1

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?

Thangachelli Debopritama
  • 433

1 Answers1

1

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, $$ \sup_{B_R} |\nabla \phi| \geq \frac1{R-\rho}. $$

ktoi
  • 7,317