# Big O function of multi variables

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

 Is the first term the Laplacian of the function $\frac{y}{\sqrt{1+|\nabla y|^2}}$? – Davide Giraudo Apr 15 '12 at 12:21 @DavideGiraudo The first term means that $\frac{\Delta y}{\sqrt{1+ |\nabla y|^2}} - \Delta y$ Here, $\Delta$ is used for the Laplacian. :) – Misaj Apr 15 '12 at 12:41 Ok, indeed it seems more logical, otherwise the $-1$ would have been useless. – Davide Giraudo Apr 15 '12 at 12:47 @DavideGiraudo I derived for the first term's result that $|\Delta y| |\frac{1}{\sqrt{1+|\nabla y|^2}} -1 | =|\Delta y| | \frac{1 - \sqrt{1+ | \nabla y|^2}}{\sqrt{1+|\nabla y|^2}}| = |\Delta y| | \frac{| \nabla y|^2}{\sqrt{1 + | \nabla y|^2} ( 1 + \sqrt{ 1 + |\nabla y|^2})}| \leqslant |\Delta y| |\nabla y|^2$ – Misaj Apr 15 '12 at 13:00
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 $$|\nabla y\cdot \nabla\left(1+|\nabla y|^2\right)^{-3/2}|\leq \left(1+|\nabla y|^2\right)^{-3/2}|\nabla y|^2\sqrt n|\nabla^2y|\leq \sqrt n|\nabla^2y|\left(1+|\nabla y|^2\right)^{-1/2}\leq \sqrt n|\nabla^2y|.$$
 Then this is sufficient to conclude that $\mathcal O (| \nabla^2 y|)..?$ – Misaj Apr 15 '12 at 13:33 If you have a time, would you solve the multivariable integral problem linked in math.stackexchange.com/questions/132026/… ? – Misaj Apr 15 '12 at 13:47 I think you have a mistake in $\partial_j\left(1+|\nabla y|^2\right)^{-1/2}=-\frac 12\cdot 2\partial_{jj}y\cdot\partial_j y\left(1+|\nabla y|^2\right)^{-3/2}$ – Misaj Apr 15 '12 at 15:06 I think $\partial_j\left(1+|\nabla y|^2\right)^{-1/2}= -\partial_j (\partial_1 y + \cdots + \partial_n y) \left(1+|\nabla y|^2\right)^{-3/2}$isn't it? – Misaj Apr 15 '12 at 15:09 because $\partial_j | \nabla y|^2 = \partial_j ( (\partial_1 y)^2 + \cdots + (\partial_n y)^2 )$. – Misaj Apr 15 '12 at 15:11