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 a9b3d8c..40f753d 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} @@ -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. @@ -1134,7 +1134,7 @@ We may suppose $\varepsilon<1$. Let $v\in E$ such that $d(v,F)=\delta>0$. Then, consider $u=\frac{v-x_0}{\norm{v-x_0}}$, where $x_0\in F$ satisfies $\delta\leq \norm{v-x_0}\leq \frac{\delta}{1-\varepsilon}$. Finally, $\forall x\in F$: $$\norm{u-x}=\frac{\norm{v-(x_0+\norm{v-x_0}x)}}{\norm{v-x_0}}\geq \frac{\delta}{\norm{v-x_0}}\geq 1-\varepsilon$$ \end{proof} - \begin{theorem}[Riesz's theorem] + \begin{theorem}[Riesz's theorem]\label{RFA:riesz_ball} Let $E$ be a normed vector space. If the unit closed sphere $\{x\in E:\norm{x}=1\}$ is compact, then $\dim E<\infty$. \end{theorem} \begin{proof} @@ -1579,7 +1579,7 @@ \end{align*} \end{proposition} \subsubsection{Duality and adjoint operator} - \begin{theorem}[Riesz representation theorem] + \begin{theorem}[Riesz representation theorem]\label{RFA:riesz_rep} Let $(H,\dotp{\cdot}{\cdot})$ be a Hilbert space. The map $$\function{}{H}{H^*}{x}{\dotp{\cdot}{x}}$$ is semilinear, bijective and isometric. \end{theorem} \begin{corollary} 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 584ae68..77bd925 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 @@ -184,11 +184,12 @@ \begin{definition} Let $E$, $F$ be Banach. We say that $F$ is \emph{embedded} in $E$ if $F\subseteq E$ and the inclusion map $i:F\hookrightarrow E$ is continuous. We say that $F$ is \emph{compactly embedded} in $E$ if $F\subseteq E$ and the inclusion map $i:F\hookrightarrow E$ is compact. \end{definition} - \begin{theorem}[Gagliardo, Nirengerg and Sobolev's inequality] + \begin{theorem}[Gagliardo, Nirengerg and Sobolev's inequality]\label{ATFAPDE:gagliardo_nirengerg_sobolev} For all $1\leq p\leq\frac{d}{m}$, there is a constant $C=C(p,m,d)$ so that $\forall u\in \mathcal{C}_0^\infty(\RR^d)$ we have: $$ \norm{u}_{q}\leq C\sum_{\abs{\alpha}= m}{\norm{\partial^\alpha u}_p}$$ where $\displaystyle\frac{1}{q}=\frac{1}{p}-\frac{m}{d}$. + That is, we have the continuous embedding $W^{m,p}(\RR^d)\hookrightarrow L^q(\RR^d)$. In particular for $m=1$, we have $W^{1,p}(\RR^d)\hookrightarrow L^{p^*}(\RR^d)$ with $\displaystyle\frac{1}{p^*}=\frac{1}{p}-\frac{1}{d}$. \end{theorem} \begin{proof} By induction, it suffices to prove the result only for $m=1$. We will prove only the case $d=2$. We start with $p=1$. Let $u\in \mathcal{C}_0^\infty(\RR^2)$. We have: 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 574ab30..e2ed222 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 @@ -1,7 +1,7 @@ \documentclass[../../../main_math.tex]{subfiles} \begin{document} -\changecolor{INLEPDE} +\changecolor{INEPDE} \begin{multicols}{2}[\section{Introduction to nonlinear elliptic PDEs}] \subsection{Introduction} \begin{definition} @@ -13,13 +13,13 @@ \begin{multline*} -\sum_{i=1}^d\pdv{}{x_i}\left(\sum_{j=1}^da_{ij}(\vf{x})\partial_ju(\vf{x})\right)+\sum_{j=1}^db_j(\vf{x})\partial_ju(\vf{x})+\\+c(\vf{x})u(\vf{x})=f(\vf{x}) \end{multline*} - then