Integrating the Derivative of the sgn Function I would like to prove the following equality 
$\int_{\mathbb{R}^N} \partial_t |u(t,x)| \text{d}x = \int_{\mathbb{R}^N} \text{sgn}(u(t,x))\partial_t u(t,x)  \ \text{d}x $.
Where $u: (0,\infty) \times \mathbb{R}^N \rightarrow \mathbb{R}$, is a function in $C([0,\infty); L^1(\mathbb{R}^N) \cap L^\infty(\mathbb{R}^N) ) \cap C([0,\infty); W^{2,p} (\mathbb{R}^N) ) \cap C^1((0,\infty); L^p(\mathbb{R}^N)) $, for all $p \in [1,\infty]$. And sgn is the sign function:
$\text{sgn}(x) = \begin{cases} 1, \ x>0 \\ 0, \ x = 0 \\ -1, \ x < 0   \end{cases}$
We know that $ \partial_t |u(t,x)| = \partial_t [u(t,x)\text{sgn}(u(t,x)] = \text{sgn}(u(t,x))\partial_t u(t,x) + u(t,x) \partial_t \text{sgn}(u(t,x))$, by the product rule. (We take derivatives here in the distribution sense.)
The goal, then is to prove that $ \int_{\mathbb{R}^N} u(t,x) \partial_t \text{sgn}(u(t,x)) \text{d}x = 0$.
My question is how we can show explicitly that the above integral is $0$.
My intuition is the following: $\partial_t \text{sgn}(u(t,x)) = \partial_u \text{sgn}(u) \partial_t u $, and we should have $\partial_u \text{sgn}(u) = 0$ almost everywhere. We only have 'jumps' for $ \text{sgn}(u) $ at $u=0$, but since we are considering the integral of $ u(t,x) \partial_t \text{sgn}(u(t,x)) $ the extra factor of $u$ should ensure we have $ u(t,x) \partial_t \text{sgn}(u(t,x)) = 0$, for all $t$, and thus $ \int_{\mathbb{R}^N} u(t,x) \partial_t \text{sgn}(u(t,x)) \text{d}x = 0$.
This intuition of course isn't a formal proof. In particular, I have been told to consider the following:
We know that: $\int_{\mathbb{R}} \phi(x)\partial_x \text{sgn}(x) \text{d}x = - \int_{\mathbb{R}} (\partial_x\phi(x)) \text{sgn}(x) \text{d}x$ 
$ = -\int_{0}^{\infty} (\partial_x\phi(x)) \text{sgn}(x) \text{d}x - \int^{0}_{-\infty} (\partial_x\phi(x)) \text{sgn}(x) \text{d}x = \phi(0) - (-\phi(0)) = 2\phi(0),$ for any differentiable function $\phi \in C(\mathbb{R}^N).$
I have been asked to show why something similar does not happen in my integral. Right away, we can see that the derivative and the integral are over different variables in my integral, but I will need a more precise answer than that...
EDIT 1
It can be shown explicitly that $ \partial_t |u(t,x)| = \text{sgn}(u(t,x))\partial_t u(t,x)$ whenever $u \neq 0$. Thus we need only show that 
$ \int_{A_t} u(t,x) \partial_t \text{sgn}(u(t,x)) \text{d}x = 0$,
where $A_t := \{ x \in \mathbb{R}^N \ | \ u(t,x) = 0 \} $.
EDIT 2
I have been told to consider approximating sgn with smooth functions.
Let us consider a set of smooth functions $ \Psi_{\epsilon}(x) =  \displaystyle \frac{\epsilon^{-1}x}{(1+\epsilon^{-2}x^2)}  $. Then, letting $\epsilon \in (0,1]$, we have $ \Psi_{\epsilon}(x) \in [0,1] $, for all $x \in \mathbb{R}$, and we have $\lim_{\epsilon \rightarrow 0} \Psi_{\epsilon}(x) = \text{sgn}(x) $, for all $x \in \mathbb{R}$. It is also easy to show that $\partial_x \Psi_{\epsilon}(x) \rightarrow \partial_x \text{sgn}(x)$, as $\epsilon \rightarrow 0$, for all $x \neq 0$. (i.e. almost everywhere). 
