From 1befcae5bfc7056d7441b871466359d023575941 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Sat, 18 Nov 2023 20:40:20 +0100 Subject: [PATCH] updated part of pdes --- .../Real_and_functional_analysis.tex | 48 ++-- ...topics_in_functional_analysis_and_PDEs.tex | 12 +- ...ntroduction_to_nonlinear_elliptic_PDEs.tex | 234 +++++++++++++++--- .../Numerical_methods_for_PDEs.tex | 2 +- 4 files changed, 226 insertions(+), 70 deletions(-) diff --git a/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex b/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex index c42e3ee..a9b3d8c 100644 --- a/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex +++ b/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex @@ -28,7 +28,7 @@ Let $\Sigma$ be a $\sigma$-algebra over a set $\Omega$. A \emph{measure} over $\Omega$ is any function $$\mu:\Sigma\longrightarrow[0,\infty]$$ satisfying the following properties: \begin{enumerate}[ref = $\sigma$-additivity] \item $\mu(\varnothing)=0$. - \item\label{RFA:sigmaadditivity} \emph{$\sigma$-additivity}: If $(A_n)\in\Sigma$ are pairwise disjoint, then: $$\mu\left(\bigsqcup_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty \mu(A_n)$$ + \item\label{RFA:sigmaadditivity} \emph{$\sigma$-additivity}: If $(A_n)\in\Sigma$ are pairwise disjoint, then: $$\mu\left(\bigsqcup_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty \mu(A_n)$$ \end{enumerate} \end{definition} \begin{definition} @@ -39,8 +39,8 @@ \begin{enumerate} \item If $A\subseteq B$, then $\mu(B\setminus A)=\mu(B)-\mu(A)$. \item If $A\subseteq B$, then $\mu(A)\leq\mu(B)$. - \item\label{RFA:incresingseq} If $A_n\nearrow A$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$. - \item\label{RFA:decresingseq} If $A_n\searrow A$ and $\mu(A_1)<\infty$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$. + \item\label{RFA:incresingseq} If $A_n\nearrow A$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$. + \item\label{RFA:decresingseq} If $A_n\searrow A$ and $\mu(A_1)<\infty$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$. \end{enumerate} \end{proposition} \begin{sproof} @@ -113,12 +113,12 @@ The outer measure has the following properties: \begin{enumerate} \item $\om{\varnothing}=0$. - \item\label{RFA:measureB} If $A\subseteq B\subseteq\RR^n$, then $\om{A}\leq \om{B}$. + \item\label{RFA:measureB} If $A\subseteq B\subseteq\RR^n$, then $\om{A}\leq \om{B}$. \item \label{RFA:measureC} If $(A_k)\subseteq \RR^n$, then: $$\om{\bigcup_{k=1}^\infty A_k}\leq \sum_{k=1}^\infty \om{A_k}$$ \item \label{RFA:measureD}If $I\subseteq \RR^n$ is an open interval and $I\subseteq A\subseteq \cl{I}$, then $\om{A}=\vol(I)$. - \item\label{RFA:measureE} If $I_1,\ldots,I_N\subseteq \RR^n$ are disjoint intervals, then: $$\displaystyle \om{\bigsqcup_{k=1}^N I_k}= \sum_{k=1}^N \vol(I_k)$$ + \item\label{RFA:measureE} If $I_1,\ldots,I_N\subseteq \RR^n$ are disjoint intervals, then: $$\displaystyle \om{\bigsqcup_{k=1}^N I_k}= \sum_{k=1}^N \vol(I_k)$$ \item If $A,B\subseteq \RR^n$ and $d(A,B):=\inf\{d(a,b):a\in A,b\in B\}>0$, then $\om{A\sqcup B}=\om{A}+\om{B}$. - \item\label{RFA:measureG} If $A\subseteq\RR^n$ and $x\in\RR^n$, then $\om{A+x}=\om{-A}=\om{A}$\footnote{Here $A+x:=\{a+x:a\in A\}$ and $-A:=\{-a:a\in A\}$}. + \item\label{RFA:measureG} If $A\subseteq\RR^n$ and $x\in\RR^n$, then $\om{A+x}=\om{-A}=\om{A}$\footnote{Here $A+x:=\{a+x:a\in A\}$ and $-A:=\{-a:a\in A\}$}. \end{enumerate} \end{theorem} \begin{sproof} @@ -666,7 +666,7 @@ \begin{enumerate}[ref = Triangular inequality] \item $\|\vf{u}\|=0\iff \vf{u}=0$ \item $\|\lambda \vf{u}\|=|\lambda|\|\vf{u}\|$ - \item\label{RFA:triangularineq} $\|\vf{u}+\vf{v}\|\leq \|\vf{u}\|+\|\vf{v}\|\quad$(\emph{triangular inequality}) + \item\label{RFA:triangularineq} $\|\vf{u}+\vf{v}\|\leq \|\vf{u}\|+\|\vf{v}\|\quad$(\emph{triangular inequality}) \end{enumerate} We define a \emph{normed vector space} as a pair $(E,\|\cdot\|)$ that satisfy the previous properties. \end{definition} @@ -802,9 +802,9 @@ \begin{definition} Let $E\subseteq\RR^n$ be a measurable set and $1\leq p<\infty$. We define: \begin{align*} - \mathcal{L}^p(E) & :=\left\{{f}:E\rightarrow \RR^n\text{ measurable}:\int_E\abs{{f}}^p<\infty\right\} \\ + \mathcal{L}^p(E) & :=\left\{{f}:E\rightarrow \RR^n\text{ measurable}:\int_E\abs{{f}}^p<\infty\right\} \\ \begin{split} - \mathcal{L}^\infty(E) &:=\left\{{f}:E\rightarrow \RR^n\text{ measurable}:\exists M>0\text{ with }\right.