evans book pde estimate question

Question

I'm reading Evans' book on PDE and I'm having troubles understanding one estimate.

He defines the fundamental solution to Laplace' equation as

$$ \Phi(x) = \begin{cases} -\frac{1}{2\pi} \, \log(|x|), \, & n=2, \\ \frac{1}{n \, (n-2) \, \omega_n} \, \frac{1}{|x|^{n-2}}, \, & n\geq 3, \end{cases} $$

where $\omega_n$ is the volume of the $n$-ball.

For the solution of Poisson's equation $ -\Delta u = f$ he computes the Laplace acting on the convolution of $f$ and $\Phi$, involving this estimate:

$$ \bigg|\int_{B(0,\varepsilon)} \Phi(y) \, \Delta_y f(x-y) \, dy \bigg| \leq C \, \lVert D^2f \rVert_{L^\infty} \int_{B(0,\varepsilon)} |\Phi(y)| \, dy \leq \begin{cases} C \, \varepsilon^2 \, |\log(\varepsilon)|, & n=2, \\ C \, \varepsilon^2, & n\geq 3. \end{cases} $$

How do you obtain the last inequality?

I'm sorry, but can I ask what the $D^2$ is? Is it the Laplacian? — Matt, Mar 10 '17 at 18:20
i think its abused notation and tells that since $f \in C^2_c$ this is bounded — augustin souchy, Mar 10 '17 at 19:03

Jacky Chong · Accepted Answer · 2017-03-11T03:48:49.840

6

Hint: use polar coordinates to integrate the green's function on the $\epsilon$ ball.

Edit: Observe for $n\geq 3$ we have \begin{align} \int_{B(0, \epsilon)} \frac{dx}{|x|^{n-2}} = \int^\epsilon_0 \int_{|x|=r} \frac{dS(x)}{|x|^{n-2}}\ dr = C\int^\epsilon_0 \frac{r^{n-1}}{r^{n-2}}\ dr = C'\epsilon^2. \end{align}

edited Mar 11 '17 at 03:48

answered Mar 10 '17 at 18:36

Jacky Chong

26,287

@JakyChong what theorem did you use for the first equality? – NatMath Dec 02 '22 at 14:40
@NatMath I only used "polar" coordinates. – Jacky Chong Dec 04 '22 at 05:42
@MathsWizzard everything is correct. I am not sure what is the confusion. – Jacky Chong Jan 18 '25 at 09:47
@JackyChong, apologies - I thought you were using a theorem from Evan's book which doesn't apply here. – Maths Wizzard Jan 19 '25 at 15:45

score 1 · Answer 2 · answered Mar 10 '17 at 21:28

following jacky chong's hint, i came up with something, and please correct me if I'm wrong.

assuming $\varepsilon < 1$, then $\big|\log |x|\big| = - \log|x|, \, \forall \, |x| \in (0,\varepsilon)$. hence integrating (by parts) we get

$$\int \big|\log|x|\big| \, dx = - C' \int \log(r) \, r \, dr = C' \, \Big( -\frac{\varepsilon^2}{2} \, \log (\varepsilon) + \frac{\varepsilon^2}{4} \big) % = C' \, \Big( \varepsilon^2 \, \big|\log (\varepsilon)\big| + \frac{\varepsilon^2}{4} \Big) \leq C \, \varepsilon^2 \, \big|\log (\varepsilon)\big| $$

since you can choose a constant $C$ such that $C \, \varepsilon^2 \, \big|\log (\varepsilon)\big| \geq \varepsilon^2$ for all $\varepsilon < 1$.

but for $n \geq 3$ i still have no clue

It sounds like you are not clear on how to integrate in polar coordinates in $n\geq 3$ dimensions. Check the appendix in Evans book for details. — Jeff, Mar 10 '17 at 23:37

evans book pde estimate question

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