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Can it be said that $$\int_{\Omega}\Delta_p u |\phi|^{p-2}\phi dx=\int_{\Omega}\Delta_p \phi |u|^{p-2}u dx\qquad\forall \phi\in C_0^2(\overline{\Omega})$$ is the generalized weak formulation of $$\int_{\Omega}\Delta u \phi dx=\int_{\Omega}\Delta \phi u dx$$ where $\Delta_p$ and $\Delta$ are the $p$-Laplacian and the Laplacian operators respectively?.

It is only an intuitive definition which I propose. Please help me to figure out an error in my definition if any.

Thank you!

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  • $\begingroup$ I have trouble making sense of this. Is your second equation supposed to be something like (a part of) a very weak formulation involving the Laplace operator, $\phi$ playing the role of a test function (and the equality sign being.. abuse of notation?), and you want to know what the analogous formulation for the $p$-Laplacian should be? In any way, your proposed "equation", however to be interpreted, is nonlinear in both variables, which kinda rules it out as a weak formulation to my taste.. $\endgroup$
    – Hannes
    Commented Sep 26, 2016 at 11:41
  • $\begingroup$ Just to be a bit more clearer - $\Delta_p u=\nabla.(|\nabla u|^{p-2})\nabla u$. Now suppose one wants to look for an existence of a 'weak' solution to $\Delta_p u=f$, $u=0$ on the boundary $\partial\Omega$ of a bounded $\Omega$. As per what I have defined above the weak formulation turns out to be $\int_{\Omega}|u|^{p-2}u\Delta_p \phi=\int_{\Omega}f|\phi|^{p-2}\phi$. What do you suggest about this?. $\endgroup$
    – Alexander
    Commented Sep 26, 2016 at 12:18
  • $\begingroup$ I do not think that your proposal is a useful weak formulation. A regular solution $u$ of your original $p$-Laplace equation should satisfy the weak formulation and vice versa (this is the justification for all we do with weak formulations, right?). I suggest you follow the usual route and multiply your $p$-Laplace equation with a test function $\phi$, integrate, use integration by parts/divergence theorem, and see what you get. $\endgroup$
    – Hannes
    Commented Sep 26, 2016 at 12:43
  • $\begingroup$ Thanks Hannes...I know that's the usual way. What if I define a `{\it modified Laplacian}'as $\Delta_{mod~Lap.}=\nabla.(|u|^{p-2}\nabla u)$?. I see that \begin{eqnarray} \int_{\Omega}(\Delta_{mod~Lap.} u) |\phi|^{p-2}\phi \nonumber\\ &=& -\int_{\Omega}|u|^{p-2}\nabla u.\nabla (|\phi|^{p-2} \phi)\nonumber\\&=&-(p-1) \int_{\Omega}|u|^{p-2}\nabla u.\nabla \phi|\phi|^{p-2}\nonumber \end{eqnarray}. Since this is symmetric in $u$, $\phi$ this implies that $\int_{\Omega}(\Delta_{mod~Lap.}u)|\phi|^{p-2}\phi=\int_{\Omega}(\Delta_{mod~Lap.}\phi)|u|^{p-2}u$. $\endgroup$
    – Alexander
    Commented Sep 26, 2016 at 13:28
  • $\begingroup$ Your "Modified Laplacian" has nothing to do with the $p$-Laplacian; doing a nonlinear change of coordinates $\tilde{u} = |u|^{p-1}$ you see that your modified Laplacian is just the Laplacian acting on $\tilde{u}$, and so the formula you derived is just Green's formula for $\tilde{u}$ and $\tilde{\phi}$. $\endgroup$
    – Willie Wong
    Commented Sep 26, 2016 at 13:33

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