\\&\hspace{4.4cm}\left.\abs{{f}(x)}\almoste{\leq} M, x\in E\right\} + \mathcal{L}^\infty(E) & :=\left\{{f}:E\rightarrow \RR^n\text{ measurable}:\exists M>0\text{ with }\right. \\&\hspace{4.4cm}\left.\abs{{f}(x)}\almoste{\leq} M, x\in E\right\} \end{split} \\ \mathcal{N}(E) & :=\{{f}:E\rightarrow \RR^n\text{ measurable}:{f}\almoste{=} 0\} \end{align*} @@ -865,11 +865,11 @@ The case of $L^\infty(E)$ is easy and the first two properties for $L^p(E)$, $p\geq 1$, too (remember \mcref{RFA:postmonotoneE}). It's missing to prove the \mref{RFA:triangularineq} (also called \emph{Minkowski inequality} in this case): $$\norm{f+g}_p\leq \norm{f}_p+\norm{g}_p$$ We have that: \begin{align*} - {\norm{f+g}_p}^p & =\int_E\abs{f+g}\abs{f+g}^{p-1} \\ - & \leq \int_E\abs{f}\abs{f+g}^{p-1}+\int_E\abs{g}\abs{f+g}^{p-1} \\ + {\norm{f+g}_p}^p & =\int_E\abs{f+g}\abs{f+g}^{p-1} \\ + & \leq \int_E\abs{f}\abs{f+g}^{p-1}+\int_E\abs{g}\abs{f+g}^{p-1} \\ \begin{split} - & \leq \left[{\left(\int_E\abs{f}^p\right)}^{1/p}+{\left(\int_E\abs{g}^p\right)}^{1/p}\right]\cdot \\ - &\hspace{2.5cm}\cdot{\left(\int_E\abs{f+g}^{(p-1)\frac{p}{p-1}}\right)}^{1-1/p} + & \leq \left[{\left(\int_E\abs{f}^p\right)}^{1/p}+{\left(\int_E\abs{g}^p\right)}^{1/p}\right]\cdot \\ + & \hspace{2.5cm}\cdot{\left(\int_E\abs{f+g}^{(p-1)\frac{p}{p-1}}\right)}^{1-1/p} \end{split} \\ & =(\norm{f}_p+\norm{g}_p)\frac{{\norm{f+g}_p}^p }{{\norm{f+g}_p}} \end{align*} @@ -1068,14 +1068,14 @@ \item\label{RFA:TcontinuousC} $T(B_E)$ is bounded on $F$, where $B_E:=\{x\in E:\norm{x}_E\leq 1\}$. \item\label{RFA:TcontinuousD} $\norm{T}<\infty$. \item\label{RFA:TcontinuousE} $\exists C\geq 0$ such that $\forall x\in E$ we have: $$\norm{Tx}_F\leq C\norm{x}_E$$ - If, moreover, $T$ is continuous, $\norm{T}$ is the least of such constants $C$. + If, moreover, $T$ is continuous, $\norm{T}$ is the least of such constants $C$. \end{enumerate} \end{theorem} \begin{sproof} \begin{enumerate}[leftmargin=1.5cm] \item[\mref{RFA:TcontinuousA}$\implies$\mref{RFA:TcontinuousB}:] Let $x\in E$ and $(x_n)\in E$ such that $\displaystyle \lim_{n\to\infty}x_n=x$. Then $\displaystyle \lim_{n\to\infty}(x_n - x)=0$ and the continuity and linearity imply $\displaystyle \lim _{n\to\infty}(Tx_n-Tx)=0$. \item[\mref{RFA:TcontinuousB}$\implies$\mref{RFA:TcontinuousC}:] The continuity at the origin of $T$ implies that given $\varepsilon =1$, $\exists \delta>0$ such that: $$T(B_E(0,\delta))\subseteq B_F(0,1)$$ - The linearity of $T$ implies that $T(B_E(0,1))\subseteq B_F(0,1/\delta)$. + The linearity of $T$ implies that $T(B_E(0,1))\subseteq B_F(0,1/\delta)$. \item[\mref{RFA:TcontinuousC}$\implies$\mref{RFA:TcontinuousD}:] Consequence of \mcref{RFA:normT}. \item[\mref{RFA:TcontinuousD}$\implies$\mref{RFA:TcontinuousE}:] By the definition of supremum we have: $$\norm{T\left(\frac{x}{\norm{x}_E}\right)}_F\leq \norm{T}$$ And so $\norm{Tx}_F\leq \norm{T}\norm{x}_E$. \item[\mref{RFA:TcontinuousE}$\implies$\mref{RFA:TcontinuousA}:] Evident. @@ -1600,7 +1600,7 @@ \item If $H_3$ is Hilbert and $S\in\mathcal{L}(H_2,H_3)$, then ${(S\circ T)}^*=T^*\circ S^*$. \end{enumerate} \end{proposition} - \begin{proposition} + \begin{proposition}\label{RFA:adjoint_im_ker} Let $H_1$, $H_2$ be Hilbert spaces and $T\in\mathcal{L}(H_1,H_2)$. Then: \begin{enumerate} \item ${(\im T)}^\perp =\ker T^*$ @@ -1619,21 +1619,7 @@ Let $H$ be a Hilbert space and $T\in\mathcal{L}(H)$ be self-adjoint. Then: $$\norm{T}=\sup\{\abs{\dotp{Tx}{x}}:\norm{x}=1\}=\max\{M(T),-m(T)\}$$ where $M(T):=\sup\{\dotp{Tx}{x}:\norm{x}=1\}$ and $m(T):=\inf\{\dotp{Tx}{x}:\norm{x}=1\}$ \end{proposition} - \subsubsection{Lax-Milgram theorem} - \begin{definition} - Let $H$ be a Hilbert space and $a:H\times H\rightarrow\RR$ be a bilinear map. We say that $a$ is \emph{continuous} if $\exists C>0$ such that $\forall u,v\in H$ we have: $$\abs{a(u,v)}\leq C\norm{u}\norm{v}$$ - \end{definition} - \begin{definition} - Let $H$ be a Hilbert space and $a:H\times H\rightarrow\RR$ be a bilinear map. We say that $a$ is \emph{coercive} if $\exists\alpha>0$ such that $\forall u\in H$ we have: $$a(u,u)\geq\alpha\norm{u}^2$$ - \end{definition} - \begin{definition} - Let $H$ be a Hilbert space and $a:H\times H\rightarrow\CC$ be a bilinear map. We say that $a$ is \emph{symmetric} if $\forall u,v\in H$ we have: $$a(u,v)=\overline{a(v,u)}$$ - \end{definition} - \begin{theorem}[Lax-Milgram theorem]\label{RFA:laxmilgram} - Let $H$ be a Hilbert space and $a:H\times H\rightarrow\RR$ be a continuous and coercive bilinear map. Then, $\forall L\in H^*$ there exists a unique $u\in H$ such that: $$a(u,v)=L(v)\quad \forall v\in H$$ - In addition, if $\mathcal{H}$ is a real Hilbert space and $a$ is symmetric, then $u$ is the unique minimizer of: - $$\min_{v\in H}\left\{\frac{1}{2}a(v,v)-L(v)\right\}$$ - \end{theorem} + \subsubsection{Orthonormal systems} \begin{definition} Let $H$ be a Hilbert space. An \emph{orthogonal system} on $H$ is a nonempty subset $E\subseteq H$ such that its vectors are pairwise orthogonal. If moreover $\norm{e}=1 $ $\forall e\in E$, we will say that $E$ is an \emph{orthonormal system}. diff --git a/Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex b/Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex index 0128b93..8ebbdaa 100644 --- a/Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex +++ b/Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex @@ -276,10 +276,10 @@ $$ By density, it is enough to prove the result for $u\in \mathcal{C}^1(\RR_{\geq 0})$. An easy check shows that $\bar{u}\in \mathcal{C}^1(\RR)$. Moreover, we have: \begin{align*} - {\norm{\bar{u}}_{W^{1,p}(\RR)}}^p & =\int_{\RR}{\abs{\bar{u}}^p}+{\abs{\bar{u}'}^p} \\ + {\norm{\bar{u}}_{W^{1,p}(\RR)}}^p & =\int_{\RR}{\abs{\bar{u}}^p}+{\abs{\bar{u}'}^p} \\ \begin{split} - &=\!\int_{\RR_{\geq 0}}\!{\abs{u}^p}\!+\!\!{\abs{u'}^p}\!+\!\int_{\RR_{\leq 0}}\![{\abs{-3u(-x)+4u(-x/2)}^p}+\\ - &\hspace{2.75cm}+{\abs{3u'(-x)-2u'(-x/2)}^p}] + & =\!\int_{\RR_{\geq 0}}\!{\abs{u}^p}\!+\!\!{\abs{u'}^p}\!+\!\int_{\RR_{\leq 0}}\![{\abs{-3u(-x)+4u(-x/2)}^p}+ \\ + & \hspace{2.75cm}+{\abs{3u'(-x)-2u'(-x/2)}^p}] \end{split} \\ & \leq C{\norm{u}_{W^{1,p}(\RR_{\geq 0})}}^p \end{align*} @@ -290,9 +290,9 @@ $$ \bar{u}(x_1,\ldots,x_d):= \begin{cases} - u(x_1,\ldots,x_d) & \text{if }x_d\geq 0 \\ + u(x_1,\ldots,x_d) & \text{if }x_d\geq 0 \\ \begin{split} - -3u(x_1,\ldots,x_{d-1}-x_d)+\\ + -3u(x_1,\ldots,x_{d-1}-x_d)+ \\ +4u(x_1,\ldots,x_{d-1}-x_d/2) \end{split} & \text{if }x_d<0 \end{cases} @@ -343,7 +343,7 @@ \end{itemize} \end{theorem} \begin{lemma} - Let $\Omega\subseteq \RR^d$ be an open set and $u\in W^{1,p}(\Omega)$ with $1\leq p\leq \infty$. Then, $\norm{\grad \abs{u}}\almoste{\leq}\norm{\grad u}$. + Let $\Omega\subseteq \RR^d$ be an open set and $u\in W^{1,p}(\Omega)$ with $1\leq p\leq \infty$. Then, $\abs{u}\in W^{1,p}(\Omega)$ and $\norm{\grad \abs{u}}\almoste{\leq}\norm{\grad u}$. \end{lemma} \begin{proof} $$ diff --git a/Mathematics/5th/Introduction_to_nonlinear_elliptic_PDEs/Introduction_to_nonlinear_elliptic_PDEs.tex b/Mathematics/5th/Introduction_to_nonlinear_elliptic_PDEs/Introduction_to_nonlinear_elliptic_PDEs.tex index 4f3d529..0cf735f 100644 --- a/Mathematics/5th/Introduction_to_nonlinear_elliptic_PDEs/Introduction_to_nonlinear_elliptic_PDEs.tex +++ b/Mathematics/5th/Introduction_to_nonlinear_elliptic_PDEs/Introduction_to_nonlinear_elliptic_PDEs.tex @@ -23,11 +23,18 @@ $$ where we have assumed that the boundary of $\Omega$ is smooth enough to define the normal vector $\vf{n}$. The condition is called \emph{homogeneous} if $g=0$. Note that if $\vf{A}=\vf{I}_d$, then the Neumann boundary condition is just $\partial_{\vf{n}} u=g$. \end{definition} + \begin{remark} + If the coefficients $a_{ij}\in\mathcal{C}^1$, then we convert the equation from non-divergence form to divergence form and viceversa. + \end{remark} \begin{definition} - Let $a_{ij},b_j,c$ be known functions on $\Omega\subseteq \RR^d$. We say that the operator $$L=-\sum_{i,j=1}^da_{ij}\partial_{ij}^2 + \sum_{j=1}^d b_j\partial_j+c$$ is \emph{uniformly elliptic} if there exists $\theta>0$ such that for all $x\in \Omega$ and all $p\in \RR^d$ we have: - \begin{equation} - Q_x(\vf{p})=\sum_{i,j=1}^da_{ij}(\vf{x})p_ip_j\geq \theta \sum_{i=1}^{d} {p_i}^2 + Let $a_{ij},b_j,c$ be known functions on $\Omega\subseteq \RR^d$. We say that the operator + \begin{equation}\label{INEPDE:operator} + L=-\sum_{i,j=1}^da_{ij}\partial_{ij}^2 + \sum_{j=1}^d b_j\partial_j+c \end{equation} + is \emph{uniformly elliptic} if there exists $\theta>0$ such that for all $x\in \Omega$ and all $\vf{p}\in \RR^d$ we have: + \begin{equation*} + Q_x(\vf{p}):=\transpose{\vf{p}}\vf{A}(\vf{x})\vf{p}=\!