$H^{-1}$ conservative gradient flow and $L^2$ projection

Question

Consider Cahn-Hilliard (see this) equation hich is known as the $H^{-1}$ gradient flow of Cahn-Hilliard energy functional, also it is easy to verify that this equation is mass preserving i.e. measure of field variable is preserved as time proceed.

To preserve mass exactly in numerical method, i want to project the gradient in to mass preserving space, by euclidean $L^2$ projection (it works for me in practice), but I do not know that does this action correct or no, more specifically, if I do $L^2$ projection of a function which is member of $H^{-1}$, is the resulted function after the $L^2$ projection is still a member of $H^{-1}$.

the projection space is convex and can be defined as follows,

$\mathcal{A} := [ u\in X\ |\ 0 \leq u \leq 1, \ \int_\Omega u\ dx = \mathtt{constant} ]$

assume $X$ as suitable Banach space.

if question is not clear please let me know to clarify.

PS: $H^{-1}$ is a member of $L^2$, right?

Answer 1

$L^2$ projection is not defined on $H^{-1}$. $H^{-1}$ is the dual of $H^1_0$, and $H^1_0$ does not contain constants. So for arbitary $u\in H^{-1}$, you cannot define the integral of $u$. By the way, $H^{-1}$ is neither a member nor a subspace of $L^2$.

Answer 2

The "$L^2$ projection" in your case is simply $$P_{\mathcal{A}} u = u + \frac { c- \int_{\Omega} u }{ | \Omega | } $$ And so it can be extended in every function space with costants

Answer 3

The answers so far are correct, however I think they miss an important point. The equation you refer to is elliptic, and it can be shown by standard means that if $u_0 \in L^2$ with $\int_{\Omega} u_0 = 1$, then $u(t) \in L^2$ for any finite time (you can say even more than this, but this seems to be all that you need).

Thus as long as your initial solution is in $L^2$, I believe you can show something along the lines of $$ \frac{dE}{dt} \leq 0,$$ where $E$ is the Cahn-Hilliard energy.

This justifies that your solution remains in $L^2$ for all times, and thus your desired projection. You don't need to rely on the $H^{-1}$ space for anything here.