1
$\begingroup$

Problem:

If $u\in H^2(R^n)$, how to prove $\|D^2u\|_{L^2}$ is equal to $\|\Delta u\|_{L^2}$ using Fourier transforms?

My first question: Is it right to prove this using integration by parts as follows? $$\|\Delta u\|_{L^2}^2=\int\Delta u \Delta u dx=\sum_{i,j=1}^{n}\int \partial_i \partial_i u \partial_j \partial_j u dx=\sum_{i,j=1}^{n}\int \partial_i \partial_j u \partial_i \partial_j u dx=\|D^2u\|_{L^2}$$ My second question: How to prove this using Fourier transform?

I use $$\hat u(\xi) =\int e^{-ix\xi}u(x)$$ as the definition of Fourier transform. $$\partial_ju = i\xi_j\hat{u}$$. Then $$\partial_k\partial_ju = i^2\xi_k\xi_j\hat{u} = \frac{\xi_k}{|\xi|}\frac{\xi_j}{|\xi|}(i^2|\xi|^2\hat{u}) = \frac{\xi_k}{|\xi|}\frac{\xi_j}{|\xi|} \widehat{\Delta u}.$$

To conlude take the $L^2$ norm on both sides and use the obvious inequality $\frac{\xi_j}{|\xi|} \le 1$.

If I take the $L^2$ norm on both sides then I get $\sum_{k,j} \int | \widehat {\partial_k \partial_j u} |^2=\sum_{k,j}\int (\frac{\xi_k \xi_j}{|\xi|^2}\widehat{\Delta u})^2$. Then how can I conclude $\|D^2u\|_{L^2} \le \|\Delta u\|_{L^2}$?

Could anyone kindly help? Thanks very much!

$\endgroup$
12
  • $\begingroup$ What is $D$ here? $\endgroup$ Mar 29 '15 at 21:07
  • 1
    $\begingroup$ The approach with Fourier transform looks basically like the approach with integration by parts: when you take the Laplacian you wind up multiplying in Fourier space by $|\xi|^2$, now rearrange the product of the sums into a sum of products so that it looks more like the Fourier result of $D^2u$. $\endgroup$
    – Ian
    Mar 29 '15 at 21:07
  • $\begingroup$ @CameronWilliams $D$ is (possibly weak) differentiation. Sherry is probably working out of Evans PDE. $\endgroup$
    – Ian
    Mar 29 '15 at 21:08
  • 1
    $\begingroup$ @user01123581321345589144... Sorry about that. I also got two downvotes. Someone seems is downvoting all questions. $\endgroup$
    – Sherry
    Mar 30 '15 at 1:33
  • 1
    $\begingroup$ It is bad form to edit your post this heavily. If you have another question, please make another post and ask it there. Do not add onto your question after receiving a legitimate answer. This is likely why you are being downvoted. $\endgroup$ Mar 30 '15 at 3:10
2
$\begingroup$

By definition $$\hat u(\xi)=\int e^{-ix\xi}u(x)dx$$ We also know $\widehat{\partial^ku(\xi)}=i^{|k|}\xi^k\widehat{u(\xi)}$ Similar to the integration by part step posted in my question, we can get $$\int(\Delta u)^2=\sum_{i,j=1}^n\int(\widehat {u_{x_i x_i}} \widehat{u_{x_j x_j}})=\sum_{i,j=1}^n\int (-\xi_i \xi_i \hat u)( -\xi_j \xi_j \hat u)=\sum_{i,j=1}^n\int (-\xi_i \xi_j \hat u)( -\xi_i \xi_j \hat u)=\sum_{i,j=1}^n\int(\widehat {u_{x_i x_j}} \widehat{u_{x_i x_j}})=\int|\widehat {D^2u|^2}$$ Thus $$\|\widehat {\Delta u}\|_{L^2}=\|\widehat {D^2 u}\|_{L^2}$$. By plancherel's theorem, $$\|\Delta u\|_{L^2}=\| {D^2 u}\|_{L^2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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