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You are looking at $$ -\Delta u + \nabla p = f \mbox{ in }\Omega $$ with $u=0$ on the boundary and $\textrm{div}(u)=0$, $f\in L^2(\Omega)$, and you know there exists such a $u\in H^1_0(\Omega,\mathbb{R}^N)$ which is divergence free and solution of this equation.

Now what you said is that $P$ is more than $L^2$ by doing the following. Take a function $\phi\in W^{2,2}(\Omega)$ such that $\phi=0$ and $\nabla \phi=0$ on $\partial\Omega$. $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi -\int_\Omega p \partial_{ii} \phi= \int_\Omega f_i \partial_i\phi $$ you integrate by parts in $j$ in the first term to obtain $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi = - \int_\Omega u_i \partial_{jij}\phi $$ which means you want $\phi\in W^{3,2}$. You then commute the derivatives and integrate in the other direction to obtain $$ \int_\Omega \partial_i u \Delta \phi - \int_\Omega p \Delta \phi = \int_\Omega f\cdot \nabla \phi $$ Using the divergence free assumption, the first term vanishes, and therefore $p$ is a (very) very weak solution of $$ -\Delta p = -\textrm{div}(f). $$ in the sense that $$ - \int_\Omega p \Delta \phi = \int_\Omega f_i \partial_i \phi \quad \forall \phi\in W^{3,2}(\Omega)\textrm{ s.t. } \phi=0,\nabla \phi=0\textrm{ on } \partial\Omega. $$ Once you are here, since your identity does not require more than two derivatives, you can lower it to $W^{2,2}$, and write it as $$ - \int_\Omega p \Delta \phi = \int_\Omega f_i \partial_i \phi \quad \forall \phi\in W^{2,2}(\Omega)\textrm{ s.t. } \phi= \partial_n\phi=0\textrm{ on } \partial\Omega. $$ So your question is: what does it say on the regularity of $p$ ? Does it imply in particular that $p$ is weakly differentiable?

The problem I see is that you need to cancel also the gradient. If you did not need the gradient to be also zero, the answer would be yes (just as outlined below), but that's not the case.

You would like to know if this implies that for dense family $\psi\in H^1(\Omega)$, there holds $$ \int_\Omega p \partial_i \psi \leq C \|\psi\|_{H^1(\Omega)}, $$ so that indeed the weak derivative exists. The natural way to use the equation is to find a function $\phi$ such that $$ \Delta \phi = \partial_i \psi \textrm{ in }\Omega, \textrm{ with } \phi=\partial_n\phi=0 \textrm{ on } \partial \Omega. $$ But of course, that's very unlikely that such a $\phi$ exists...

You are looking at $$ -\Delta u + \nabla p = f \mbox{ in }\Omega $$ with $u=0$ on the boundary and $\textrm{div}(u)=0$, $f\in L^2(\Omega)$, and you know there exists such a $u\in H^1_0(\Omega,\mathbb{R}^N)$ which is divergence free and solution of this equation.

Now what you said is that $P$ is more than $L^2$ by doing the following. Take a function $\phi\in W^{2,2}(\Omega)$ such that $\phi=0$ and $\nabla \phi=0$ on $\partial\Omega$. $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi -\int_\Omega p \partial_{ii} \phi= \int_\Omega f_i \partial_i\phi $$ you integrate by parts in $j$ in the first term to obtain $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi = - \int_\Omega u_i \partial_{jij}\phi $$ which means you want $\phi\in W^{3,2}$. You then commute the derivatives and integrate in the other direction to obtain $$ \int_\Omega \partial_i u \Delta \phi - \int_\Omega p \Delta \phi = \int_\Omega f\cdot \nabla \phi $$ Using the divergence free assumption, the first term vanishes, and therefore $p$ is a (very) very weak solution of $$ -\Delta p = -\textrm{div}(f). $$ in the sense that $$ - \int_\Omega p \Delta \phi = \int_\Omega f_i \partial_i \phi \quad \forall \phi\in W^{3,2}(\Omega)\textrm{ s.t. } \phi=0,\nabla \phi=0\textrm{ on } \partial\Omega. $$ Once you are here, since your identity does not require more than two derivatives, you can lower it to $W^{2,2}$, and write it as $$ - \int_\Omega p \Delta \phi = \int_\Omega f_i \partial_i \phi \quad \forall \phi\in W^{2,2}(\Omega)\textrm{ s.t. } \phi= \partial_n\phi=0\textrm{ on } \partial\Omega. $$ So your question is: what does it say on the regularity of $p$ ? Does it imply in particular that $p$ has a weak derivative in $L^2$ ?

