# Integration by parts with few regularity

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 $$\int_0^T \int_\Omega\partial_t s \psi(s)\xi^2 = -\int_0^T \int_\Omega \Psi(s) \partial_t \xi^2$$

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.

$$lhs \leftarrow \int\int \frac{s(t+h)-s(t)}{h}\ \psi(s(t)) \xi^2(t)$$ 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?

• 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) – Quickbeam2k1 Dec 17 '13 at 10:54
• 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. – Stephen Montgomery-Smith Dec 18 '13 at 0:47
• 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? – Quickbeam2k1 Dec 18 '13 at 9:08

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.
• 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. – Quickbeam2k1 Dec 18 '13 at 13:16