3
$\begingroup$

In Evans' PDE book, the author claims (p.352) that the uniform ellipticity condition \begin{equation} \label{e:1} \sum_{i,j=1}^n a^{ij}(x)\xi_i \xi_j \ge \theta|\xi|^2 \end{equation}

implies that $$ \sum_{i,j,k,\ell=1}^n a^{ij} a^{k\ell}v_{x_i x_k}v_{x_j x_\ell} \ge \theta^2 |D^2 u|^2. $$ This is very believable to me, but I cannot write down a proof. Why is it true? It seems to me that the most obvious thing to try would be to rearrange the sum $$ \sum_{i,j,k,\ell=1}^n a^{ij} a^{k\ell}v_{x_i x_k}v_{x_j x_\ell} = \sum_{i,j=1}^n a^{ij} \sum_{k,\ell=1}^n a^{k\ell}v_{x_i x_k}v_{x_j x_\ell} $$ and try to apply uniform ellipticity to the inner sum first, followed by the outer sum. However, the inner sum here is not quite in the right form. (What would $\xi$ be? $v_{x_i x_k}$ and $v_{x_j x_\ell}$ are the $k$th and $\ell$th components of two different vectors, namely $Du_{x_i}$ and $Du_{x_j}$, respectively.) Am I missing something easy here?

$\endgroup$

1 Answer 1

1
$\begingroup$

Suppose we fix a value of $x$. Since $a^{ij}$ (evaluated at our fixed choice of $x$) is a real symmetric matrix, is it not the case that we can rotate the coordinate axes so that $a^{ij}$ becomes a diagonal matrix? Of course, our choice of rotation will vary from one value of $x$ to another, but as far as I can tell, this doesn't matter here...

So suppose we've done this rotation, and suppose the diagonal entries of $a^{ij}(x)$ are $(\theta_1, \dots, \theta_n)$. By the ellipticity condition, we must have $\theta_i \geq \theta > 0$ for each $i$. Then it is clear that $$ \sum_{i,j,k,l}a^{ij}a^{kl}v_{ik}v_{jl} = \sum_{ik} \theta_i \theta_k v_{ik}v_{ik} \geq \theta^2 \sum_{ik} v_{ik} v_{ik} .$$

Finally, notice that $\sum_{ik} v_{ik} v_{ik}$ is invariant under rotations of coordinate axes, so this quantity is the same as the $|D^2 v|^2$ evaluated in the original coordinates.

$\endgroup$
1
  • $\begingroup$ Thanks! I like this argument. Also, thanks for pointing out the typo in my original question--it should now read correctly. $\endgroup$
    – 1234
    Jun 2, 2017 at 2:51

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.