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I'm having problems proving an integration by parts formula presented in the work of Alt and DiBenedetto on porous media flow (Remark 3.4.2).

Essentially, the problem is the following. Let $s\in L^\infty(\Omega\times(0,T),[0,1])$ with $\partial_t s \in L^2(0,T; H^{-1}(\Omega))$. $\Omega$ is a bounded domain in $\mathbb{R}^n$. Let $\psi$ be any Lipschitz function with $\psi'=0$ in a neighborhood of one. The one concludes that $\psi(s)\in L^2(0,T;H^1(\Omega))$. Note that there is no spatial regularity on $s$ itself. (But loosely speaking one can prove that $s$ away from $1$ has an integrable gradient).

Then for $\xi\in C_c^\infty(\Omega\times(0,T))$ holds \begin{equation} \int_0^T \int_\Omega\partial_t s \psi(s)\xi^2 = -\int_0^T \int_\Omega \Psi(s) \partial_t \xi^2 \end{equation}

there $\Psi$ is the primitive of $\psi$. Note that the left hand side is well defined if intepreted as a duality pairing. The right hand side is clearly well defined.

I can prove this formula if $\psi$ is monotone. If I lack monotonicity I run into the following problem: We use the strong convergence of the discrete difference quotient.

\begin{equation} lhs \leftarrow \int\int \frac{s(t+h)-s(t)}{h}\ \psi(s(t)) \xi^2(t) \end{equation} Now $(s(t+h)-s(t))\psi(s(t)) \approx \Psi(s(t+h))-\Psi(s(t))$. But proving the convergence seems to fail since $(s(t+h)-s(t))/h \to \partial_t s$ in $L^2(0,T;H^{-1}(\Omega))$ acts on something that does converge to zero but not in $L^2(0,T;H^1_0(\Omega))$ (After of course inserting a zero). In The paper mentioned above the formula is stated as well know. Any ideas, comments or suggestions?

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  • $\begingroup$ What I should add, a proof along the lines of Evans Theorem 3 in Chapter 5.9 fails, since the regularisations in time do not converge in $L^2(0,T;H^1(\Omega))$. In particular one does no know a priori if $\psi(s_\varepsilon)\to\psi(s)$ remains bounded, where the $\varepsilon$ denotes regularization in time. If one could show that the proof would follow (but I don't see how this should be true) $\endgroup$ – Quickbeam2k1 Dec 17 '13 at 10:54
  • $\begingroup$ Before starting this problem, let me try something easy. You say you can prove it if $\psi$ is monotone. But since $\psi$ is Lipschitz, it is of bounded variation, and hence is the difference of two monotone functions. And each of those monotone functions can be taken to be Lipschitz and satisfy $\psi' = 0$ in a neighbourhood of one. $\endgroup$ – Stephen Montgomery-Smith Dec 18 '13 at 0:47
  • $\begingroup$ Hey, that really works. I didn't think on bounded variation. Could you post your comment as an answer so that you can get the bounty? $\endgroup$ – Quickbeam2k1 Dec 18 '13 at 9:08
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You say you can prove it if $\psi$ is monotone. But since $\psi$ is Lipschitz, it is of bounded variation, and hence is the difference of two monotone functions. And each of those monotone functions can be taken to be Lipschitz and satisfy $\psi′=0$ in a neighbourhood of one.

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  • $\begingroup$ To add some detail. Assume the result holds for nondecreasing $\psi$. Then $\psi=\psi_1 -\psi_2$ where each $\psi_j$ is nondecreasing. Then $\int <\partial_t s , \psi(s)\xi^2> = \int_0^T <\partial_t s , \psi_1(s)\xi^2> + \int_0^T <\partial_t s , \psi_2(s)\xi^2>$. For each term we can apply the result for the monotone functions and the primitives can also be decomposed in a similar way. $\endgroup$ – Quickbeam2k1 Dec 18 '13 at 13:16

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