\sum_{i,j=1}^da_{ij}(\vf{x})p_ip_j\geq \theta \sum_{i=1}^{d} {p_i}^2=\theta \norm{\vf{p}}^2 + \end{equation*} \end{definition} \begin{remark} Geometrically speaking, this implies that the sets @@ -36,6 +43,23 @@ $$ are ellipsoids. \end{remark} + \begin{definition} + Consider the problem + $$ + \mathcal{D}_f:=\begin{cases} + Lu=f & \text{in }\Omega \\ + u=0 & \text{on }\partial\Omega + \end{cases} + $$ + where $L$ is as in \mcref{INEPDE:operator}. The \emph{weak formulation} (or \emph{variational formulation}) of the problem is: + \begin{equation*} + {\langle \grad u,\grad v\rangle}_2+{\langle \vf{b}\cdot \grad u,v\rangle}_2 + {\langle cu,v\rangle}_2={\langle f,v\rangle}_2\quad \forall v\in H_0^1(\Omega) + \end{equation*} + A solution of such problem is called a \emph{weak solution} of $\mathcal{D}_f$. + \end{definition} + \begin{definition} + If the weak solution $u_f$ of the problem $\mathcal{D}_f$ is in $H^1_0(\Omega)\cap W^{2,p}(\Omega)$ for some $p\in [1,\infty)$, then $u_f$ is called a \emph{strong solution} of $\mathcal{D}_f$. If $u_f\in \mathcal{C}^2(\Omega)\cap H^1_0(\Omega)$, then we say that $u_f$ is a \emph{classical solution} of $\mathcal{D}_f$. + \end{definition} \begin{proposition} Let $H$ be Hilbert and $K:H\to H$ be a continuous linear operator. Then, the following are equivalent: \begin{enumerate} @@ -46,30 +70,155 @@ \end{proposition} \subsection{Hilbert space methods for divergence form linear PDEs} In this section, we will assume that $\Omega\subset\RR^d$ is an open, bounded subset, $a_{ij}=a_{ji}$ and $a_{ij},b_j,c\in L^\infty(\Omega)$. - \begin{theorem}[Abstract Fredholm alternative] - Let $H$ be Hilbert and $K:H\to H$ be a compact linear operator. Then: + \subsubsection{Lax-Milgram theorem} + \begin{remark} + Instead of the usual norm for $H_0^1(\Omega)$, here we will use the following one: + $$ + \norm{u}_{H_0^1(\Omega)}^2=\norm{\grad u}_{L^2(\Omega)}^2 + $$ + \end{remark} + \begin{definition} + Let $H$ be a Hilbert space and $a:H\times H\rightarrow\RR$ be a bilinear map. We say that $a$ is \emph{continuous} if $\exists C>0$ such that $\forall u,v\in H$ we have: $$\abs{a(u,v)}\leq C\norm{u}\norm{v}$$ + \end{definition} + \begin{definition} + Let $H$ be a Hilbert space and $a:H\times H\rightarrow\RR$ be a bilinear map. We say that $a$ is \emph{coercive} if $\exists\alpha>0$ such that $\forall u\in H$ we have: $$a(u,u)\geq\alpha\norm{u}^2$$ + \end{definition} + \begin{definition} + Let $H$ be a Hilbert space and $a:H\times H\rightarrow\CC$ be a bilinear map. We say that $a$ is \emph{symmetric} if $\forall u,v\in H$ we have: $$a(u,v)=\overline{a(v,u)}$$ + \end{definition} + \begin{theorem}[Lax-Milgram theorem]\label{INEPDE:laxmilgram} + Let $H$ be a Hilbert space and $a:H\times H\rightarrow\RR$ be a continuous and coercive bilinear map. Then, $\forall f\in H^*$ $\exists! u_f\in H$ such that: $$a(u_f,v)=f(v)\quad \forall v\in H$$ + In addition, if ${H}$ is a real Hilbert space and $a$ is symmetric, then $u$ is the unique minimizer of: + $$\min_{v\in H}\left\{\frac{1}{2}a(v,v)-f(v)\right\}$$ + \end{theorem} + \begin{proposition} + Consider the problem: + $$ + \begin{cases} + L u=f & \text{in }\Omega \\ + u=0 & \text{on }\partial\Omega + \end{cases} + $$ + with $L=-\sum_{i,j=1}^d\partial_i(a_{ij}\partial_j)$ and $f\in L^2(\Omega)$. Then, the problem has a unique weak solution $u\in H_0^1(\Omega)$ and + $$ + \norm{u}_{H_0^1(\Omega)}\leq C\norm{f}_{L^2(\Omega)} + $$ + \end{proposition} + \begin{proof} + Consider the bilinear form $$ + a(u,v):=\int_\Omega\sum_{i,j=1}^da_{ij}\partial_iu\partial_jv + $$ + We check the hypotheses of \mnameref{INEPDE:laxmilgram}: \begin{enumerate} - \item $\ker(\id-K)$ and $\ker(\id-K^*)$ are both finite dimensional and they have the same dimension. - \item $\im(\id-K)={(\ker(\id-K^*))}^\perp$. In particular, $\im(\id-K)$ is closed. - \item Either $\ker(\id-K)\ne\{0\}$ or $\id -K$ is and isomorphism. + \item $a$ is continuous: + \begin{align*} + \abs{a(u,v)} & \leq \sum_{i,j=1}^d\norm{a_{ij}}_{\infty}\norm{\grad u}_{2}\norm{\grad v}_{2} \\ + & \leq C\norm{u}_{H_0^1(\Omega)}\norm{v}_{H_0^1(\Omega)} + \end{align*} + \item $a$ is coercive: + \begin{align*} + a(u,u) & =\int_\Omega\sum_{i,j=1}^da_{ij}\partial_iu\partial_ju \\ + & \geq \theta \int_\Omega\sum_{i=1}^d{\abs{\partial_iu}}^2 \\ + & =\theta {\norm{u}_{H_0^1(\Omega)}}^2 + \end{align*} + by the uniform ellipticity of $L$ and the \mnameref{ATFAPDE:poincare_ineq}. \end{enumerate} - \end{theorem} - \begin{definition} - Consider the problem + Moreover, since $ a(u,u)={\langle f,u\rangle}_2$ we have that: + \begin{align*} + \theta {\norm{u}_{H_0^1(\Omega)}}^2\leq {\langle f,u\rangle}_2\leq \norm{f}_2\norm{u}_{2}\leq C \norm{f}_2\norm{u}_{H_0^1(\Omega)} + \end{align*} + again by the \mnameref{ATFAPDE:poincare_ineq}. + \end{proof} + \subsubsection{Abstract Fredholm alternative} + \begin{remark} + One can check that if we try to apply \mnameref{INEPDE:laxmilgram} to the problem: $$ - \mathcal{D}_f:=\begin{cases} + \begin{cases} Lu=f & \text{in }\Omega \\ u=0 & \text{on }\partial\Omega \end{cases} $$ - where $L=-\laplacian+\vf{b}\cdot \grad$. Its \emph{weak formulation} is: - \begin{equation*} - \langle \grad u,\grad v\rangle+\langle \vf{b}\cdot \grad u,v\rangle=\langle f,v\rangle\quad \forall v\in H_0^1(\Omega) - \end{equation*} - We define the \emph{formal adjoint} of $L$ as: + with $L=-\sum_{i,j=1}^d\partial_i(a_{ij}\partial_j)+\sum_{j=1}^db_j\partial_j$, it fails due to the coercivity condition. + \end{remark} + \begin{proposition} + Consider the problem: + $$ + \begin{cases} + L_\mu u=f & \text{in }\Omega \\ + u=0 & \text{on }\partial\Omega + \end{cases} + $$ + with $L_\mu=-\sum_{i,j=1}^d\partial_i(a_{ij}\partial_j)+\sum_{j=1}^db_j\partial_j+\mu c$. Then, if $\mu>0$ is large enough, the problem has a unique weak solution in $H_0^1(\Omega)$ + \end{proposition} + \begin{sproof} + Taking the natural bilinear map $a$, the coercivity condition becomes: $$ - L^*v=-\laplacian v-\div(\vf{b}v) + a_\mu(u,u)\geq \theta \norm{u}_{H_0^1(\Omega)}^2-C\norm{u}_{H_0^1(\Omega)}\norm{u}_2 + \mu \norm{u}_2^2 $$ + which is for $\mu$ large enough it is bigger than $\delta \norm{u}_{H_0^1(\Omega)}^2$ for some $\delta>0$. + \end{sproof} + \begin{lemma}\label{INEPDE:lemma1_fredholm} + Let $H$ be Hilbert and $K:H\to H$ be a compact linear operator. Then, $\dim \ker(\id-K)<\infty$. + \end{lemma} + \begin{proof} + We first prove that $\dim\ker(\id-K)<\infty$. If $\dim\ker(\id-K)=\infty$, then $\exists (u_n)\in \ker(\id-K)$ orthonormal, and thus bounded. In particular, $u_n=Ku_n$ and since $K$ is compact, we have that $(Ku_n)$ has a convergent subsequence. But: + \begin{align*} + 0 & =\lim_{k\to\infty}\norm{Ku_{n_k}-Ku_{n_{k+1}}}^2 \\ + & =\lim_{k\to\infty}\norm{u_{n_k}-u_{n_{k+1}}}^2 \\ + & =\lim_{k\to\infty}\norm{u_{n_k}}^2+\norm{u_{n_{k+1}}}^2 \\ + & =2 + \end{align*} + by \mnameref{RFA:pythagorean}. + \end{proof} + \begin{lemma}\label{INEPDE:lemma2_fredholm} + Let $H$ be Hilbert and $K:H\to H$ be a compact linear operator. Then, $\exists c>0$ such that $\forall u\in {\ker(\id-K)}^{\perp}$ we have $\norm{u-Ku}\geq c\norm{u}$. + \end{lemma} + \begin{proof} + We proceed by contradiction. Suppose we have a sequence $(u_n)\in {\ker(\id-K)}^{\perp}$ with $\norm{u_n}=1$ such that $\norm{u_n-Ku_n}\to 0$. Since $(u_n)$ is bounded, we have that $(u_n)$ has a weakly convergent subsequence $(u_{n_k})$ to $u\in H$. Since $K$ is compact, we have that $Ku_{n_k}\to Ku$, and thus by continuity of the norm, $u=Ku$. Thus $u\in \ker(\id-K)$ and $u\in {\ker(\id-K)}^{\perp}$, which implies $u=0$, a contraction with $\norm{u}=1$. + \end{proof} + \begin{lemma}\label{INEPDE:lemma3_fredholm} + Let $H$ be Hilbert and $K:H\to H$ be a compact