From 4c2ea119a5b5042cd4fd5b2ee8f0ebe30808276f Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Wed, 3 Jan 2024 11:49:15 +0100 Subject: [PATCH] updated mor pdes --- ...ntroduction_to_nonlinear_elliptic_PDEs.tex | 37 ++++++++++++------- 1 file changed, 23 insertions(+), 14 deletions(-) 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 8a32197..c8efd7f 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 @@ -763,18 +763,27 @@ $$ is compact. Thus, a Hilbert basis $(u_n)$ of $K$ with $Ku_n=\mu_n u_n$, $\mu_n>0$ with $\mu_n\to 0$ as $n\to\infty$ exists (we may assume $\mu_n\searrow 0$). Thus, $Lu_n=\lambda_nu_n$ with $\lambda_n=\frac{1}{\mu_n}\nearrow+\infty$. Then: $$ - \lambda_1=\min_{\substack{u\in H_0^1(\Omega) \\ \norm{u}_{L^2(\Omega)}=1}}\beta(u,u)=\min_{u\in H_0^1(\Omega)\setminus\{0\}}\frac{\beta(u,u)}{\norm{u}_{L^2(\Omega)}^2} + \lambda_1=\!\!\min_{\substack{u\in H_0^1(\Omega) \\ \norm{u}_{L^2(\Omega)}=1}}\!\!{\langle Lu,u\rangle}_{H^{-1}\times H_0^1}=\!\!\min_{u\in H_0^1(\Omega)\setminus\{0\}}\!\!\frac{{\langle Lu,u\rangle}_{H^{-1}\times H_0^1}}{\norm{u}_{L^2(\Omega)}^2} $$ And: \begin{align*} - \lambda_n & =\min_{\substack{u\in H_0^1(\Omega) \\u\in\langle u_1,\dots,u_{n-1}\rangle^{\perp_{L^2}}\\ \norm{u}_{L^2(\Omega)}=1}}\beta(u,u)\\ - & = \min_{\substack{u\in H_0^1(\Omega)\setminus\{0\} \\ u\in\langle u_1,\dots,u_{n-1}\rangle^{\perp_{L^2}}}}\frac{\beta(u,u)}{\norm{u}_{L^2(\Omega)}^2}\\ - & =\min_{\substack{V\text{ subspace of }H_0^1(\Omega) \\ \dim(V)=n}}\max_{\substack{u\in V\setminus\{0\} \\ \norm{u}_{L^2(\Omega)}=1}}\beta(u,u)\\ - & = \max_{\substack{W\text{ subspace of } H_0^1(\Omega) \\ \codim(W)=n-1}}\min_{\substack{u\in W\setminus\{0\} \\ \norm{u}_{L^2(\Omega)}=1}}\beta(u,u) + \lambda_k & =\min_{\substack{u\in H_0^1(\Omega) \\u\in\langle u_1,\dots,u_{k-1}\rangle^{\perp_{L^2}}\\ \norm{u}_{L^2(\Omega)}=1}}{\langle Lu,u\rangle}_{H^{-1}\times H_0^1}\\ + & = \min_{\substack{u\in H_0^1(\Omega)\setminus\{0\} \\ u\in\langle u_1,\dots,u_{k-1}\rangle^{\perp_{L^2}}}}\frac{{\langle Lu,u\rangle}_{H^{-1}\times H_0^1}}{\norm{u}_{L^2(\Omega)}^2}\\ + & =\min_{\substack{V\text{ subspace of }H_0^1(\Omega) \\ \dim(V)=k}}\max_{\substack{u\in V\setminus\{0\} \\ \norm{u}_{L^2(\Omega)}=1}}{\langle Lu,u\rangle}_{H^{-1}\times H_0^1}\\ + & = \max_{\substack{W\text{ subspace of } H_0^1(\Omega) \\ \codim(W)=k-1}}\min_{\substack{u\in W\setminus\{0\} \\ \norm{u}_{L^2(\Omega)}=1}}{\langle Lu,u\rangle}_{H^{-1}\times H_0^1} \end{align*} \end{theorem} \begin{proof} - TODO + We only prove some of them. + Recall that $H_0^1=\overline{\bigoplus_{n\in\NN} \langle u_n\rangle}^{{}_{H_0^1}}$ and $L^2=\overline{\bigoplus_{n\in\NN} \langle u_n\rangle}^{{}_{L^2}}$. Take $u\in H_0^1(\Omega)\setminus\{0\}$ and write $u=\sum_{n\in\NN}\alpha_n u_n$, which converges in both $L^2$ and $H_0^1$. We have: + $$ + {\langle Lu,u\rangle}_{H^{-1}\times H_0^1}= \sum_{n\in\NN}\lambda_n{\alpha_n}^2\geq \lambda_1\sum_{n\in\NN}{\alpha_n}^2=\lambda_1\norm{u}_{L^2(\Omega)}^2 + $$ + So the first equality holds since the lower bound is attained by $u=u_1$. Now take $u\perp_{L^2}\langle u_1,\dots,u_{k-1}\rangle$. Then, $\alpha_1=\dots=\alpha_{k-1}=0$ and so: + $$ + {\langle Lu,u\rangle}_{H^{-1}\times H_0^1}= \sum_{n\geq