we say that the equation is in \emph{divergence form}. Together with the PDE we usually impose boundary conditions on $\partial\Omega$. The \emph{Dirichlet boundary condition} is: + then we say that the equation is in \emph{divergence form}. Together with the PDE we usually impose boundary conditions on $\Fr{\Omega}$. The \emph{Dirichlet boundary condition} is: $$ - u|_{\partial\Omega}=g + u|_{\Fr{\Omega}}=g $$ and it is called \emph{homogeneous} if $g=0$. The \emph{Neumann boundary condition} is: $$ - \langle \vf{n},\vf{A} \grad u\rangle|_{\partial\Omega}=g + \langle \vf{n},\vf{A} \grad u\rangle|_{\Fr{\Omega}}=g $$ 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} @@ -47,8 +47,8 @@ Consider the problem $$ \mathcal{D}_f:=\begin{cases} - Lu=f & \text{in }\Omega \\ - u=0 & \text{on }\partial\Omega + Lu=f & \text{in }\Omega \\ + u=0 & \text{on }\Fr{\Omega} \end{cases} $$ where $L$ is as in \mcref{INEPDE:operator}. The \emph{weak formulation} (or \emph{variational formulation}) of the problem is: @@ -95,8 +95,8 @@ Consider the problem: $$ \begin{cases} - L u=f & \text{in }\Omega \\ - u=0 & \text{on }\partial\Omega + L u=f & \text{in }\Omega \\ + u=0 & \text{on }\Fr{\Omega} \end{cases} $$ 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 @@ -134,8 +134,8 @@ One can check that if we try to apply \mnameref{INEPDE:laxmilgram} to the problem: $$ \begin{cases} - Lu=f & \text{in }\Omega \\ - u=0 & \text{on }\partial\Omega + Lu=f & \text{in }\Omega \\ + u=0 & \text{on }\Fr{\Omega} \end{cases} $$ 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. @@ -144,8 +144,8 @@ Consider the problem: $$ \mathcal{D}_{\mu,f}:=\begin{cases} - L_\mu u=f & \text{in }\Omega \\ - u=0 & \text{on }\partial\Omega + L_\mu u=f & \text{in }\Omega \\ + u=0 & \text{on }\Fr{\Omega} \end{cases} $$ with $L_\mu=-\sum_{i,j=1}^d\partial_i(a_{ij}\partial_j)+\sum_{j=1}^db_j\partial_j+\mu$. Then, if $\mu>0$ is large enough, the problem has a unique weak solution in $H_0^1(\Omega)$ @@ -223,8 +223,8 @@ The \emph{homogeneous adjoint problem} $$ \mathcal{D}_0^*:=\begin{cases} - L^*v=0 & \text{in }\Omega \\ - v=0 & \text{on }\partial\Omega + L^*v=0 & \text{in }\Omega \\ + v=0 & \text{on }\Fr{\Omega} \end{cases} $$ whose weak formulation is @@ -244,15 +244,12 @@ $$ By \mcref{INEPDE:fredholm}, we have that $(\id-K)w=f$ has a solution if and only if $(\id-K^*)h=0\implies \langle f,h\rangle_{L^2} =0$ for all $h\in L^2$. But ${L_{\mu_0}}^*$ is an isomorphism, so $(\id-K^*)h=0\iff {L_0}^*h=0$. \end{proof} - \begin{proposition} - - \end{proposition} \begin{definition} We define the following problem: $$ \mathcal{N}_f:=\begin{cases} - -\laplacian u=f & \text{in }\Omega \\ - \pdv{u}{\vf{n}}=0 & \text{on }\partial\Omega + -\laplacian u=f & \text{in }\Omega \\ + \pdv{u}{\vf{n}}=0 & \text{on }\Fr{\Omega} \end{cases} $$ and $\mathcal{N}_f^*=\mathcal{N}_f$. The weak formulation of the problem is: @@ -266,7 +263,7 @@ \subsubsection{Spectrum of compact operators} In