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I want to prove that

$ \Delta y \left ( \frac{1}{\sqrt{1+ |\nabla y|^2}}-1 \right) + \nabla y \cdot \nabla \frac{1}{\sqrt{1 + |\nabla y|^2}} = \mathcal O ((| \nabla y| + |\nabla^2 y|)^3) $ as $ | \nabla y| + |\nabla^2 y| \to 0$.

Here, $ \nabla = \left ( \frac{\partial}{\partial x_1}, \cdots , \frac{\partial}{\partial x_n} \right) $ $ \nabla^j = \left ( \frac{\partial^j}{\partial x_1^j}, \cdots , \frac{\partial^j}{\partial x_n^j} \right) \mbox{ for } j \in \mathbb N $ $ \Delta = \sum_{j=1}^n \frac{\partial^2}{\partial x_j^2}.$

Please help me!

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    @DavideGiraudo I derived for the first term's result that $|\Delta y| |\frac{1}{\sqrt{1+|\nabla y|^2}} -1 | =|\Delta y| | \frac{1 - \sqrt{1+ | \nabla y|^2}}{\sqrt{1+|\nabla y|^2}}| = |\Delta y| | \frac{| \nabla y|^2}{\sqrt{1 + | \nabla y|^2} ( 1 + \sqrt{ 1 + |\nabla y|^2})}| \leqslant |\Delta y| |\nabla y|^2 $2012-04-15

1 Answers 1

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We have $|\Delta y|=\left|\sum_{j=1}^n\partial_{jj}y\right|=\left|\sum_{j=1}^n(\nabla^2y)_j\right|\leq \sum_{j=1}^n|(\nabla^2y)_j|\leq \sqrt n|\nabla^2y|$ so $|\Delta y|\left|1-\frac 1{\sqrt{1+|\nabla y|^2}}\right|\leq \sqrt n|\nabla^2y|\frac{|\nabla y|^2}{\sqrt{1+|\nabla y|^2}(1+\sqrt{1+|\nabla y|^2})}\leq \sqrt n|\nabla^2y|.$ For the second term $\partial_j\left(1+|\nabla y|^2\right)^{-1/2}=-\frac 12\cdot 2\partial_{jj}y\cdot\partial_j y\left(1+|\nabla y|^2\right)^{-3/2}=-\partial_{jj}y\cdot\partial_j y\left(1+|\nabla y|^2\right)^{-3/2}$ hence $\nabla y\cdot \nabla\left(1+|\nabla y|^2\right)^{-3/2}=-\sum_{j=1}^n\partial_{jj}y(\partial_jy)^2\left(1+|\nabla y|^2\right)^{-3/2}$ and $|\nabla y\cdot \nabla\left(1+|\nabla y|^2\right)^{-3/2}|\leq \left(1+|\nabla y|^2\right)^{-3/2}|\nabla y|^2\sqrt n|\nabla^2y|\leq \sqrt n|\nabla^2y|\left(1+|\nabla y|^2\right)^{-1/2}\leq \sqrt n|\nabla^2y|.$

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    You are right, I made an error. We get mixed derivatives, so it's more complicated.2012-04-15