-2
$\begingroup$

$A$ is a bounded domain in $\mathbb{R}^N$ with $\partial A$ of class $C^2$;

$u,v:\overline{A}\to\mathbb{R}$ - we write $u\leq v$ if $u(x)\leq v(x)$ for a.e. $x\in A$ and $u(x)<v(x)$ if $u(x)\leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure;

$\|\nabla u\|_\infty<1$, $\|\nabla v\|_\infty<1$;

$u,v\in C^{0,1}(\overline{A})$ and $u=0$ on $\partial A$ and $v\leq 0$ on $\partial A$.

Why $\nabla(u-v)=0$ in the set $\{u<v\}$ imply that $u\geq v$?

$\endgroup$
  • 1
    $\begingroup$ Please clarify what is meant with ${u<v}$. Also, what are $u$ and $v$? Are those functions or variables? $\endgroup$ – Mefitico Jul 16 '18 at 19:16
  • $\begingroup$ Assuming that $u$ and $v$ are functions of $x,y,z$ in the $3$-dimensional real space and that $\{u<v\}$ is the set of $(x,y,z)$ such that $u(x,y,z)<v(x,y,z)$, what about $u=0$ and $v=1$ (i.e, they are constant functions)? $\endgroup$ – Batominovski Jul 16 '18 at 19:19
  • $\begingroup$ $A$ is a bounded domain in $\mathbb{R}^N$ with $\partial A$ of class $C^2$; $u,v:\overline{A}\to\mathbb{R}$ - we write $u\leq v$ if $u(x)\leq v(x)$ for a.e. $x\in A$ and $u(x)<v(x)$ if $u(x)\leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure. $\endgroup$ – BlackHawk Jul 16 '18 at 19:26
  • 1
    $\begingroup$ I think then u and v must vanish at the boundary then. @BlackHawk please give the precise space for u and v $\endgroup$ – Calvin Khor Jul 16 '18 at 21:05
  • $\begingroup$ There must be more information missing. I can imagine either $u=v$ on the boundary, or $u$ is greater than $v$ somewhere in the domain... $\endgroup$ – Jeff Jul 16 '18 at 21:05
0
$\begingroup$

Write $w=(v-u)_+$, where $t_+=\max\{t,0\}$. Then $w\in C^{0,1}(\bar{A})$, $\nabla w=0$ on $A$ and $w\leq 0$ on $\partial A$. It follows that $w\leq 0$ on $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.