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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 $f(1,2)$

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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?

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  • $\begingroup$ what do you mean with $f-F$? The way you define $F$ is the same as $f$. $\endgroup$
    – wanymose
    Commented Apr 10, 2023 at 23:09
  • $\begingroup$ 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 $f(1,2)=-1$. $\endgroup$
    – wanymose
    Commented Apr 10, 2023 at 23:25
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    $\begingroup$ @wanymose: yes, exactly. $\endgroup$
    – Martin R
    Commented Apr 11, 2023 at 2:38
  • $\begingroup$ @wanymose: Does this answer your question? Please let me know if anything is still unclear. $\endgroup$
    – Martin R
    Commented Apr 13, 2023 at 5:09
  • $\begingroup$ Yes, everything is clear. Thank you very much! $\endgroup$
    – wanymose
    Commented Apr 14, 2023 at 0:11

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