1
$\begingroup$

Could someone explain what the mean value property means, and how I can apply it to a question?

Mean value property: Let $u$ be a harmonic function in a disk $D$, and continuous in its closure $D$, then the value of $u$ at the center of the disk is equal to the average of $u$ on its circumference.

It is 10.1 in this link. http://www.math.ucsb.edu/~grigoryan/124B/lecs/lec10.pdf

How would I apply it to the following?

If $u$ is harmonic function in disk $D=\{r<2\}$ and $u(\theta)=3\sin(2\theta) +1$ for $r = 2$, without finding the solution calculate $u$ at the origin.

Thanks!

$\endgroup$
  • $\begingroup$ Isn't that a direct application?? Calculate the average of $u$ along $\{r=2\}$. $\endgroup$ – user99914 Apr 25 '14 at 20:03
2
$\begingroup$

Basically if a function is harmonic, then its value at a point is equal to the integral over a sphere of any radius (or its boundary) centred at that point.

I.e. $u(x_0)=\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x_0,r)}u(x)dS(x)$

Now in your case, we consider the disk of radius $2$, so $r=2$, and in polar coordinates, you have $u(\theta)=3\sin(2\theta)+1$ So we have:

$u(0,0)=\frac{1}{4\pi}\int_0^{2\pi}3\sin(2\theta)+1d\theta=\frac{1}{4\pi}(\theta-\frac{3}{2}\cos(2\theta)|^{2\pi}_0)=\frac{2\pi}{4\pi}=\frac{1}{2}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.