Question 12.20: Let E(x) =1/σn(2-n)|x|^n-2 for n ≥ 3, n where σn = 2π^n / 2/......
Let
E(x)=\frac{1}{\sigma_n(2-n)|x|^{n-2}} \text { for } n \geq 3
where \sigma_n=2 \pi^{n / 2} / \Gamma(n / 2).
Prove that
a) E is a locally integrable function;
b) E is the fundamental solution for the Laplace operator Δ.
a) There exist M > 0 and a < n such that
|E(x)| \leq \frac{M}{|x|^a},
where the integral \int_{B(0, s)} \frac{d x}{x^a} converges f or arbitrary s > 0, since
b) By definition of the distributional derivative and Lebesgue convergence theorem we have for every \varphi \in \mathcal {D} \text { and } 0<\varepsilon<s
<\Delta E, \varphi>=<E, \Delta \varphi>=\int_{ \mathbf{R} ^n} E \Delta \varphi d x=\lim _{\varepsilon \rightarrow 0+} \int_{\varepsilon \leq|x| \leq s} E(x) \Delta \varepsilon(x) d x
and supp \varphi \subset B(0, s). Applying the classical symmetric Green formula on the region B(0, s) \backslash \overline{B(0, \varepsilon)}, \text { since } E \in C^{\infty}(\{x|| x \mid \geq \varepsilon\}), we obtain
Therefore we have
\int_{\varepsilon \leq|x| \leq s} E \Delta \varphi d x=\int_{\varepsilon \leq|x| \leq s} \varphi \Delta E d x+\int_{\partial B(0, \varepsilon)}\left(E \frac{\partial \varphi}{\partial n }-\varphi \frac{\partial E}{\partial n }\right) d S (12.38)
since \varphi=0 \text { on } \partial B(0, s) \text {. Let now }|x|=r. Then
\frac{\partial}{\partial x_i}\left(\frac{1}{r^{n-2}}\right)=(2-n) x_i r^{-n}
and
\frac{\partial^2}{\partial x_i{ }^2}\left(\frac{1}{r^{n-2}}\right)=(2-n) r^{-n}+(2-n) x_i\left(-\frac{n}{2}\right) 2 x_i r^{-n-2} .
Therefore we have
\Delta\left((2-n) \sigma_n E\right)=(2-n) n r^{-n}-n(2-n)\left(\overset{n}{\underset{i=1}{\sum}} x_i^2\right) r^{-n-2}=0
Then we obtain by (12.38)
<\Delta E, \varphi>=\lim _{\varepsilon \rightarrow 0+} \int_{\partial B(0, \varepsilon)}\left(E \frac{\partial \varphi}{\partial r}-\varphi \frac{\partial E}{\partial r}\right) d S_{\varepsilon} .
We shall prove that the last limit is equal \varphi(0), \text { i.e., }\langle\delta, \varphi> . Using the polar coordinates we have that d S_{\varepsilon}=\varepsilon^{n-1} d S_1 and so
Since
\left|\frac{\partial \varphi}{\partial r}\right| \leq \overset{n}{\underset{i=1}{\sum}} \sup \left|\frac{\partial \varphi}{\partial x_i}\right|<M
we obtain \int_{r=\varepsilon} \varepsilon \frac{\partial \varphi}{\partial r} d S_1 \rightarrow 0 \text { as } \varepsilon \rightarrow 0+ We have by Lebesgue theorem on convergence
\int_{r=\varepsilon} \varphi\left(\varepsilon, \theta_1, \ldots, \theta_{n-1}\right) d S_1 \rightarrow \sigma_n \varphi(0)
as \varepsilon \rightarrow 0+\text {. Therefore letting } \varepsilon \rightarrow 0+ in (12.39) we obtain
<\Delta E, \varphi>=\lim _{\varepsilon \rightarrow 0+} \int_{\partial B(0, \varepsilon)}\left(E \frac{\partial \varphi}{\partial r}-\varphi \frac{\partial E}{\partial r}\right) d S_{\varepsilon}=\varphi(0)=<\delta, \varphi>.