# 线性偏微分方程|Linear Partial Differential Equations MATH0070

Newtonian potential. Let $f \in \mathcal{D}^{\prime}$. The convolutions
$$V_{n}=\frac{1}{|x|^{n-1}} * f, \quad n \geq 3 ; \quad V_{2}=\ln \frac{1}{|x|} * f, \quad n=2$$
(if they exist) are called the Newtonian (for $n=2$, the logarithmic) potential with density $f$.
The potential $V_{n}$ satisfies the Poisson equation
$$\Delta V_{n}=-(n-2) \sigma_{n} f, \quad n \geq 3 ; \quad \Delta V_{2}=-2 \pi f .$$

Indeed, using the formula of Sec. $2.4$ and, we obtain, for $n \geq 3$,
\begin{aligned} \Delta V_{n} &=\Delta\left(\frac{1}{|x|^{n-1}} * f\right)=\Delta \frac{1}{|x|^{n-2}} * f \ &=-(n-2) \sigma_{n} \delta * f=-(n-2) \sigma_{n} f \end{aligned}
We proceed in similar fashion in the case of $n=2$ as well.

\begin{aligned} &a(x)=\sum_{|\alpha| \leq m} a_{\alpha} \partial^{\alpha} \delta(x) \ &a * u=\sum_{|\alpha| \leq m} a_{\alpha} \partial^{\alpha} u(x) \end{aligned}
\begin{aligned} a(x) &=\sum_{\alpha} a_{\alpha} \delta\left(x-x_{\alpha}\right), \ a * u &=\sum_{\alpha} a_{\alpha} u\left(x-x_{\alpha}\right) \end{aligned}