The problem I see is that you need to cancel also the gradient. If you did not need the gradient to be also zero, the answer would be yes (just as outlined below), but that's not the case.

You would like to know if this implies that for dense family $\psi\in H^1(\Omega)$, there holds $$ \int_\Omega p \partial_i \psi \leq C \|\psi\|_{H^1(\Omega)}, $$ so that indeed the weak derivative exists. The natural way to use the equation is to find a function $\phi$ such that $$ \Delta \phi = \partial_i \psi \textrm{ in }\Omega, \textrm{ with } \phi=\partial_n\phi=0 \textrm{ on } \partial \Omega. $$ But of course, that's very unlikely that such a $\phi$ exists...

This is just a rephrasing of the question, not an answer, just to put it in a familiar context (for me and hopefully others). You are looking at $$ -\Delta u + \nabla p = f \mbox{ in }\Omega $$ with $u=0$ on the boundary and $\textrm{div}(u)=0$, $f\in L^2(\Omega)$, and you know there exists such a $u\in H^1_0(\Omega,\mathbb{R}^N)$ which is divergence free and solution of this equation.

Now what you said is that $P$ is more than $L^2$ by doing the following. Take a function $\phi\in W^{2,2}(\Omega)$ such that $\phi=0$ and $\nabla \phi=0$ on $\partial\Omega$. $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi -\int_\Omega p \partial_{ii} \phi= \int_\Omega f_i \partial_i\phi $$ you integrate by parts in $j$ in the first term to obtain $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi = - \int_\Omega u_i \partial_{jij}\phi $$ which means you want $\phi\in W^{3,2}$. You then commute the derivatives and integrate in the other direction to obtain $$ \int_\Omega \partial_i u \Delta \phi - \int_\Omega p \Delta \phi = \int_\Omega f\cdot \nabla \phi $$ Using the divergence free assumption, the first term vanishes, and therefore $p$ is a (very) very weak solution of $$ -\Delta p = -\textrm{div}(f). $$ in the sense that $$ - \int_\Omega p \Delta \phi = \int_\Omega f_i \partial_i \phi \quad \forall \phi\in W^{3,2}(\Omega)\textrm{ s.t. } \phi=0,\nabla \phi=0\textrm{ on } \partial\Omega. $$ Once you are here, since your identity does not require more than two derivatives, you can lower it to $W^{2,2}$. So your question is: what does it say on the regularity of $p$ ? Does it imply that $p$ is weakly differentiable?

The problem I see is that you cancel also the gradient. If you did not need the gradient to be also zero, the answer would be yes (and I can provide a proof if required, but that does not answer the question).

You are looking at $$ -\Delta u + \nabla p = f \mbox{ in }\Omega $$ with $u=0$ on the boundary and $\textrm{div}(u)=0$, $f\in L^2(\Omega)$, and you know there exists such a $u\in H^1_0(\Omega,\mathbb{R}^N)$ which is divergence free and solution of this equation.

Now what you said is that $P$ is more than $L^2$ by doing the following. Take a function $\phi\in W^{2,2}(\Omega)$ such that $\phi=0$ and $\nabla \phi=0$ on $\partial\Omega$. $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi -\int_\Omega p \partial_{ii} \phi= \int_\Omega f_i \partial_i\phi $$ you integrate by parts in $j$ in the first term to obtain $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi = - \int_\Omega u_i \partial_{jij}\phi $$ which means you want $\phi\in W^{3,2}$. You then commute the derivatives and integrate in the other direction to obtain $$ \int_\Omega \partial_i u \Delta \phi - \int_\Omega p \Delta \phi = \int_\Omega f\cdot \nabla \phi $$ Using the divergence free assumption, the first term vanishes, and therefore $p$ is a (very) very weak solution of $$ -\Delta p = -\textrm{div}(f). $$ in the sense that $$ - \int_\Omega p \Delta \phi = \int_\Omega f_i \partial_i \phi \quad \forall \phi\in W^{3,2}(\Omega)\textrm{ s.t. } \phi=0,\nabla \phi=0\textrm{ on } \partial\Omega. $$ Once you are here, since your identity does not require more than two derivatives, you can lower it to $W^{2,2}$, and write it as $$ - \int_\Omega p \Delta \phi = \int_\Omega f_i \partial_i \phi \quad \forall \phi\in W^{2,2}(\Omega)\textrm{ s.t. } \phi= \partial_n\phi=0\textrm{ on } \partial\Omega. $$ So your question is: what does it say on the regularity of $p$ ? Does it imply in particular that $p$ is weakly differentiable?

