Let $\Omega \subset \mathbb{R}^n$ be a bounded domain and let $u \in L^2(0,T;H^1(\Omega))$ with $u_t \in L^2(0,T;H^{-1}(\Omega))$. Define the truncation function$$T_\epsilon(x) = \begin{cases} -\epsilon &: x \in (-\infty, -\epsilon]\\ x & x \in (-\epsilon, \epsilon)\\ \epsilon &: x \in [\epsilon, \infty) \end{cases}.$$ According to the second to last page in this paper,

clearly $$\lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \langle u_t(t), T_\epsilon(u(t)) \rangle_{H^{-1}(\Omega), H^1(\Omega)} = \int_\Omega |u(T)| - \int_\Omega |u(0)|$$

but I don't see this. We can write the LHS as $$\lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \langle u_t(t), T_\epsilon(u(t)) \rangle_{H^{-1}(\Omega), H^1(\Omega)} = \lim_{\epsilon \to 0}\frac{1}{\epsilon} \int_0^T \left(\frac{d}{dt}\int uT_\epsilon(u) - \langle (T_\epsilon u)_t, u \rangle_{H^{-1}(\Omega), H^1(\Omega)}\right)$$ and the first term on the RHS is good (after using the FTOC) and we want to show the second term on the RHS vanishes. But I am not sure that the second term even makes sense since $u_t$ is a distribution not a function.

I tried to do this via a density argument (so let $u_n \to u$ where $u_n$ are smooth) but I was unable to pass to the limit on the left hand side of the desired equation.

Crossposted from MSE (https://math.stackexchange.com/questions/843760/want-to-show-lim-epsilon-to-0-frac1-epsilon-int-0t-langle-u-tt-t). There is an answer there but it is incorrect since the answerere assumes $u_t(t) \in L^2(\Omega)$ which is not true.


1 Answer 1


This is a classical trick: the function $\frac{T_{\epsilon}(.)}{\epsilon}$ is an approximation to the sign function so at least formally the result is clear.

In order to work ou the details, define $$ S_{\epsilon}(z)=\int_0^z\frac{T_{\epsilon}(s)}{\epsilon}ds. $$ You can compute explicitly $S_{\epsilon}(z)$ so I won't do it here, but once you do that you realize $S_{\epsilon}(.)$ is an approximation to $|.|$ (with uniform convergence when $\epsilon\to 0$). Now for fixed $\epsilon>0$ you have that $T_{\epsilon}$ is bounded and Lipschitz, and $S_{\epsilon}$ is $C^1$ and uniformly Lipschitz (actually linear with slope $\pm 1$ outside $(-\epsilon,\epsilon)$). Since you're assuming that $u \in L^2(0,T;H^1(\Omega))$ and $u_t \in L^2(0,T;H^{-1}(\Omega))$ you can legitimately write for fixed $\epsilon >0$ \begin{align*} \frac{1}{\epsilon} \int_0^T \langle u_t(t), T_\epsilon(u(t)) \rangle_{H^{-1}(\Omega), H^1(\Omega)} & = \int_0^T\frac{d}{dt}\left(\int_{\Omega}S_{\epsilon}\circ u\,dx\right)\,dt\\ & =\int_\Omega S_{\epsilon}(u(T))\,dx - \int_\Omega S_{\epsilon}(u(0))\,dx \end{align*} (you have all you need to justify the chain rule $\partial_t \left(S_{\epsilon}(u)\right)=S_{\epsilon}'(u)\,\partial_t u=\frac{T_{\epsilon}(u)}{\epsilon}\partial_t u$ with the usual duality pairings etc...). Now since $S_{\epsilon}(.)\to |.|$ uniformly when $\epsilon\to 0$ the convergence of the r.h.s. should be a simple exercise (you may want to use here $L^2(\Omega)\subset L^1(\Omega)$ for bounded $\Omega$).

  • $\begingroup$ Thanks for answering. It is precisely justifying the chain rule that I cannot see. Presumably we want something like $\frac{d}{dt}\int S_\epsilon(u) = \langle (S_\epsilon(u))', 1 \rangle = \langle S_\epsilon'(u)u_t, 1 \rangle := \langle u_t, S_\epsilon'(u) \rangle$ and this kinda makes sense since $S_\epsilon'(u) \in H^1$. But this formula needs to be justified right? Maybe via some density argument. $\endgroup$
    – riem
    Jun 23, 2014 at 12:43
  • 1
    $\begingroup$ It's really like saying that $\frac{d}{dt}|u|^2_{L^2}=2<u_t,u>_{H^{-1},H^1_0}$. Indeed approximation should work but this is classical. The key argument is that $T_{\epsilon}$ is lipschitz so with your assumptions $S_{\epsilon}'(u)=T_{\epsilon}(u)/\epsilon$ is in $L^2(0,T;H^1_0)$. But note carefully that, in my chain rule, I'm not integrating by parts in time as you wrote in your OP, this is the point (apply the chain rule to $S_{\epsilon}$, not $T_{\epsilon}$). $\endgroup$ Jun 23, 2014 at 12:49

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