}
3. 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\|

4. 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})

Second part: 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

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*}

1.  First we define T:X\to X be the operator:

   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).

2. 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).

3. (Homework 1)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.

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

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

1. Linearity, f(a g_1+bg_2)=af(g_1)+bf(g_2) .
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.

(Homework 2)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

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*}

(Homework 3) 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

(Homework 4)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)

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

This means D^{\alpha}f(\varphi-\varphi_n)|\le\sup_{x\in K}|\varphi-\varphi_n|\left|\int_{K}fdx\right|\to0

Here we will show two examples of special generalized function.

(Homework 5(a))Let h(x) be that h(x)=\left\{ \begin{array}{rl} 0 \quad (x\le 0)\\ 1 \quad (x>0) \end{array} \right.

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)

(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

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}}.

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

1. \|x\|\ge 0,\|x\|=0 iff x=0 .
2. \|\lambda x\|=|\lambda|\|x\| .
3. \|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

(Homework 6)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

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.

1. \exists f\in L^{1}(-1,1) that f is not weak differentiable
   (Homework 7)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
2. 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.

3. (Homework 8)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.

4. (Homework 9)
   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:

1. Sobolev norm is really a norm.
2. 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 .
3. 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

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))

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