0
$\begingroup$

Caccioppoli's inequality states that the solution $u$ of the equation $-\nabla\cdot(A\nabla u)=0$ in some bounded domain $\Omega$ satisfies $$\int_{B(0,\rho)}|\nabla u|^2dy\leq \frac{C}{(R-\rho)^2}\int_{B(0,R)}|u|^2~dy,$$ for $0<\rho<R$, $2R$ should be smaller than diameter of $\Omega$, etc. In the proof of Caccioppoli's inequality, a cut-off function $\phi$ is constructed on $B(0,R)$ satisfying $|\nabla \phi|<\frac{C}{R-\rho}$.

Is it possible to construct a cut-off function satisfying $|\nabla \phi|<\frac{C}{(R-\rho)^2}$ or with higher powers of $R-\rho$? What is the limit on this?

$\endgroup$
1
$\begingroup$

In general no, you can't expect to improve the exponent. Given such a $\phi,$ we have $\phi(\rho e_1) = 1$ and $\phi(R e_1) = 0,$ where $e_1 = (1,0,\dots,0)$ is the first of the standard basis vectors. Thus applying the mean value theorem to $t \mapsto \phi(te_1)$ we obtain $\xi \in (\rho, R)$ such that, $$ \frac{\partial \phi}{\partial x_1}(\xi e_1) = \frac1{R-\rho}. $$ In particular, we see that, $$ \sup_{B_R} |\nabla \phi| \geq \frac1{R-\rho}. $$

$\endgroup$

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.