We now consider another kind of problem involving biharmonic operator. Let us assume 
in 
Theorem. The following identity
holds.
Ngô Quốc Anh
![\displaystyle\begin{gathered} \frac{3}{2}\left[ {\int_{{B_r}(0)} {uQ(x)f(u)dx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } } \right] + \hfill \\ \frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma } + \int_{{B_r}(0)} {(x\cdot\nabla (\Delta u))\Delta udx} - \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x\cdot\nabla u)}}{{\partial \nu }} - (x\cdot\nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}} \right]d\sigma } \hfill \\ \qquad\qquad= \int_{\partial {B_r}(0)} {ruQ(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left( {Q(x)f(u) + (x\cdot\nabla Q)f(u) + (x\cdot\nabla u)Q(x)f'(u)} \right)dx}\hfill \\ \end{gathered}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Bgathered%7D+%5Cfrac%7B3%7D%7B2%7D%5Cleft%5B+%7B%5Cint_%7B%7BB_r%7D%280%29%7D++%7BuQ%28x%29f%28u%29dx%7D+-+%5Cint_%7B%5Cpartial+%7BB_r%7D%280%29%7D+%7Bu%5Cfrac%7B%7B%5Cpartial+%5CDelta++u%7D%7D%7B%7B%5Cpartial+%5Cnu+%7D%7Dd%5Csigma+%7D+%2B+%5Cint_%7B%5Cpartial+%7BB_r%7D%280%29%7D++%7B%5Cfrac%7B%7B%5Cpartial+u%7D%7D%7B%7B%5Cpartial+%5Cnu+%7D%7D%5CDelta+ud%5Csigma+%7D+%7D+%5Cright%5D+%2B++%5Chfill+%5C%5C+%5Cfrac%7B1%7D%7B2%7D%5Cint_%7B%5Cpartial+%7BB_r%7D%280%29%7D+%7Br%7B%7B%28%5CDelta+u%29%7D%5E2%7Dd%5Csigma+%7D++%2B+%5Cint_%7B%7BB_r%7D%280%29%7D+%7B%28x%5Ccdot%5Cnabla+%28%5CDelta+u%29%29%5CDelta+udx%7D+-++%5Cint_%7B%5Cpartial+%7BB_r%7D%280%29%7D+%7B%5Cleft%5B+%7B%5CDelta+u%5Cfrac%7B%7B%5Cpartial+%28x%5Ccdot%5Cnabla++u%29%7D%7D%7B%7B%5Cpartial+%5Cnu+%7D%7D+-+%28x%5Ccdot%5Cnabla+u%29%5Cfrac%7B%7B%5Cpartial+%5CDelta++u%7D%7D%7B%7B%5Cpartial+%5Cnu+%7D%7D%7D+%5Cright%5Dd%5Csigma+%7D+%5Chfill+%5C%5C+%5Cqquad%5Cqquad%3D++%5Cint_%7B%5Cpartial+%7BB_r%7D%280%29%7D+%7BruQ%28x%29f%28u%29d%5Csigma+%7D+-+%5Cint_%7B%7BB_r%7D%280%29%7D+%7Bu%5Cleft%28++%7BQ%28x%29f%28u%29+%2B+%28x%5Ccdot%5Cnabla+Q%29f%28u%29+%2B+%28x%5Ccdot%5Cnabla+u%29Q%28x%29f%27%28u%29%7D++%5Cright%29dx%7D%5Chfill+%5C%5C+%5Cend%7Bgathered%7D&bg=ffffff&fg=333333&s=0 "\displaystyle\begin{gathered} \frac{3}{2}\left[ {\int_{{B_r}(0)} {uQ(x)f(u)dx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } } \right] + \hfill \\ \frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma } + \int_{{B_r}(0)} {(x\cdot\nabla (\Delta u))\Delta udx} - \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x\cdot\nabla u)}}{{\partial \nu }} - (x\cdot\nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}} \right]d\sigma } \hfill \\ \qquad\qquad= \int_{\partial {B_r}(0)} {ruQ(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left( {Q(x)f(u) + (x\cdot\nabla Q)f(u) + (x\cdot\nabla u)Q(x)f'(u)} \right)dx}\hfill \\ \end{gathered}")
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