The problem I see is that you need to cancel also the gradient. If you did not need the gradient to be also zero, the answer would be yes (just as outlined below), but that's not the case.

You would like to know if this implies that for dense family $\psi\in H^1(\Omega)$, there holds $$ \int_\Omega p \partial_i \psi \leq C \|\psi\|_{H^1(\Omega)}, $$ so that indeed the weak derivative exists. The natural way to use the equation is to find a function $\phi$ such that $$ \Delta \phi = \partial_i \psi \textrm{ in }\Omega, \textrm{ with } \phi=\partial_n\phi=0 \textrm{ on } \partial \Omega. $$ But of course, that's very unlikely that such a $\phi$ exists...

This is just a rephrasing of the question, not an answer, just to put it in a familiar context (for me and hopefully others). You are looking at $$ -\Delta u + \nabla p = f \mbox{ in }\Omega $$ with $u=0$ on the boundary and $\textrm{div}(u)=0$, $f\in L^2(\Omega)$, and you know there exists such a $u\in H^1_0(\Omega,\mathbb{R}^N)$ which is divergence free and solution of this equation.

Now what you said is that $P$ is more than $L^2$ by doing the following. Take a function $\phi\in W^{2,2}(\Omega)$ such that $\phi=0$ and $\nabla \phi=0$ on $\partial\Omega$. $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi -\int_\Omega p \partial_{ii} \phi= \int_\Omega f_i \partial_i\phi $$ you integrate by parts in $j$ in the first term to obtain $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi = - \int_\Omega u_i \partial_{jij}\phi $$ which means you want $\phi\in W^{3,2}$. You then commute the derivatives and integrate in the other direction to obtain $$ \int_\Omega \partial_i u \Delta \phi - \int_\Omega p \Delta \phi = \int_\Omega f\cdot \nabla \phi $$ Using the divergence free assumption, the first term vanishes, and therefore $p$ is a (very) very weak solution of $$ -\Delta p = -\textrm{div}(f). $$ in the sense that $$ - \int_\Omega p \Delta \phi = \int_\Omega f_i \partial_i \phi \quad \forall \phi\in W^{3,2}(\Omega)\textrm{ s.t. } \phi=0,\nabla \phi=0\textrm{ on } \partial\Omega. $$ So your question is: what does it say on the regularity of $p$ ? In particular, does it imply that $p$ is in fact a weak solution of the same equation, which would give you the local $H^1$ regularity you want?

This is just a rephrasing of the question, not an answer, just to put it in a familiar context (for me and hopefully others). You are looking at $$ -\Delta u + \nabla p = f \mbox{ in }\Omega $$ with $u=0$ on the boundary and $\textrm{div}(u)=0$, $f\in L^2(\Omega)$, and you know there exists such a $u\in H^1_0(\Omega,\mathbb{R}^N)$ which is divergence free and solution of this equation.

Now what you said is that $P$ is more than $L^2$ by doing the following. Take a function $\phi\in W^{2,2}(\Omega)$ such that $\phi=0$ and $\nabla \phi=0$ on $\partial\Omega$. $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi -\int_\Omega p \partial_{ii} \phi= \int_\Omega f_i \partial_i\phi $$ you integrate by parts in $j$ in the first term to obtain $$ \int_\Omega \partial_j u_i \cdot \partial_{ij}\phi = - \int_\Omega u_i \partial_{jij}\phi $$ which means you want $\phi\in W^{3,2}$. You then commute the derivatives and integrate in the other direction to obtain $$ \int_\Omega \partial_i u \Delta \phi - \int_\Omega p \Delta \phi = \int_\Omega f\cdot \nabla \phi $$ Using the divergence free assumption, the first term vanishes, and therefore $p$ is a (very) very weak solution of $$ -\Delta p = -\textrm{div}(f). $$ in the sense that $$ - \int_\Omega p \Delta \phi = \int_\Omega f_i \partial_i \phi \quad \forall \phi\in W^{3,2}(\Omega)\textrm{ s.t. } \phi=0,\nabla \phi=0\textrm{ on } \partial\Omega. $$ Once you are here, since your identity does not require more than two derivatives, you can lower it to $W^{2,2}$. So your question is: what does it say on the regularity of $p$ ? Does it imply that $p$ is weakly differentiable?

The problem I see is that you cancel also the gradient. If you did not need the gradient to be also zero, the answer would be yes (and I can provide a proof if required, but that does not answer the question).