Time-derivative of norm of solution to the Porous Media Equation Let $u$ be a solution of the $d$-dimensional porous media equation$$
\frac{d}{dt}u=\Delta(u^m)
$$
where $m\geq 1$ and assuming decay at infinity for any order derivative of the solution $u(x,t)$, with $x\in\mathbb{R}^d$. I'm trying to use integration by parts to get a simplified expression for the time derivative of the $L^p$-norm of the $k$th derivative of the solution, i.e,$$
\frac{d}{dt}\int |D^ku|^p \,dx.
$$
So far I'm getting a big expression and I haven't been able to simplify much. Any ideas?
 A: From the chain rule we can compute $\def\p{\partial}\def\d#1#2{\left\langle#1,#2\right\rangle}$$$\p|T|^p = p|T|^{p-2} \d{T}{\p T} \tag{1}$$ for any tensor $T$; so applying this with $T = D^k u$ we find $$\begin{align}
\frac d{d t}\int|D^ku|^p &= p \int |D^ku|^{p-2}\d{D^k u}{\p_t D^k u} \\
&=p\int\d{|D^ku|^{p-2}D^k u}{\Delta D^k (u^m)},
\end{align}$$
where we used the fact that $D, \partial_t, \Delta$ all commute in flat space. Since $\Delta D^k(u^m) = \mathrm{div}(D^{k+1}(u^m)),$ we can integrate by parts to get $$\begin{align}
\frac d{d t}\int|D^ku|^p &= -p \int \d{D\left(|D^k u|^{p-2} D^k u\right)}{D^{k+1}(u^m)} \tag{2}.
\end{align}$$
Expanding the left side of the inner product with the product rule and $(1)$ is easy. The ugly part is expanding $D^{k+1}(u^m)$ in terms of derivatives of $u$ so that we can extract a good $-\int|D^{k+1} u|^2$ term. I won't try to write down exactly what we get (since it will be a combinatorial nightmare), but (assuming you're just trying to get an estimate) there are two important features you should be able to verify: 


*

*The only term involving $k+1$ derivatives is $m u^{m-1} D^{k+1} u$, which will contribute to $(2)$ a good term something like  $$-C\int u^{m-1}|D^ku|^{p-2} |D^{k+1} u|^2;$$ and 

*the remaining terms are multiples of $u^{m-a} D^{i_1} u \otimes \cdots \otimes D^{i_a} u$ where $1 \le i_j \le k$ satisfy $\sum_{j=1}^a i_j = k+1.$ Here $a > 1$ is just the number of times we differentiate $u$ when expanding with the product rule. (The good term is also of this form, but with $a=1, i_1 = k+1.$)
I don't think you should expect things to simplify any further - as you say, if you tried to write out the whole thing you get a very big expression. For the purposes of obtaining estimates, this kind of understanding is typically good enough: if you can bound an arbitrary bad term, then you can bound all of them by just increasing your constant (in this case, multiplying it by something like $2^m$).
