Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Define a function $\Gamma(\cdot)$ as $$ \Gamma(x-y)=\frac{1}{2\pi}\log\|x-y\|,\quad x\neq y $$ where $x=(x_1,x_2),y=(y_1,y_2)\in R^2$, and $\|x-y\|^2=(x_1-y_1)^2+(x_2-y_2)^2$. Note that $\Gamma(x-y)$ is the fundamental solution of $\Delta u=0$.

  1. If $f(y)$ is an integrable (Riemannian) function in the unit ball $B_1$, then show that $$ w(x)\overset{def}{=}\int_{B_1}\Gamma(x-y)f(y)d y, $$ is well defined function on $R^2$, moreover, $w(x)$ is continuous on $R^2$;
  2. If $w_\epsilon(x)$ is defined as $$ w_\epsilon(x)\overset{def}{=}\int_{B_1}\eta\left(|x-y|/\epsilon\right)\Gamma(x-y)f(y)dy, $$ where $\eta(t)\in C^1(R)$ with $0\leq\eta\leq1$, and $\eta(t)=0$ if $t\leq 1$, $\eta(t)=1$ if $t\geq2$, then, show that $w_\epsilon(x)\in C^1(R^2)$, i.e., it is continuous differentiable on $R^2$.
share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.