linear operator. Then, $\im(\id-K)$ is closed. + \end{lemma} + \begin{proof} + Let $(v_n)\in \im(\id-K)$ be such that $v_n\to v\in H$. Then, $\exists (u_n)\in H$ such that $v_n=(\id-K)u_n$. Thanks to \mnameref{RFA:projection}, we can write $u_n=u_n^{\text{ker}}+ u_n^{\text{ker}^\perp}$, where $u_n^{\text{ker}}\in \ker(\id-K)$ and $u_n^{\text{ker}^\perp}\in {\ker(\id-K)}^{\perp}$. Thus, $v_n=(\id-K)u_n^{\text{ker}^\perp}$ and by \mcref{INEPDE:lemma2_fredholm}, we have: + $$ + \norm{v_n-v_m}\geq c\norm{u_n^{\text{ker}^\perp}-u_m^{\text{ker}^\perp}} + $$ + Since $(v_n)$ is Cauchy, so it is $(u_n^{\text{ker}^\perp})$, and thus $(u_n^{\text{ker}^\perp})$ converges to some $u\in {\ker(\id-K)}^{\perp}$. Thus, $v=(\id-K)u\in \im(\id-K)$. + \end{proof} + % \begin{lemma} + % Let $H$ be Hilbert and $K:H\to H$ be a compact linear operator. Then, $\ker(\id-K)=\{0\}\iff \ker(\id-K^*)=\{0\}$. + % \end{lemma} + % \begin{proof} + % The argument is symmetric since $K^{**}=K$ and $K$ is compact $\iff K^*$ is compact. So suppose $\ker(\id-K)=\{0\}$. Then, $\id -K$ is injective. + % \end{proof} + \begin{theorem}[Abstract Fredholm alternative] + Let $H$ be Hilbert and $K:H\to H$ be a compact linear operator. Then: + \begin{enumerate} + \item $\ker(\id-K)$ and $\ker(\id-K^*)$ are both finite dimensional, and they have the same dimension. + \item $\im(\id-K)={\ker(\id-K^*)}^\perp$. In particular, $\im(\id-K)$ is closed. + \item Either $\ker(\id-K)\ne\{0\}$ or $\id -K$ is an isomorphism. + \end{enumerate} + \end{theorem} + \begin{proof} + We organize the proof in several steps: + \begin{enumerate} + \setcounter{enumi}{1} + \item From \mcref{RFA:adjoint_im_ker} we have that $\overline{\im A}={\ker A^*}^\perp$ for any general operator $A$ between Hilbert spaces. Thus, $\im(\id-K)={\ker(\id-K^*)}^\perp\iff \im(\id-K)$ is closed, which reduces to \mcref{INEPDE:lemma3_fredholm}. + \item We first show that $\ker(\id-K)=\{0\}\iff \ker(\id-K^*)=\{0\}$. The argument is symmetric since $K^{**}=K$ and the fact that $K$ is compact $\iff K^*$ is compact. So suppose $\ker(\id-K)=\{0\}$. Then, $\id -K$ is injective. Assume $\ker(\id-K^*)\ne\{0\}$. Then, $\im (\id-K)=\ker(\id-K)^\perp\ne H$ and so $\im({(\id-K)}^2)\subsetneq \im(\id-K)$. Indeed, if we had equality, then for any $u\in H$, we would have ${(\id-K)u}\in \im({(\id-K)}^2)$, and thus $\exists v\in H$ such that ${(\id-K)u}={(\id-K)}^2v$, which implies $u={(\id-K)}v$ because $\ker (\id-K)=\{0\}$. Recursively, we have an infinite sequence $\im({(\id-K)}^{n+1})\subsetneq \im({(\id-K)}^n)$, which implies that $\forall n$ $\exists u_n\in \im({(\id-K)}^n)\cap\im({(\id-K)}^{n+1})^\perp$ with $\norm{u_n}=1$. Thus, $\langle u_n,u_m\rangle=\delta_{n,m}$. But $u_n-Ku_n\in \im({(\id-K)}^{n+1})$ so, $u_n-Ku_n\perp u_n$. This implies, by \mnameref{RFA:pythagorean}, that $\norm{Ku_n}=\norm{u_n-Ku_n}+\norm{u_n}\geq 1$, which is a contradiction with the compactness of $K$ because any orthonormal sequence always converges weakly to zero (and so $Ku_n\to 0$). So either $\ker(\id-K)\ne\{0\}$ or $\id-K$ is bijective. + + To finish this point, we need to prove that if $\ker(\id-K)$, then ${(\id-K)}^{-1}$ is a bounded linear operator. But this is a consequence of \mcref{INEPDE:lemma2_fredholm}: if $u\in H$, then $u\in \ker {(\id-K)}^\perp$ and thus $\norm{(\id-K)u}\geq c\norm{u}$, which implies that $\norm{v}\geq c \norm{{(\id-K)}^{-1}v}$ taking $v=(\id-K)u$. + \setcounter{enumi}{0} + \item Assume without loss of generality that $\dim\ker(\id-K)<\dim \ker(\id-K^*)$. Then, there exists a linear injective map $A:\ker(\id-K)\to \ker(\id-K^*)=\im(\id-K)^\perp$. Let $\tilde{K}$ be the operator defined by $\tilde{K}u=Ku+Au^{\text{ker}}$, where $u^{\text{ker}}$ is the projection of $u$ onto $\ker(\id-K)$. Then, $\tilde{K}$ is compact (because $K$ is compact and so is $A$, because it has finite range). Moreover, if $u\in \ker(\id-\tilde{K})$, then $(\id-K)u+Au^{\text{ker}}=0$, which since $(\id-K)u\in \im(\id-K)$ and $Au^{\text{ker}}\in \im(\id-K)^\perp$ implies that both terms are zero. So $u=u^{\text{ker}}\in \ker(\id-K)$ and since $A$ is injective, $u=u^{\text{ker}}=0$. Thus, $\ker(\id-\tilde{K})=\{0\}$ and by the previous point, $\id-\tilde{K}$ is an isomorphism from $H$ to itself. So, for every $w\in \ker(\id-K^*)$, $\exists u\in H$ such that $w=(\id-\tilde{K})u$. Projecting both side onto $\ker(\id-K^*)=\im(\id-K)^\perp$, we have $w=-Au^{\text{ker}}$, which implies that $A$ is onto, and so $\dim\ker(\id-K)=\dim\ker(\id-K^*)$. \mcref{INEPDE:lemma1_fredholm} finishes the proof. + \end{enumerate} + \end{proof} + \begin{definition} + Consider the operator $L$ as in \mcref{INEPDE:operator}. We define the \emph{formal adjoint} of $L$ as: + \begin{align*} + L^*v & :=-\sum_{i,j=1}^d\partial_i(a_{ij}\partial_jv)-\sum_{j=1}^d\partial_j(b_jv)+c v \\ + & =\sum_{i,j=1}^d\partial_i(a_{ij}\partial_jv)-\sum_{j=1}^db_j\partial_jv+ \left(c-\sum_{j=1}^d\partial_jb_j\right)v + \end{align*} + It satisfies $\langle Lu,v\rangle=\langle u,L^*v\rangle$ for all $u,v\in H_0^1(\Omega)$. \end{definition} \begin{proposition} The \emph{homogeneous adjoint problem} @@ -81,9 +230,12 @@ $$ whose weak formulation is \begin{equation*} - \langle \grad v,\grad w\rangle+\langle \vf{b}\cdot \grad v,w\rangle=0\quad \forall w\in H_0^1(\Omega) + {\langle \grad v,\grad w\rangle}_2+{\langle \vf{b}\cdot \grad v,w\rangle}_2=0\quad \forall w\in H_0^1(\Omega) \end{equation*} has a finite dimensional solution space $W_0$, as well as the space $V_0$ of solutions of $\mathcal{D}_0$, and $\dim W_0=\dim V_0$. Moreover, if $f\in L^2(\Omega)$, $\mathcal{D}_f$ is solvable if and only if $\langle f,v\rangle=0$ for all $v\in W_0$. + \end{proposition} + \begin{proposition} + \end{proposition} \begin{definition} We define the following problem: @@ -114,18 +266,33 @@ $$ \end{definition} \begin{theorem} - Let $H$ be a Hilbert space and $K:H\to H$ be a compact operator. Then, $0\in \sigma(K)$ and $\sigma(K)$ is closed and at most countable. Moreover, if $\lambda\in \sigma(K)\setminus\{0\}$, then $\lambda$ is an eigenvalue of $K$ and: + Let $H$ be an infinite-dimensional Hilbert space and $K:H\to H$ be a compact operator. Then, $0\in \sigma(K)$ and $\sigma(K)$ is closed and at most countable. Moreover, if $\lambda\in \sigma(K)\setminus\{0\}$, then $\lambda$ is an eigenvalue of $K$ and: $$ \dim\left(\bigcup_{p\geq 1}\ker{(\lambda\id-K)}^p\right)<\infty $$ If $\sigma(K)\cap\RR^*$ is infinite, then it is of the form $\{\lambda_n\}_{n\in \NN}$ with $\lambda_n\to 0$. \end{theorem} + \begin{proof} + \begin{enumerate} + \item Assume $0\notin \sigma(K)$. Then, $K$ is bijective and so $\id = K\circ K^{-1}$ is compact, as it is the composition of a compact operator and a bounded operator. But this is a contradiction with \mcref{INEPDE:lemma1_fredholm}. + \end{enumerate} + \end{proof} \begin{lemma} - Let $H$ be a Hilbert space and $K:H\to H$ be a continuous self-adjoint operator. Then: + Let $H$ be a Hilbert space and $T:H\to H$ be a continuous self-adjoint operator. Then: $$ - \norm{K}=\sup_{\norm{x}=1}\langle x,Kx\rangle + \norm{T}=\sup_{\norm{x}=1}\abs{\langle x,Tx\rangle} $$ \end{lemma} + \begin{proof} + Clearly $\alpha:=\sup_{\norm{x}=1}\abs{\langle x,Tx\rangle}\leq \norm{T}$. For the converse, it suffices to show that $\abs{\langle Tx,y\rangle}\leq \alpha$ for all $\norm{x}=\norm{y}=1$. We have: + $$ + \langle Tx,y\rangle = \frac{1}{4}\left(\langle T(x+y),x+y\rangle-\langle T(x-y),x-y\rangle\right) + $$ + And then, by \mnameref{RFA:parallelogram}: + $$ + \abs{\langle Tx,y\rangle}\leq \frac{\alpha}{4}\left(\norm{x+y}^2+\norm{x-y}^2\right)= \alpha + $$ + \end{proof} \begin{lemma} Let $H\ne\{ 0\}$ be Hilbert and $K:H\to H$ be a compact and self-adjoint operator. Then: $$ @@ -135,15 +302,21 @@ \end{lemma} \subsubsection{Regularity theorems for weak solutions of divergence-form elliptic PDEs} \begin{theorem}[Inner regularity] - Assume, in addition to the usual assumptions, that $a_{ij}\in \mathcal{C}^1(\Omega)$. Let $f\in L^2(\Omega)$ and $u$ be a weak solution of $Lu=f$. Then, $u\in H^2_{\text{loc}}(\Omega)$ and for any compact $\omega\subset\subset \Omega$, meaning that $\overline{\omega}\subset\Omega$ compact, we have $u\in H^2(\omega)$ and: + Assume, in addition to the usual assumptions, that $a_{ij}\in \mathcal{C}^1(\Omega)$. Let $f\in L^2(\Omega)$ and $u\in H^1(\Omega)$ be a weak solution of $Lu=f$. Then, $u\in H^2_{\text{loc}}(\Omega)$ and for any compact embedding $\omega\subset\subset \Omega$, meaning that $\overline{\omega}\subset\Omega$ compact, we have $u\in H^2(\omega)$ and: $$ \norm{u}_{H^2(\omega)}\leq C\left(\norm{f}_{L^2(\Omega)}+\norm{u}_{L^2(\Omega)}\right) $$ \end{theorem} \begin{corollary} - Assume that $a_{ij}\in\mathcal{C}^m(\Omega)$ for some $m\geq 2$, and $b_j,c\in \mathcal{C}^{m-1}(\Omega)$. Let $f\in H^{m-1}(\Omega)$ and $u\in H^1_{\text{loc}}(\Omega)\cap L^2(\Omega)$ be a weak solution of $Lu=f$. Then, $u\in H^{m+1}_{\text{loc}}(\Omega)$ and for any $\omega\subset\subset \Omega$ we have $u\in H^{m+2}(\omega)$ and: + Assume that $a_{ij}\in\mathcal{C}^{m+1}(\Omega)$ for some $m\in\NN$, and $b_j,c\in \mathcal{C}^{m}(\Omega)$. Let $f\in H^{m}(\Omega)$ and $u\in H^1$ be a weak solution of $Lu=f$. Then, $u\in H^{m+2}_{\text{loc}}(\Omega)$ and for any $\omega\subset\subset \Omega$ we have $u\in H^{m+2}(\omega)$ and: + $$ + \norm{u}_{H^{m+2}(\omega)}\leq C\left(\norm{f}_{H^{m}(\Omega)}+\norm{u}_{L^2(\Omega)}\right) + $$ + \end{corollary} + \begin{corollary} + Assume $a_{ij},b_j,c,f\in\mathcal{C}^\infty(\Omega)$. Let $u\in H^1(\Omega)$ be a weak solution of $Lu=f$. Then, $u\in \mathcal{C}^\infty(\Omega)$ and $\forall m\in \NN$: $$ - \norm{u}_{H^{m+2}(\omega)}\leq C\left(\norm{f}_{H^{m-1}(\Omega)}+\norm{u}_{L^2(\Omega)}\right) + \norm{u}_{H^{m}(\Omega)}\leq C\left(\norm{f}_{H^{m}(\Omega)}+\norm{u}_{L^2(\Omega)}\right) $$ \end{corollary} \begin{theorem}[Regularity up to the boundary] @@ -153,15 +326,15 @@ $$ \end{theorem} \begin{corollary} - Assume that $\Fr{\Omega}$ is $\mathcal{C}^{m+1}$, $m\geq 2$, and that $a_{ij}\in \mathcal{C}^m(\overline{\Omega})$, $b_j,c\in \mathcal{C}^{m-1}(\overline{\Omega})$. Let $f\in H^{m-1}(\Omega)$ and $u\in H^1_0(\Omega)$ be a weak solution of $\mathcal{D}_f$. Then, $u\in H^{m+2}(\Omega)$ and: + Assume that $\Fr{\Omega}$ is $\mathcal{C}^{m}$, $m\in\NN$, and that $a_{ij}\in \mathcal{C}^{m+1}(\overline{\Omega})$, $b_j,c\in \mathcal{C}^{m}(\overline{\Omega})$. Let $f\in H^{m}(\Omega)$ and $u\in H^1_0(\Omega)$ be a weak solution of $\mathcal{D}_f$. Then, $u\in H^{m+2}(\Omega)$ and: $$ - \norm{u}_{H^{m+2}(\Omega)}\leq C\left(\norm{f}_{H^{m-1}(\Omega)}+\norm{u}_{L^2(\Omega)}\right) + \norm{u}_{H^{m+2}(\Omega)}\leq C\left(\norm{f}_{H^{m}(\Omega)}+\norm{u}_{L^2(\Omega)}\right) $$ \end{corollary} \begin{corollary} - Assume that $\Fr{\Omega}$ is $\mathcal{C}^\infty$ and that $a_{ij},b_j,c,f\in \mathcal{C}^\infty(\overline{\Omega})$. Let $u\in H^1_0(\Omega)$ be a weak solution of $\mathcal{D}_f$. Then, $u\in \mathcal{C}^\infty(\Omega)$ and $\forall m\geq 1$: + Assume that $\Fr{\Omega}$ is $\mathcal{C}^\infty$ and that $a_{ij},b_j,c,f\in \mathcal{C}^\infty(\overline{\Omega})$. Let $u\in H^1_0(\Omega)$ be a weak solution of $\mathcal{D}_f$. Then, $u\in \mathcal{C}^\infty(\Omega)$ and $\forall m\in \NN$: $$ - \norm{u}_{H^{m+1}(\Omega)}\leq C\left(\norm{f}_{H^{m-1}(\Omega)}+\norm{u}_{L^2(\Omega)}\right) + \norm{u}_{H^{m}(\Omega)}\leq C\left(\norm{f}_{H^{m}(\Omega)}+\norm{u}_{L^2(\Omega)}\right) $$ \end{corollary} \subsubsection{Weak maximum principle for weak solutions of divergence-form elliptic PDEs} @@ -220,9 +393,6 @@ \begin{corollary} For each $f\in L^2(\Omega)$, the problem $\mathcal{D}_f$ has a unique weak solution $u_f$. Moreover, if $\Fr{\Omega}\in\mathcal{C}^1$, then $u_f\in H^2(\Omega)$ and $f\mapsto u_f$ is a bounded linear operator from $L^2(\Omega)$ to $H^2(\Omega)$. If $\Fr{\Omega}\in\mathcal{C}^{m+1}$, $b_j\in\mathcal{C}^{m-1}$ and $f\in H^{m-1}(\Omega)$, then $u_f\in H^{m+1}(Omega)$ and $f\mapsto u_f$ is a bounded linear operator from $H^{m-1}(\Omega)$ to $H^{m+1}(\Omega)$. \end{corollary} - \begin{definition} - If the weak solution $u_f$ of the problem $\mathcal{D}_f$ is in $H^1_0(\Omega)\cap W^{2,p}(\Omega)$ for some $p\in [1,\infty)$, then we say that $u_f$ is called a \emph{strong solution} of $\mathcal{D}_f$. If $u_f\in \mathcal{C}^2(\Omega)\cap H^1_0(\Omega)$, then we say that $u_f$ is a \emph{classical solution} of $\mathcal{D}_f$. - \end{definition} \begin{theorem} Let $10$ such that: