From 90c749b14876b59bdc6c8a68a469098807400aa8 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Fri, 19 Jan 2024 21:26:28 +0100 Subject: [PATCH] updated small typos pdes --- ...topics_in_functional_analysis_and_PDEs.tex | 8 ++-- ...ntroduction_to_nonlinear_elliptic_PDEs.tex | 45 ++++++++++--------- 2 files changed, 27 insertions(+), 26 deletions(-) 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 77bd925..6bbff84 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 @@ -13,18 +13,18 @@ We denote this by $x_n\rightharpoonup x$. \end{definition} \begin{definition} - Let $E$ be a Banach space and ${(x_n)}_{n\in\NN}\in E$. We say that ${(x_n)}_{n\in\NN}$ \emph{converges strongly} to $x\in E$ if $\forall f\in E^*$ we have: + Let $E$ be a Banach space and ${(x_n)}_{n\in\NN}\in E$. We say that ${(x_n)}_{n\in\NN}$ \emph{converges strongly} to $x\in E$ if we have: $$ \lim_{n\to\infty}{\norm{x_n-x}}=0 $$ We denote this by $x_n\to x$. \end{definition} \begin{definition} - Let $E$ be a Banach space and ${(x_n)}_{n\in\NN}\in E$. Assume $E=F^*$, where $F$ is a Banach space. Then, we say that ${(x_n)}_{n\in\NN}$ \emph{converges weakly-*} to $x\in E$ if $\forall f\in F$ we have: + Let $E$ be a Banach space and ${(L_n)}_{n\in\NN}\in E$. Assume $E=F^*$, where $F$ is a Banach space. Then, we say that ${(L_n)}_{n\in\NN}$ \emph{converges weakly-*} to $L\in E$ if $\forall x\in F$ we have: $$ - \lim_{n\to\infty}{x_n(f)}=x(f) + \lim_{n\to\infty}{L_n(x)}=L(x) $$ - We denote this by $x_n\overset{*}\rightharpoonup x$. + We denote this by $L_n\overset{*}\rightharpoonup L$. \end{definition} \begin{theorem} Let $E$ be a Banach space and ${(x_n)}_{n\in\NN}\in E$. Then: 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 ba3d635..9425046 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 @@ -121,13 +121,13 @@ & \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}. + by the uniform ellipticity of $L$. \end{enumerate} 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}. + by the \mnameref{ATFAPDE:poincare_ineq}. \end{proof} \subsubsection{Abstract Fredholm alternative} \begin{remark} @@ -155,7 +155,7 @@ $$ a_\mu(u,u)\geq \theta \norm{u}_{H_0^1(\Omega)}^2-C\norm{u}_{H_0^1(\Omega)}\norm{u}_2 + \mu \norm{u}_2^2 $$ - which is for $\mu$ large enough it is bigger than $\delta \norm{u}_{H_0^1(\Omega)}^2$ for some $\delta>0$. + which 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, $\dim \ker(\id-K)<\infty$. @@ -203,7 +203,7 @@ \begin{proof} \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 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 because $K^{**}=K$ 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 $\ker (\id-K)=\{0\}$. Now, 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 $\norm{Ku_n}^2=\norm{u_n-Ku_n}^2+\norm{u_n}^2\geq 1$, 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 $\ker(\id-K)=\{0\}$, 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 $v=(\id-K)u$. @@ -219,7 +219,7 @@ \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} + \begin{proposition}\label{INEPDE:adjoint_im_ker} The \emph{homogeneous adjoint problem} $$ \mathcal{D}_0^*:=\begin{cases} @@ -238,7 +238,7 @@ $$ f= (L_{\mu_0}-\mu_0){L_{\mu_0}}^{-1}w=w-\mu_0 {L_{\mu_0}}^{-1}w=(\id-K)w $$ - with $K=\mu_0 {L_{\mu_0}}^{-1}$. We claim that $K:L^2(\Omega)\to L^2(\Omega)$ is compact. Note that $K=\iota_{H_0^1\hookrightarrow L^2}\circ {L_{\mu_0}}^{-1}\circ \iota_{L^2\hookrightarrow H^{-1}}$, so since ${L_{\mu_0}}^{-1}$ and $\iota_{L^2\hookrightarrow H^{-1}}$ are bounded, and we have a compact embedding $H_0^1\hookrightarrow L^2$, we have that $K$ is compact. Finally, one can check that: + with $K=\mu_0 {L_{\mu_0}}^{-1}$. We claim that $K:L^2(\Omega)\to L^2(\Omega)$ is compact. Note that $K=\mu_0\iota_{H_0^1\hookrightarrow L^2}\circ {L_{\mu_0}}^{-1}\circ \iota_{L^2\hookrightarrow H^{-1}}$, so since ${L_{\mu_0}}^{-1}$ and $\iota_{L^2\hookrightarrow H^{-1}}$ are bounded, and we have a compact embedding $H_0^1\hookrightarrow L^2$, we have that $K$ is compact. Finally, one can check that: $$ \id-K^*={(\id-K)}^*={\left(L_0{L_{\mu_0}}^{-1}\right)}^*={({L_{\mu_0}}^*)}^{-1}{L_0}^{*} $$ @@ -426,20 +426,20 @@ \begin{proof} 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} \\ + 0 & \leq \theta \norm{\grad v_m}_{L^2}^2-d\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: + 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}} + \norm{\grad v_{m_0-\varepsilon}}_{L^2}\leq C \delta {S_\varepsilon}^{1-\frac{2}{2^*}}\norm{\grad v_{m_0-\varepsilon}}_{L^2}\leq \frac{1}{2}\norm{\grad v_{m_0-\varepsilon}}_{L^2} $$ - 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$. + by choosing $\varepsilon$ small enough. Thus, $\norm{\grad v_{m_0-\varepsilon}}_{L^2}=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: @@ -462,7 +462,7 @@ 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}. + \mnameref{INEPDE:fredholm} applied to this problem (check \mcref{INEPDE:adjoint_im_ker}) 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 $10$ 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 + Let $u\in \mathcal{C}^2(\Omega)$ be such that $Lu\leq 0$ and suppose that 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 $$ @@ -614,34 +614,35 @@ $$ \begin{cases} Lu=f & \text{in }\Omega \\ - u|_{\Fr{\Omega}}=0 & \text{on }\Fr{\Omega} + u|_{\Fr{\Omega}}=h & \text{on }\Fr{\Omega} \end{cases} $$ with $f,a_{ij},b_j,c\in\mathcal{C}^{0,\alpha}( \overline{\Omega})$ and $h\in\mathcal{C}^{0,\alpha}(\Fr{\Omega})$. Then, there exists a solution to this problem in $\mathcal{C}^{2,\alpha}(\overline{\Omega})$. \end{theorem} \begin{proof} + We will do it for $h=0$. Let $t\in [0,1]$ and consider the problem: $$ \mathcal{D}_t:= \begin{cases} - L_tu=f & \text{in }\Omega \\ - u|_{\Fr{\Omega}}=ht & \text{on }\Fr{\Omega} + L_tu=f & \text{in }\Omega \\ + u|_{\Fr{\Omega}}=0 & \text{on }\Fr{\Omega} \end{cases}$$ - with $L_t=tL-(1-t)\laplacian$. We know that $\mathcal{D}_0$ has a unique weak solution $u_0\in H_0^1(\Omega)$. The idea of the Continuation method is that if $\mathcal{D}_t$ is solvable for all $f$, then for $h>0$ small enough, $\mathcal{D}_{t+h}$ is solvable for all $f$ too. We will do it for $h=0$. Rewrite $\mathcal{D}_{t+h}$ as: + with $L_t=tL-(1-t)\laplacian$. We know that $\mathcal{D}_0$ has a unique weak solution $u_0\in H_0^1(\Omega)$. The idea of the continuation method is that if $\mathcal{D}_t$ is solvable for all $f$, then for $k>0$ small enough, $\mathcal{D}_{t+k}$ is solvable for all $f$ too. Rewrite $\mathcal{D}_{t+k}$ as: $$ \begin{cases} - L_{t}u=f-h(L+\laplacian)u & \text{in }\Omega \\ + L_{t}u=f-k(L+\laplacian)u & \text{in }\Omega \\ u|_{\Fr{\Omega}}=0 & \text{on }\Fr{\Omega} \end{cases} $$ We need to solve the fixed point problem $u=\phi(u)$, with $$ - \function{\phi}{\mathcal{C}^{2,\alpha}}{\mathcal{C}^{2,\alpha}}{u}{\phi(u)={L_t}^{-1}f - h{L_t}^{-1}(L+\laplacian)u} + \function{\phi}{\mathcal{C}^{2,\alpha}}{\mathcal{C}^{2,\alpha}}{u}{\phi(u)={L_t}^{-1}f - k{L_t}^{-1}(L+\laplacian)u} $$ - From \mnameref{INEPDE:schauder_estimates} and \mnameref{INEPDE:apriori} we deduce that $\norm{u}_{\mathcal{C}^{2,\alpha}(\overline{\Omega})}\leq C\norm{f}_{\mathcal{C}^{0,\alpha}(\overline{\Omega})}$. So $\forall \varphi\in \mathcal{C}^{0,\alpha}(\overline{\Omega})$ we have $\norm{{L_t}^{-1}\varphi}_{\mathcal{C}^{2,\alpha}(\overline{\Omega})}\leq C\norm{\varphi}_{\mathcal{C}^{0,\alpha}(\overline{\Omega})}$ (it can be seen that the constant does not depend on $t$). We will show that $\phi$ is a contraction for $h$ small enough. Let $u,v\in \mathcal{C}^{2,\alpha}(\overline{\Omega})$. Then: + From \mnameref{INEPDE:schauder_estimates} and \mnameref{INEPDE:apriori} we deduce that $\norm{u}_{\mathcal{C}^{2,\alpha}(\overline{\Omega})}\leq C\norm{f}_{\mathcal{C}^{0,\alpha}(\overline{\Omega})}$. So $\forall \varphi\in \mathcal{C}^{0,\alpha}(\overline{\Omega})$ we have $\norm{{L_t}^{-1}\varphi}_{\mathcal{C}^{2,\alpha}(\overline{\Omega})}\leq C\norm{\varphi}_{\mathcal{C}^{0,\alpha}(\overline{\Omega})}$ (it can be seen that the constant does not depend on $t$). We will show that $\phi$ is a contraction for $k$ small enough. Let $u,v\in \mathcal{C}^{2,\alpha}(\overline{\Omega})$. Then: \begin{align*} - \norm{\phi(u)-\phi(v)}_{\mathcal{C}^{2,\alpha}(\overline{\Omega})} & \leq Ch\norm{(L+\laplacian)(u-v)}_{\mathcal{C}^{0,\alpha}(\overline{\Omega})} \\ - & \leq \tilde{C}h\norm{u-v}_{\mathcal{C}^{2,\alpha}(\overline{\Omega})} + \norm{\phi(u)-\phi(v)}_{\mathcal{C}^{2,\alpha}(\overline{\Omega})} & \leq Ck\norm{(L+\laplacian)(u-v)}_{\mathcal{C}^{0,\alpha}(\overline{\Omega})} \\ + & \leq \tilde{C}k\norm{u-v}_{\mathcal{C}^{2,\alpha}(\overline{\Omega})} \end{align*} - So take $h\leq \frac{1}{2\tilde{C}}$. Repeating this argument a finite number of times ($\tilde{C}$ does not depend on $t$) we conclude that $\mathcal{D}_1$ is solvable. + So take $k\leq \frac{1}{2\tilde{C}}$. Repeating this argument a finite number of times ($\tilde{C}$ does not depend on $t$) we conclude that $\mathcal{D}_1$ is solvable. \end{proof} \subsection{Existence theorems for nonlinear elliptic PDEs by fixed point methods} In this section we will mostly consider almost linear elliptic PDEs of the form: