20
$\begingroup$

Suppose $\Omega\subset\mathbb{R}^n$ is a bounded domain. Let $u\in H_0^1(\Omega)$ and $\tilde{u}\in L^2(\Omega)$ satisfies $$\int_\Omega\nabla u\nabla v=\int_\Omega \tilde{u}v,\ \forall\ v\in C_0^{\infty}(\Omega)$$

Suppose that there exist a set $\Gamma$ of positive measure such that $\nabla u=0, a.e.\ x\in\Gamma$. How can one show that $\tilde{u}=0,\ a.e.\ x\in\Gamma$?

Any help is appreciate. I have to solve this problem without considering the fact that $\tilde{u}=-\Delta u$. I know that i have to take some "nice" $v\in C_0^{\infty}(\Omega)$, but what $v$?

Edit 1: Why am i trying to solve this problem? Let $p\geq 1$. Suppose $u\in W_0^{1,p}(\Omega)$ and $\tilde{u}\in L^q(\Omega)$ $(\frac{1}{p}+\frac{1}{q}=1)$ satisfies $$\int_\Omega|\nabla u|^{p-2} \nabla u\nabla v=\int_\Omega \tilde{u}v,\ \forall\ v\in C_0^{\infty}(\Omega)$$

In this case we dont have enough regularity to show that $\tilde{u}=-\Delta_p u$, so i need a more direct aproach, and consequently i think that this aproach is the same for both cases.

Lastly we can have the $\Phi$-laplacian too and considering the spaces where it makes sense, we have $$\int_\Omega\Phi(|\nabla u|)\nabla u\nabla v=\int_\Omega \tilde{u}v,\ \forall\ v\in C_0^{\infty}(\Omega)$$

Edit 2: In general, consider $\sigma$ a vector valued function. Suppose $\sigma$ is in a convenient space and $$\int_\Omega \sigma\cdot\nabla v=\int_\Omega \tilde{u}v,\ \forall\ v\in C_0^{\infty}(\Omega)$$

where, $\cdot$ stands for inner product. Now i want to conclude the same thing for all the cases above.

Note: $\tilde{u}$ is called weak divergence of $\sigma$.

Edit 3: In any open set $U$ contained in $\Gamma$, we can take functions with compact supoort in $U$ and conclude that $\tilde{u}=0,\ a.e.\ x\in U$, but in the general case, how to proceed? Any opinion?

The question is: If closure of $\Gamma$ have interior empty, what we have to do?

Thanks

$\endgroup$
1
  • 1
    $\begingroup$ Take a small ball around an arbitrary point in $\Gamma$. Then the integral must be small by Lebesgue differentiation theorem? $\endgroup$
    – timur
    Oct 30, 2012 at 17:31

2 Answers 2

0
+100
$\begingroup$

If $\Gamma$ is a bounded domain then considering $v \in C_0^\infty(\Gamma) \subset C_0^\infty(\Omega)$ we have $$ 0= \int_\Gamma \nabla u \cdot \nabla v =\int_\Gamma \tilde u v$$ Therefore $\tilde u=0$ in $L^2(\Gamma)$. I don't know what to do if $\Gamma$ is less regular.

$\endgroup$
0
$\begingroup$

This follows from the following fact: $\DeclareMathOperator{\dv}{div}$

Fact. If $-\dv V = \bar{u}$ weakly (and both $V,\bar{u}$ are integrable), then $\bar{u} = 0$ a.e. on the set $\{ V = 0 \}$.

To see why, just take $V = \nabla u$, so that $- \dv V = - \Delta u = \bar{u}$ weakly. Since $\Gamma$ is a subset of $\{ V = 0 \}$, we have $\bar{u} = 0$ a.e. on $\Gamma$ as well.


Consider a related fact: $\nabla u = 0$ a.e. on the set $\{ u = 0 \}$ (if both $u, \nabla u$ are integrable). Its proof can be found here, and it can be easily adapted. The outline is as follows:

  • Show that if $\psi \colon \mathbb{R}^n \to \mathbb{R}^n$ is $C^1$ (with suitable growth), then the composition $\psi \circ V$ satisfies the chain rule: $$ \dv(\psi \circ V) = \sum_{i,j} (\partial_j \psi^i \circ V) \cdot \partial_i V^j \quad \text{in the weak sense.} $$ To this end, one considers a mollification $V_\varepsilon$ of $V$ and takes the limit $\varepsilon \to 0$.
  • Take $\psi$ satisfying $$ |\psi(p)| \le |p| \text{ for } p \in \mathbb R^n, \quad \psi(p) = p \text{ for } |p| \ge 1, \quad D\psi(0) = 0 $$ and consider $V_\varepsilon(x) := \varepsilon \cdot \psi(V(x)/\varepsilon)$. By the previous bullet, $\dv V_\varepsilon = 0$ a.e. on the set $\{ V = 0 \}$, just because the derivatives of $\psi$ vanish there. Looking at the limit $\varepsilon \to 0$ more carefully, we can see that also $\dv V = 0$ a.e. on $\{ V = 0 \}$.
$\endgroup$

You must log in to answer this question.

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