Let $\Delta u=0$ for $0<r<R$. If $u\geq0$, prove that $u=\alpha\ln(r)+v$ where $r=\|x\|$ and $v$ is harmonic (i.e. $\Delta v=0$) for $r<R$ and $α\leq0$.
Hint: Consider $\displaystyle\int_{0}^{2\pi}u(r,\theta)(1\pm{\cos(n\theta))d\theta}$ and $\displaystyle\int_{0}^{2\pi}u(r,\theta)(1\pm{\sin(n\theta))d\theta}$
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Hi. I thought in using polar coordinates but the condition $u\geq 0$ confuses me a bit. Anyway, I don't see how the hint could help. I noticed the integrands are by assumption nonnegative, but nothing more, so I'm a bit lost.
Thanks in advance.
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I have the following result now: in $\mathbb{R}^2$, if $u$ is harmonic in the disk $0\leq{R_1}<r<R_2$ then $u$ can be written in the form: $$u(r,\theta)=\displaystyle\sum_{n=1}^{\infty}{r^n(A_n\cos(n\theta)+B_n\sin(n\theta))}+\displaystyle\sum_{n=1}^{\infty}{\displaystyle\frac{1}{r^n}(A'_n\cos(n\theta)+B'_n\sin(n\theta))}+\alpha \ln(r)+\beta$$ where the first series converge uniformly in $r\leq{R'_2}<R_2$ and the second one in $r\geq{R'_1}>R_1$.
I think the integrals of the hint can be written in the form $$\displaystyle\int_{0}^{2\pi}u(r,\theta)(1\pm{\cos(m\theta)})d\theta=\pm{}A_mr^m\pm{}\displaystyle\frac{A'_m}{r^m}+\alpha \pi \ln(r)+\pi \beta$$
Best Regards.
Once you have $$\displaystyle\int_{0}^{2\pi}u(r,\theta)(1\pm{\cos(m\theta)})d\theta=\pm{}A_mr^m\pm{}\displaystyle\frac{A'_m}{r^m}+\alpha \pi \ln(r)+\pi \beta$$ let $r\to 0$ and recall that the integral is nonnegative for either choice of the sign, since $u\ge 0$ and $(1\pm{\cos(m\theta)})\ge 0$. The coefficient $A_m'$ must be zero, for otherwise the term with it would dominate the rest of the expression and thus determine its sign.
Doing the same with $(1\pm{\sin(m\theta)}) $ yields $B_m'=0$. So you are left with $$u(r,\theta)=\displaystyle\sum_{n=1}^{\infty}{r^n(A_n\cos(n\theta)+B_n\sin(n\theta))}+ \alpha \ln(r)+\beta$$ Since $$ \int_{0}^{2\pi}u(r,\theta) d\theta = 2\pi(\alpha\ln(r)+\beta)$$ also cannot be negative, letting $r\to 0$ yields $\alpha\le 0$. $\quad\Box$