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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!

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  • $\begingroup$ Isn't that a direct application?? Calculate the average of $u$ along $\{r=2\}$. $\endgroup$
    – user99914
    Apr 25, 2014 at 20:03

1 Answer 1

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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:

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

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