this section $\KK$ will denote either $\RR$ or $\CC$. \begin{definition} - Let $H$ be a $\KK$-Hilbert space and $K:H\to H$ be a compact operator. We define the \emph{resolvent set} of $K$ as: + Let $H$ be a $\KK$-Hilbert space and $K:H\to H$ be a bounded operator. We define the \emph{resolvent set} of $K$ as: $$ \rho(K)=\{\lambda\in \KK: \lambda\id-K \text{ is invertible}\} $$ @@ -275,16 +272,31 @@ \sigma(K)=\KK\setminus \rho(K) $$ \end{definition} + \begin{proposition} + Let $H$ be a $\KK$-Hilbert space and $T:H\to H$ be a bounded operator. Then, $\sigma(T)$ is closed. + \end{proposition} + \begin{proof} + Note that $\rho(K)$ is open because if $\lambda\in \rho(T)$, then $\exists\varepsilon\in\RR$ such that $\abs{\varepsilon} < \norm{{(\lambda\id-K)}^{-1}}$. And so $(\lambda+\varepsilon)\id-K$ is invertible. Thus, $\sigma(K)$ is closed. + \end{proof} \begin{theorem} - 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 $\lambda_n\to 0$. + Let $H$ be an infinite-dimensional separable Hilbert space and $K:H\to H$ be a compact operator. Then: + \begin{enumerate} + \item $0\in \sigma(K)$. + \item\label{INEPDE:item2_spectrum} If $\lambda\in \sigma(K)\setminus\{0\}$, then $\lambda$ is an eigenvalue of $K$. + \item $\sigma(K)$ is closed and at most countable. + \item If $\sigma(K)\cap\RR$ is infinite, then $\sigma(K)\setminus\{0\}$ is of the form $\{\lambda_n\}_{n\in \NN}$ with $\lambda_n\to 0$. + \item If $\lambda\in \sigma(K)\setminus\{0\}$, then: + $$ + \dim\left(\bigcup_{p\geq 1}\ker{(\lambda\id-K)}^p\right)<\infty + $$ + \end{enumerate} \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}. + \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 \mnameref{RFA:riesz_ball} because the image of any bounded set under a compact operator is relatively compact (or \emph{precompact}). + \item If $\ker(\lambda\id-K)=\{0\}$, then by \mnameref{INEPDE:fredholm}, $\lambda\id-K$ is an isomorphism, and thus $\lambda\in \rho(K)$. + \setcounter{enumi}{3} + % \item If $(\lambda_n)\in \sigma(K)\setminus\{0\}$ is an infinite sequence, then by \mcref{INEPDE:item2_spectrum} we have that $\exists (u_n)\in H$ with $\norm{u_n}=1$ such that $Ku_n=\lambda_nu_n$. \end{enumerate} \end{proof} \begin{lemma} @@ -294,7 +306,7 @@ $$ \end{lemma} \begin{proof} - Clearly $\alpha:=\sup_{\norm{x}=1}\abs{\langle x,Tx\rangle}\leq \norm{T}$. For the converse, it suffices to show that $\abs{\langle Tx,y\rangle}\leq \alpha$ for all $\norm{x}=\norm{y}=1$. We have: + Clearly $\displaystyle\alpha:=\sup_{\norm{x}=1}\abs{\langle x,Tx\rangle}\leq \norm{T}$. For the converse, it suffices to show that $\abs{\langle Tx,y\rangle}\leq \alpha$ for all $\norm{x}=\norm{y}=1$. We have: $$ \langle Tx,y\rangle = \frac{1}{4}\left(\langle T(x+y),x+y\rangle-\langle T(x-y),x-y\rangle\right) $$ @@ -310,6 +322,13 @@ $$ where $\lambda$ is the largest eigenvalue of $K$. \end{lemma} + \begin{proof} + Let $(x_n)$ be a maximizing sequence with $\norm{x_n}=1$. After extraction, we can assume that $x_n\rightharpoonup x_*$ and so $Kx_n\to