Let $x\in \mathbb R^n$ and $B_\varepsilon = \{x:\|x\|\leq \varepsilon\}$. In the proof of the solution for a Poisson equation $ -\Delta u = f $ there is written that $ \int\limits_{B_\varepsilon}|\Phi(y)|dy\leq C\varepsilon^2 $ for $n\geq 3$. Here $\Phi(y)$ is a fundamental solution of Laplace equation. Since $\Phi(y) = k |y|^{2-n}$ I used to think that for $n>4$ the integral will diverge since $\int\limits_{0}^\varepsilon y^{-2}dy = \infty$. Do I have the wrong reasoning?
Estimation of an integral
1
$\begingroup$
real-analysis
pde
1 Answers
6
The fundamental solution of the Laplace equation is $|y|^{2-n}$ (exponent $2-n$, not $n-2$). In your last integral, you're missing the factor from the Jacobian determinant. The correct integral is
$\iint_0^\epsilon|y|^{2-n}|y|^{n-1}\mathrm d|y|\mathrm d\Omega=\iint_0^\epsilon|y|\mathrm d|y|\mathrm d\Omega= \frac12\epsilon^2\int\mathrm d\Omega=C\epsilon^2\;.$
-
0If by "area" you mean "surface area", yes, the surface area of $B_\epsilon$ is proportional to $\epsilon^{n-1}$, which is one factor of $\epsilon$ more in the numerator than $y^{2-n}$ has in the denominator. – 2011-08-01