principiu tare de maxim; Alexandrov-Bakelman
Posted: Tue Mar 09, 2010 12:42 pm
Presupunem ca \( u\in C^2(\Omega)\cap C^0(\overline{\Omega}) \) satisface
\( Lu(x):=\sum_{i, j=1}^{d}a_{ij}(x)\frac{\partial u}{\partial x_{i}\partial x_{j}}\geq f(x), \)
unde matricea \( a_{ij}(x) \) este pozitiv definita, simetrica pentru orice \( x\in\Omega \). Mai mult, presupunem ca
\( \int_{\Omega}\frac{|f(x)|^d}{\det (a_{ij}(x)}dx<\infty. \)
Atunci are loc
\( \sup_{\Omega}u\leq\max_{\partial\Omega}u+\frac{diam(\Omega)}{d\omega_d^{1/d}}\left(\int_{\Omega}\frac{|f(x)|^d}{\det (a_{ij}(x)}dx\right)^{1/d} \).
\( Lu(x):=\sum_{i, j=1}^{d}a_{ij}(x)\frac{\partial u}{\partial x_{i}\partial x_{j}}\geq f(x), \)
unde matricea \( a_{ij}(x) \) este pozitiv definita, simetrica pentru orice \( x\in\Omega \). Mai mult, presupunem ca
\( \int_{\Omega}\frac{|f(x)|^d}{\det (a_{ij}(x)}dx<\infty. \)
Atunci are loc
\( \sup_{\Omega}u\leq\max_{\partial\Omega}u+\frac{diam(\Omega)}{d\omega_d^{1/d}}\left(\int_{\Omega}\frac{|f(x)|^d}{\det (a_{ij}(x)}dx\right)^{1/d} \).