Kx_*$. Thus, $\langle x_n,Kx_n\rangle\to \langle x_*,Kx_*\rangle$. So $x_*$ is a maximizer. Now, take $h\perp x_*$ and $\norm{h}=1$. Then, $x_t:=\frac{x_*+th}{\sqrt{1+t^2}}$ satisfies $\norm{x_t}=1$ and: + $$ + \langle x_t,Kx_t\rangle=\langle x_*,Kx_*\rangle+2t\langle h,Kx_*\rangle+\o{t}\leq \langle x_*,Kx_*\rangle + $$ + because of the maximality. So we must have $\langle h,Kx_*\rangle=0$, which implies $Kx_*\in {{({\langle x_*\rangle}^\perp)}^\perp}=\langle x_*\rangle$. Thus, $Kx_*=\lambda x_*$. + \end{proof} \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 $u\in H^1(\Omega)$ 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: @@ -376,40 +395,75 @@ Let $(u_n)\in H^1(\Omega)$ be such that $u_n\overset{H^1(\Omega)}{\longrightarrow} u$. Then, ${u_n}^\pm\overset{H^1(\Omega)}{\longrightarrow} u^\pm$. \end{lemma} \begin{corollary} - Let $u\in H^1(\Omega)$. Then, $\Tr_{\partial\Omega}(u^\pm)={(\Tr_{\partial\Omega}u)}^\pm$. + Let $u\in H^1(\Omega)$. Then, $\Tr_{\Fr{\Omega}}(u^\pm)={(\Tr_{\Fr{\Omega}}u)}^\pm$. \end{corollary} - \begin{lemma} - Let $\Omega\subseteq\RR^d$ open with $\mathcal{C}^1$ boundary, $u\in H^1(\Omega)$ and $\Tr_{\partial\Omega}u\almoste{\leq}0$. Then, $u^+\in H^1_0(\Omega)$. + \begin{lemma}\label{INEPDE:lemma3_weak_max} + Let $\Omega\subseteq\RR^d$ open with $\mathcal{C}^1$ boundary, $u\in H^1(\Omega)$ and $\Tr_{\Fr{\Omega}}u\almoste{\leq}0$. Then, $u^+\in H^1_0(\Omega)$. \end{lemma} \begin{theorem}[Weak maximum principle] Let $\Omega\subseteq\RR^d$ open and bounded with $\mathcal{C}^1$ boundary, $a_{ij}=a_{ji},c\in L^\infty(\Omega)$, $c\almoste{\geq}0$, $L=-\sum_{i,j=1}^d\partial_i(a_{ij}\partial_j)+c$ be elliptic and $f\in L^2(\Omega)$ with $f\almoste{\leq} 0$. Let $u\in H^1(\Omega)$ be such that: \begin{itemize} \item $\displaystyle \int_\Omega\left[\sum_{i,j=1}^da_{ij}\partial_iu\partial_jv+cuv\right]=\int_\Omega fv$ $\forall v\in H^1_0(\Omega)$ - \item $\Tr_{\partial\Omega}u\almoste{\leq}0$ + \item $\Tr_{\Fr{\Omega}}u\almoste{\leq}0$ \end{itemize} Then, $u\almoste{\leq}0$. \end{theorem} \begin{proof} - Take $v=u^+\in H^1_0(\Omega)$ by \mcref{INEPDE:lemma1_weak_max}. Then, we have: + Take $v=u^+\in H^1_0(\Omega)$ by \mcref{INEPDE:lemma3_weak_max}. Then, we have: $$ 0\leq \theta {\norm{\grad u^+}_{L^2}}^2\leq\!\! \int_{\{u>0\}}\!\!\sum_{i,j=1}^d a_{ij}\partial_iu\partial_ju+cu^2=\!\!\int_{\{u>0\}} \!\!fu\leq 0 $$ where in the second inequality we used the ellipticity of $L$. Thus, we must have $\grad u^+=0$ a.e. in $\Omega$, which implies $u^+=0$ a.e. in $\Omega$, because $u^+|_{\Fr{\Omega}}=0$. \end{proof} - \begin{theorem}[Weak maximum principle] + \begin{theorem}[Weak maximum principle]\label{INEPDE:weak_max_principle} Let $\Omega\subseteq\RR^d$ open and bounded with $\mathcal{C}^1$ boundary, $a_{ij}=a_{ji},b_j,c\in L^\infty(\Omega)$, $c \almoste{\geq}0$, $L=-\sum_{i,j=1}^d\partial_i(a_{ij}\partial_j)+\sum_{j=1}^db_j\partial_j+c$ be elliptic and $f\in L^2(\Omega)$ with $f\almoste{\leq} 0$. Let $u\in H^1(\Omega)$ be such that: \begin{itemize} \item $\displaystyle \int_\Omega\left[\sum_{i,j=1}^da_{ij}\partial_iu\partial_jv+ \sum_{j=1}^db_jv\partial_ju+cuv\right]=\int_\Omega fv$ $\forall v\in H^1_0(\Omega)$ - \item $\Tr_{\partial\Omega}u\almoste{\leq}0$ + \item $\Tr_{\Fr{\Omega}}u\almoste{\leq}0$ \end{itemize} Then, $u\almoste{\leq}0$. \end{theorem} \begin{proof} - TODO + Let $m\geq 0$ and $v_m={(u-m)}^+$. Proceeding as in the previous proof, we have: + \begin{align*} + 0 & \leq \theta \norm{\grad v_m}_{L^2}^2-\norm{\vf{b}}_\infty\norm{\grad v_m}_{L^2}\norm{v_m}_{L^2} \\ + & \leq \int_{\{u>m\}}\sum_{i,j=1}^d a_{ij}\partial_iu\partial_jv_m+\sum_{j=1}^db_j\partial_ju v_m+cu v_m \\ + & =\int_{\{u>m\}}fv_m\leq 0 + \end{align*} + Thus, $\norm{\grad v_m}_{L^2}\leq C \norm{v_m}_{L^2}$, with $C$ independent of $m$. Note that since $\Omega$ is bounded, $\displaystyle\lim_{m\to\infty}\abs{\{u>m\}}=0$ by\mnameref{RFA:dominated}, and so $\displaystyle\lim_{m\to\infty}\supp v_m=0$ as well since $\supp v_m\subseteq \{u>m\}$. We now continue the proof for $d\geq 3$. By \mnameref{ATFAPDE:gagliardo_nirengerg_sobolev} we have a continuous embedding $H_0^1(\Omega)\hookrightarrow L^{2^*}$ with $\frac{1}{2^*}=\frac{1}{2}-\frac{1}{d}$. So, $\norm{v_m}_{L^{2^*}}\leq \delta \norm{\grad v_m}_{L^2}$. Thus: + \begin{multline*} + \norm{\grad v_m}_{L^2}\leq C \norm{v_m}_{L^2}\leq C{\abs{\supp v_m}}^{1-\frac{2}{2^*}}\norm{v_m}_{L^{2^*}}\leq \\ + \leq C\delta {\abs{\supp v_m}}^{1-\frac{2}{2^*}}\norm{\grad v_m}_{L^2}\leq \frac{1}{2} \norm{\grad v_m}_{L^2} + \end{multline*} + where in the second inequality we used \mnameref{RFA:holder} and the last one is valid for $m\geq m_0$ large enough. Thus, $\norm{\grad v_m}_{L^2}=0$ for $m\geq m_0$, which implies $u\almoste{\leq}m_0$. This means that $\abs{\{u>m_0\}}=0$ and that $\forall \varepsilon >0$, $\abs{\{u\geq m_0-\varepsilon\}}>0$. Suppose now that $m_0>0$ and let $S_\varepsilon=\abs{\{u> m_0-\varepsilon\}}$. Again by \mnameref{RFA:dominated}, $\displaystyle\lim_{\varepsilon\to 0}S_\varepsilon=0$. But then, proceeding as in the previous step: + $$ + \norm{\grad v_{m_0-\varepsilon}}\leq C \delta {S_\varepsilon}^{1-\frac{2}{2^*}}\norm{\grad v_{m_0-\varepsilon}}\leq \frac{1}{2}\norm{\grad v_{m_0-\varepsilon}} + $$ + by choosing $\varepsilon$ small enough. Thus, $\norm{\grad v_{m_0-\varepsilon}}=0$ and so $u\almoste{\leq}m_0-\varepsilon$, which is a contradiction. Thus, $m_0= 0$ (because $m_0\geq 0$ from the beginning) and so $u\almoste{\leq}0$. + \end{proof} + \begin{theorem}[Weak minimum principle]\label{INEPDE:weak_min_principle} + Let $\Omega\subseteq\RR^d$ open and bounded with $\mathcal{C}^1$ boundary, $a_{ij}=a_{ji},b_j,c\in L^\infty(\Omega)$, $c \almoste{\geq}0$, $L=-\sum_{i,j=1}^d\partial_i(a_{ij}\partial_j)+\sum_{j=1}^db_j\partial_j+c$ be elliptic and $f\in L^2(\Omega)$ with $f\almoste{\geq} 0$. Let $u\in H^1(\Omega)$ be such that: + \begin{itemize} + \item $\displaystyle \int_\Omega\left[\sum_{i,j=1}^da_{ij}\partial_iu\partial_jv+ \sum_{j=1}^db_jv\partial_ju+cuv\right]=\int_\Omega fv$ $\forall v\in H^1_0(\Omega)$ + \item $\Tr_{\Fr{\Omega}}u\almoste{\geq}0$ + \end{itemize} + Then, $u\almoste{\geq}0$. + \end{theorem} + \begin{sproof} + Apply \mnameref{INEPDE:weak_max_principle} to $u\mapsto -u$ with $f\mapsto -f$. + \end{sproof} + \begin{corollary}\label{INEPDE:corollary_max_min} + If $u$ is a weak solution of $\mathcal{D}_0$ with $c\geq 0$, then $u\almoste{=}0$. + \end{corollary} + \begin{proof} + If $u$ is a weak solution of $\mathcal{D}_0$, then $u$ is a super- (that is, $Lu\geq 0$) and sub-solution (that is $Lu\leq 0$) of $\mathcal{D}_0$. Thus, using \mnameref{INEPDE:weak_max_principle} and \mnameref{INEPDE:weak_min_principle} we conclude that $u\almoste{=}0$. \end{proof} \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 $H^{m+1}(\Omega)$. + 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 $u_f\in H^2(\Omega)\cap H_0^1(\Omega)$ and $f\mapsto u_f$ is a bounded linear operator from $L^2(\Omega)$ to $H^2(\Omega)\cap H_0^1(\Omega)$. If $\Fr{\Omega}\in\mathcal{C}^{m+1}$, $b_j\in\mathcal{C}^{m-1}$ and $f\in H^{m-1}(\Omega)$, then $u_f\in H^{m+1}(\Omega)\cap H_0^1(\Omega)$ and $f\mapsto u_f$ is a bounded linear operator from $H^{m-1}(\Omega)$ to $H^{m+1}(\Omega)\cap H_0^1(\Omega)$. \end{corollary} + \begin{sproof} + \mnameref{INEPDE:fredholm} applied to this problem tells us that either there is a nonzero weak solution to $\mathcal{D}_0$ or $\mathcal{D}_f$ is solvable for all $f\in L^2(\Omega)$. But the first case is impossible by \mcref{INEPDE:corollary_max_min}. + \end{sproof} \begin{theorem} Let $1
0$ such that if $u\in\mathcal{C}^2(\Omega)\cap \mathcal{C}^0(\overline{\Omega})$ solves $Lu=f$, with $f\in\mathcal{C}^{0,\alpha}(\overline{\Omega})$, then $u\in \mathcal{C}^{2,\alpha}(\overline{\Omega})$ and:
+ \begin{theorem}[Schauder estimates]\label{INEPDE:schauder_estimates}
+ Let $\Omega\subset \RR^d$ be open and bounded with $\Fr{\Omega}\in\mathcal{C}^{2,\alpha}$ for some $0<\alpha<1$. In the elliptic operator $L$ assume that $a_{ij},b_j,c\in\mathcal{C}^{0,\alpha}(\overline{\Omega})$. Then, $\exists C>0$ such that if $u\in\mathcal{C}^2(\Omega)\cap \mathcal{C}^0(\overline{\Omega})$ solves $Lu=f$, with $f\in\mathcal{C}^{0,\alpha}(\overline{\Omega})$, then $u\in \mathcal{C}^{2,\alpha}(\overline{\Omega})$ and:
$$
\norm{u}_{\mathcal{C}^{2,\alpha}(\overline{\Omega})}\leq C\left(\norm{f}_{\mathcal{C}^{0,\alpha}(\overline{\Omega})}+\norm{u}_{\mathcal{C}^{1,\alpha}(\overline{\Omega})}\right)
$$
@@ -455,69 +509,182 @@
$$
\end{corollary}
\subsubsection{Maximum and comparison principles}
- \begin{theorem}[Weak maximum principle]
+ \begin{lemma}\label{INEPDE:lemma_matrices_max_principle}
+ If $\vf{A},\vf{B}\in \mathcal{M}_d(\RR)$ are symmetric and $\vf{A}, \vf{B}\geq 0$, then $\tr(\vf{A}\vf{B})\geq 0$.
+ \end{lemma}
+ \begin{theorem}[Weak maximum principle]\label{INEPDE:weak_max_principle_nondiv}
Let $u\in \mathcal{C}^2(\Omega)$ be such that $Lu\leq 0$. Then:
\begin{itemize}
- \item If $c=0$, $\displaystyle \max_{\overline{\Omega}}u=\max_{\partial\Omega}u$.
