Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

1 Answer 1

up vote 0 down vote accepted

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 $$|\nabla y\cdot \nabla\left(1+|\nabla y|^2\right)^{-3/2}|\leq \left(1+|\nabla y|^2\right)^{-3/2}|\nabla y|^2\sqrt n|\nabla^2y|\leq \sqrt n|\nabla^2y|\left(1+|\nabla y|^2\right)^{-1/2}\leq \sqrt n|\nabla^2y|.$$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.