n}\lambda_n{\alpha_n}^2\geq \lambda_n\sum_{n\geq n}{\alpha_n}^2=\lambda_n\norm{u}_{L^2(\Omega)}^2 + $$ + So the third equality holds since the lower bound is attained by $u=u_k$. \end{proof} \subsubsection{Nonlinear case without constraints} \begin{definition} @@ -987,7 +996,7 @@ In general, it suffices to have $f:\Omega\times \RR\to\RR$ Carathéodory with: \begin{itemize} \item $\abs{f(x,t)}\leq C(1+\abs{t}^\theta)$ $\forall (x,t)\in \Omega\times \RR$ with $1\leq \theta\leq \frac{d+2}{d-2}$ (if $d\geq 3$) and $1\leq \theta<\infty$ (if $d=2$). - \item $f(x,t)t\geq C'\min_{\alpha,\beta>0}\{\abs{t}^\alpha,\abs{t}^\beta\}$ $\forall (x,t)\in \Omega\times \RR$, $2\leq \alpha\leq \theta+1$. + \item $\displaystyle f(x,t)t\geq C'\min_{2\leq \alpha,\beta\leq \theta+1}\{\abs{t}^\alpha,\abs{t}^\beta\}$ $\forall (x,t)\in \Omega\times \RR$. \end{itemize} \end{remark} \begin{remark} @@ -1044,14 +1053,14 @@ TODO \end{proof} \begin{corollary}[Mountain pass theorem] - Let $E$ be a Banach space and $I\in\mathcal{C}^1(E,\RR)$ satisfying the Palais-Smale condition at level $c\in\RR$. Assume that $\exists a\ne b\in E$ such that + Let $E$ be a Banach space and $I\in\mathcal{C}^1(E,\RR)$. Assume that $\exists a\ne b\in E$ such that $$ c:=\inf_{\gamma\in \Gamma}\max_{t\in [0,1]}I(\gamma(t))>\max\{I(a),I(b)\} $$ with $$ \Gamma:=\{\gamma\in \mathcal{C}([0,1],E):\gamma(0)=a,\gamma(1)=b\} $$ - Then, $\exists u_*\in E$ such that $I(u_*)=c$ and $\dd{I(u_*)}=0$. + If, moreover, $I$ satisfies the Palais-Smale condition at level $c$, then $\exists u_*\in E$ such that $I(u_*)=c$ and $\dd{I(u_*)}=0$. \end{corollary} \begin{proposition} Let $f:\Omega\times \RR\to\RR$ be Carathéodory satisfying: @@ -1067,11 +1076,11 @@ Then, ??????????? \end{proposition} \begin{proof} - From the superquadradicity condition, for $t>0$, the function $\abs{t}^{-p}F(x,t)$ is nondecreasing (the derivative is nonnegative). So, for $0\leq t\leq 1$ we have $F(x,t)\leq \abs{t}^p F(x,1)$. Similarly, for $-1\leq t\leq 0$ we have $F(x,t)\leq \abs{t}^p F(x,-1)$. Using the upper estimate we get, for $\abs{t}\geq 1$, $\abs{F(x,t)}\leq \overline{\overline{C}} \abs{t}^{p_1}$ and so - $\abs{F(x,t)}\leq C'(\abs{t}^p+\abs{t}^{p_1})$ $\forall t$ (the positivity of $F(x,1)$ and $F(x,-1)$ follows from the third hypothesis on $f$). So: - % $$ - % \int_\Omega \abs{F(x,u)}\leq C'\left(\norm{u}_{ - % $$ + First of all, note that the thrid hypothesis on $f$ implies that $F(x,t)\geq 0$ $\forall (x,t)\in \Omega\times \RR$. From the superquadradicity condition, for $t>0$, the function $\abs{t}^{-p}F(x,t)$ is nondecreasing (the derivative is nonnegative). So, for $0\leq t\leq 1$ we have $F(x,t)\leq \abs{t}^p F(x,1)$. Similarly, for $-1\leq t\leq 0$ the function is nonincreasing and so we have $F(x,t)\leq \abs{t}^p F(x,-1)$. Using the upper estimate we get, for $\abs{t}\geq 1$, $\abs{F(x,t)}\leq \overline{\overline{C}} \abs{t}^{p_1}$ and so + $\abs{F(x,t)}\leq C'(\abs{t}^p+\abs{t}^{p_1})$ $\forall t$. So: + $$ + \int_\Omega \abs{F(x,u)}\leq C'\left(\norm{u}_{L^{p}}^p+\norm{u}_{L^{p_1}}^{p_1}\right)\leq C''\left(\norm{\grad u}_{L^2}^p+\norm{\grad u}_{L^2}^{p_1}\right) + $$ \end{proof} \end{multicols} \end{document} \ No newline at end of file