# Exercise regarding the mean value theorem for harmonic functions

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 $$f(1,2)$$

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?
• what do you mean with $f-F$? The way you define $F$ is the same as $f$. Commented Apr 10, 2023 at 23:09
• Do you mean that $0=max_{\partial \Omega}=max_{\overline{ \Omega}}$ and $0=min_{\partial \Omega}=min{\overline{ \Omega}}$, thus $f=F$ in $B_4(0,0)$ and further $f(1,2)=-1$. Commented Apr 10, 2023 at 23:25