- \item If $c\geq 0$, $\displaystyle \max_{\overline{\Omega}}u^+\leq \max_{\partial\Omega}u^+$.
+ \item If $c=0$, then $\displaystyle \max_{\overline{\Omega}}u=\max_{\Fr{\Omega}}u$.
+ \item If $c\geq 0$, then $\displaystyle \max_{\overline{\Omega}}u\leq \max_{\Fr{\Omega}}u^+$.
+ \end{itemize}
+ \end{theorem}
+ \begin{proof}
+ Assume first that $Lu<0$ and $c=0$. Suppose $\exists x_0\in \Omega$ such that $u(x_0)=\max_{\overline{\Omega}}u$. Then, $\grad u(x_0)=0$ and $\vf{H}u(x_0)\leq0$, that is, $\sum_{i,j=1}^d\frac{\partial^2u}{\partial x_i\partial x_j}(x_0)p_ip_j\leq 0$ $\forall \vf{p}\in\RR^d$. On the other hand:
+ \begin{align*}
+ \trace(\vf{A}(x_0)\vf{H}u(x_0)) & = \sum_{i,j=1}^d a_{ij}(x_0)\frac{\partial^2u}{\partial x_i\partial x_j}(x_0) \\
+ & =-Lu(x_0)>0
+ \end{align*}
+ The ellipticity of $L$ implies that $\vf{A}>0$, but this is a contradiction with \mcref{INEPDE:lemma_matrices_max_principle} because $\vf{H}u(x_0)\leq 0$. If we now have $c\geq 0$, assume that $\displaystyle \max_{\overline{\Omega}}u^+> \max_{\Fr{\Omega}}u^+$. Then, $\exists x_0\in \Omega$ such that $u(x_0)>0$ and $u(x_0)=\max_{\overline{\Omega}}u^+$. Similarly, we have:
+ \begin{align*}
+ \trace(\vf{A}(x_0)\vf{H}u(x_0)) & = \sum_{i,j=1}^d a_{ij}(x_0)\frac{\partial^2u}{\partial x_i\partial x_j}(x_0) \\
+ & =-Lu(x_0) + c(x_0)u(x_0)\geq 0
+ \end{align*}
+ which again leads to a contraction. Now assume $Lu\leq 0$. Take $u_\varepsilon=u+\varepsilon \exp{\lambda x_1}$, with $\varepsilon>0$ and $\lambda>0$ yet to be chosen. An easy computation shows that:
+ \begin{align*}
+ Lu_\varepsilon & \leq \exp{\lambda x_1}[-\lambda^2a_11+b_1\lambda+c] \\
+ & \leq \exp{\lambda x_1}[-\lambda^2a_11+\norm{\vf{b}}_\infty\lambda+c]<0
+ \end{align*}
+ for $\lambda$ large enough. We do here the case $c=0$ (the other is analogous). From what we have previously seen, $\exists y_\varepsilon\in\Fr{\Omega}$ such that $u(x)\leq u_\varepsilon(x)\leq u_\varepsilon(y_\varepsilon)$. And so we can find a sequence $y_{\varepsilon_n}$ that converges to some $y_0\in\Fr{\Omega}$ (because $\Fr{\Omega}$ is compact) as $\varepsilon_n\to 0$, which implies $u(x)\leq u(y_0)$.
+ \end{proof}
+ \begin{theorem}[Weak minimum principle]
+ Let $u\in \mathcal{C}^2(\Omega)$ be such that $Lu\geq 0$. Then:
+ \begin{itemize}
+ \item If $c=0$, then $\displaystyle \min_{\overline{\Omega}}u=\min_{\Fr{\Omega}}u$.
+ \item If $c\geq 0$, then $\displaystyle \min_{\overline{\Omega}} u \geq -\max_{\Fr{\Omega}}u^-$.
\end{itemize}
\end{theorem}
- \begin{lemma}[Hopf's lemma]\label{INLEPDE:Hopf}
- Let $u\in \mathcal{C}^2(\Omega)$ be such that $Lu\leq 0$ and suppose the region $\Omega$ is connected and that satisfies the \emph{interior ball condition}: for any $x\in \partial\Omega$ there exists $r>0$ and $y\in \Omega$ such that $B(y,r)\subset \Omega$ and $\overline{B(y,r)}\cap \Fr{\Omega}=\{x\}$. Suppose in addition that $c=0$ and $x_0\in\Fr\Omega$ be such that $u(x_0)=\max_{\overline{\Omega}}u$. Then, either $u$ is constant in $\Omega$ or
+ \begin{sproof}
+ Apply \mnameref{INEPDE:weak_max_principle_nondiv} to $-u$ using that ${(-u)}^+=u^-$.
+ \end{sproof}
+ \begin{remark}
+ Nothing can be said if $c<0$. For example, consider $-u''-u=0$, which has $u(x)=\sin(x)$ as a solution, and take $\Omega=(0,\pi)$.
+ \end{remark}
+ \begin{lemma}[Hopf's lemma]\label{INEPDE:Hopf}
+ Let $u\in \mathcal{C}^2(\Omega)$ be such that $Lu\leq 0$ and suppose the region $\Omega$ is connected and that satisfies the \emph{interior ball condition}: for any $x\in \Fr{\Omega}$ there exists $r>0$ and $y\in \Omega$ such that $B(y,r)\subset \Omega$ and $\overline{B(y,r)}\cap \Fr{\Omega}=\{x\}$. Suppose in addition that $c=0$ and $x_0\in\Fr\Omega$ is such that $\displaystyle u(x_0)=\max_{\overline{\Omega}}u$. Then, either $u$ is constant in $\Omega$ or
$$
\liminf_{t\to 0^+}\frac{u(x_0) - u(x_0+t\vf{n})}{t}>0
$$
for any vector $\vf{n}$ of the form $\vf{n}=\frac{x_0-y_0}{\norm{x_0-y_0}}$ with $B(y_0,r)\subset \Omega$ and $\overline{B(y_0,r)}\cap \Fr{\Omega}=\{x_0\}$.
\end{lemma}
\begin{remark}
- In particular, if $\Fr\Omega\in\mathcal{C}^1$ and $u\in \mathcal{C}^1(\overline{\Omega})$, then \mnameref{INLEPDE:Hopf} implies that either $u$ is constant in $\Omega$ or $\partial_{\vf{n}}u(x_0)=\grad u(x_0)\cdot \vf{n}>0$.
+ In particular, if $\Fr\Omega\in\mathcal{C}^1$ and $u\in \mathcal{C}^1(\overline{\Omega})$, then \mnameref{INEPDE:Hopf} implies that either $u$ is constant in $\Omega$ or $\partial_{\vf{n}}u(x_0)=\grad u(x_0)\cdot \vf{n}>0$.
\end{remark}
- \begin{theorem}[Strong maximum principle]
- Let $\Omega\subset\RR^d$ be open, bounded and connected, and $u\in \mathcal{C}^2(\Omega)$ be such that $Lu\leq 0$ with $c=0$. If $\exists x_0\in\Omega$ such that $u(x_0)\geq u(x)$ $\forall x\in\Omega$, then $u$ is constant in $\Omega$. Similarly, if $c\geq 0$ and $\exists x_0\in\Omega$ such that $u(x_0)\geq 0$ and $u(x_0)\geq u(x)$ $\forall x\in\Omega$, then $u$ is constant in $\Omega$.
+ \begin{theorem}[Strong maximum principle]\label{INEPDE:strong_max_principle}
+ Let $\Omega\subset\RR^d$ be open, bounded and connected, and $u\in \mathcal{C}^2(\Omega)$ be such that $Lu\leq 0$. Then:
+ \begin{enumerate}
+ \item If $c=0$ and $\exists x_0\in\Omega$ such that $u(x_0)\geq u(x)$ $\forall x\in\Omega$, then $u=\const$ in $\Omega$.
+ \item If $c\geq 0$ and $\exists x_0\in\Omega$ such that $u(x_0)\geq 0$ and $u(x_0)\geq u(x)$ $\forall x\in\Omega$, then $u=\const$ in $\Omega$.
+ \end{enumerate}
\end{theorem}
- \begin{theorem}
- Suppose that $c\geq 0$ and $u\in \mathcal{C}^2(\Omega)$ is a solution to
+ \begin{proof}
+ Assume $c=0$, the other case is similar. Let $\displaystyle M=\max_{\overline{\Omega}}u$, $C:= \{u=M\}$ and $V:=\{u