I am learning about harmonic functions and its proterpties. To understand them better I am doing some exercises.
Let $f \in C^2(\mathbb{R})^2$ be a harmonic function with $f(x_1,x_2)=x_1-x_2$ on the set {$(x_1,x_2) \in \mathbb{R}^2: x_1^2+x_2^2=16$}
Calculate f(1,2)
My attempt:
The mean value theorem for harmonic functions says that
$f(x)=\frac{1}{nV_nr^{n-1}} \int_{\partial B_r(x)} f(y) dy$
So first I tried to calculate the integral:
$\partial B_r$ is in my case the curve $\gamma(t)=4(sin(t),cos(t))$
Thus,
$\int_{\partial B_r(x)} f(y) dy=\int_0^{2 \pi} f(\gamma(t)) |\gamma'(t)| dt=4 \int_0^{2 \pi} sin(t) – cos(t) dt=0$
By the mean value theorem $f(x_1,x_2)=0$ in $B_4$ but $f(4,0)=4$ which would mean that $f$ is not continuous, which is not possible because $f$ is harmonic.
Question: Where is my mistake? Can someone show me how to calculate $f(1,2)$
Best Answer
The center of the circle $ x_1^2+x_2^2=16$ is the point $(0, 0)$, so what you calculated is that $f(0, 0) = 0$, and not that $f(x_1, x_2) = 0$ in $B_4(0, 0)$.
But it is not difficult to verify that $F(x_1, x_2) = x_1 - x_2$ is harmonic in $\Bbb R^2$, so that $f-F$ is harmonic in $\Bbb R^2$ and zero on the boundary of $B_4(0, 0)$. What can you conclude?