0
$\begingroup$

I'm reading Linear and Nonlinear Functional Analysis by Phillipe Ciarlet, my aim in this book is the Minty-Browder Theorem and their applications. The author shows the following problem

$$\left\{\begin{array}{c} −\Delta_{p} u = f(x) & \text{on }\Omega \\ u(x)=0, \partial\Omega \end{array}\right.$$

The main idea to solve this problem is use the Minty-Browder Theorem for the operator $\Delta_{p}: W_{0}^{1,p}(\Omega) \rightarrow (W_{0}^{1,p}(\Omega))'$, where $\Delta_{p}(u):= \displaystyle\sum_{i=1}^{N}\dfrac{\partial}{\partial x_{i}}\left(|\nabla u|^{p-2}\dfrac{\partial u}{\partial x_{i}}\right)$. Then, the author says that the duality of p-Laplacian is given by:

$$<\Delta_{p} u, w > = -\displaystyle\int_{\Omega} |\nabla u|^{p-2}u . w $$

My question is: Why? How can I prove thle last equality?

Thanks in advance.

$\endgroup$
2
  • $\begingroup$ By definition it looks to be $-\int_\Omega |\nabla u|^{p-2} \nabla u \cdot \nabla w$ istead. $\endgroup$
    – user99914
    Sep 1, 2016 at 3:56
  • $\begingroup$ @ArcticChar I can't even show that $\Delta_{p}(u) \in (W^{1,p}_{0}(\Omega))' = W^{-1,q}(\Omega)$, I'm really a bit confused. $\endgroup$
    – BBVM
    Sep 1, 2016 at 15:51

1 Answer 1

1
$\begingroup$

The duality is just integration by parts

$$\langle \Delta_p u,w\rangle = \int_\Omega \text{div} (|\nabla u|^{p-2}\nabla u)\, w \, dx = -\int_\Omega |\nabla u|^{p-2} \nabla u \cdot \nabla w \, dx$$

provided $w\in W_0^{1,p}$. The identity above and Holder's inequality show that

$$|\langle \Delta_p u,w\rangle| \leq C\|\nabla w\|_{L^p(\Omega)} \leq C\|w\|_{W^{1,p}_0},$$

where $C=\||\nabla u|^{p-2}\nabla u\|_{L^q(\Omega)}$ and $\frac{1}{p} + \frac{1}{q} = 1$. Thus $\Delta_p u$ is a bounded linear operator on $W^{1,p}_0$, hence $\Delta_p u \in (W^{1,p}_0(\Omega))'$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .