Hi I was trying to understand the solution given by Ray Yang in the post
Question $5.9$ – Evans PDE $2$nd edition
It gets to the sort of things I am quite bad at…
When I get to the point
$$\int_{\Omega}|Du|^p\,dx=-\sum_{i=1}^n\int_{\Omega}u\left[(\frac{\partial^2 u}{\partial x_i^2}|Du|^{p-2})+\frac{\partial}{\partial x_i}(u)\frac{\partial}{\partial x_i}|Du|^{p-2}\right]dx,$$
I deal with the claim,
$$ \frac{\partial}{\partial x_i}|Du|^{p-2}=(p-2)|Du|^{p-4}\sum_j\frac{\partial^2 u}{\partial x_i\partial x_j}\frac{\partial u}{\partial x_j},$$
I feel I start making mistakes, what I did was…I have
$$ \frac{\partial}{\partial x_i}|Du|^{p-2}=(p-2)|Du|^{p-3}\frac{\partial}{\partial x_i}(|Du|)$$,
Treat $Du$ as y, I have
$$(p-2)|Du|^{p-3}\frac{Du}{|Du|}\frac{\partial}{\partial x_i}(Du)$$,
What I could have done wrong?
Next, putting things together to get the second term, could anyone show me how to get from
$$ -\sum_{i=1}^n\int_{\Omega} u\frac{\partial u}{\partial x_i}\frac{\partial}{\partial x_i}|Du|^{p-2} dx$$
to
$$-\int_{\Omega}(p-2)(\nabla u^T\cdot D^2 u\nabla u)|Du|^{p-4}?$$, for this term, what I have was
$$=-\sum_{i=1}^n\int_{\Omega} u\frac{\partial u}{\partial x_i}(p-2)|Du|^{p-4} Du\frac{\partial}{\partial x_i}(Du).$$
I think my $\frac{\partial u}{\partial x_i}$ corresponds to $\nabla u^T$, $\frac{\partial}{\partial x_i}(Du)$ corresponds to $D^2 u$, $Du$ corresponds $\nabla u$ in the solution posted.
Could anyone help?
Best Answer
In the first part, you haven't done anything wrong. Your answer, $$(p-2)|Du|^{p-3}\frac{Du}{|Du|}\frac{\partial}{\partial x_i}(Du)$$ is the same as $$(p-2)|Du|^{p-4}\sum_j\frac{\partial^2 u}{\partial x_i\partial x_j}\frac{\partial u}{\partial x_j}$$ (I'll remark that often people differentiate $|F|^p$ by writing it as $(|F|^2)^{p/2}$, so that the derivative is $\frac{p}{2}|F|^{p-2} (|F|^2)'$ where the last part is easy to deal with since it's just the sum of squares, with no square root. This is probably what Ray Yang did.)
Similarly, your result in the second part is what's in Ray Yang's answer, except you forgot there is a factor of $u$ outside of parentheses there.