It seems that integrating with the derivative of $\Psi$ runs into the exact same problems as before? Does anyone know why this is helpful?
EDIT 3
It can be shown, via the monotone convergence theorem, that
$\int_{\mathbb{R}} \partial_x \text{sgn}(x) \text{d}x = \lim_{\epsilon \rightarrow 0} \int_{\mathbb{R}} \partial_x \Psi_{\epsilon}(x) \text{d}x = 2.$
The question then is whether we can extend this result in a useful way, perhaps 
$\int_{\mathbb{R}} u \partial_t u \partial_u \text{sgn}(u) \text{d}x = \lim_{\epsilon \rightarrow 0} \int_{\mathbb{R}} u \partial_t u \partial_u \Psi_{\epsilon}(u) \text{d}x$.
And if so, hopefully this integral can be found to be 0 somehow. 
 A: We will use the following, which is exercise 17, page 291 in "Partial Differential Equations" by Evans: if $U\subseteq\mathbb R^{N+1}$ is bounded, $f\in W^{1,p}(U)$ for some $p>1$ and $f^+,f^-$ denote the positive and negative parts of $f$, respectively, then $$\partial_tf^+=\partial_tf\cdot\chi_{[f>0]},\quad \partial_tf^-=-\partial_tf\cdot\chi_{[f<0]},$$ almost everywhere in $U$, and also $\partial_tf=0$ almost everywhere in $[f=0]$. Hence, considering a sequence that exhausts $\mathbb R^{N+1}$, the assumptions on $u$ show that $$\partial_tu^+=\partial_tu\cdot\chi_{[u>0]},\quad \partial_tu^-=-\partial_tu\cdot\chi_{[u<0]},$$ almost everywhere in $\mathbb R^{N+1}$, and $\partial_tu=0$ almost everywhere in $[u=0]$.
Now, consider a smooth, nonegative cutoff function $\phi$, supported in $[-1,1]$. Then, \begin{align*}\int_{\mathbb R^{N+1}}\partial_t|u(t,x)|\cdot \phi(t)\,dxdt&=\int_{\mathbb R^{N+1}}\partial_t(u^++u^-)\cdot\phi\\
&=\int_{\mathbb R^{N+1}}\partial_tu^+\cdot\phi+\int_{\mathbb R^{N+1}}\partial_tu^-\cdot\phi\\
&=\int_{[u>0]}\partial_tu\cdot\phi-\int_{[u<0]}\partial_tu\cdot\phi=\int_{\mathbb R^{N+1}}\partial_tu\cdot\text{sgn}(u)\phi.\end{align*} Letting $\phi$ be an approximation of the identity at $0$ then completes the proof.
Edit: to go around the boundedness issue, let $B_M$ be the ball with radius $M$ in $\mathbb R^n$, and set $U_M=B_M\times[-1,1]$. Then, $$\partial_tu^+=\partial_tu\cdot\chi_{[u>0]},\quad \partial_tu^-=-\partial_tu\cdot\chi_{[u<0]},$$ almost everywhere in $U_M$, and $\partial_tu=0$ almost everywhere in $[u=0]\cap U_M$. Then, proceeding as above, you obtain that $$\int_{\mathbb R^{N+1}}\partial_t|u|\cdot \phi\chi_{U_M}=\int_{\mathbb R^{N+1}}\partial_tu\cdot\text{sgn}(u)\phi\chi_{U_M},$$ and letting $M\to\infty$, the dominated convergence theorem shows that $$\int_{\mathbb R^{N+1}}\partial_t|u|\cdot \phi=\int_{\mathbb R^{N+1}}\partial_tu\cdot\text{sgn}(u)\phi.$$ Proceeding as above then finishes the proof.
Edit 2: the above shows the result for $t=0$, but the proof for other $t$ is similar.
