I am self-studying this class notes on MIT OCW https://ocw.mit.edu/courses/mathematics/18-112-functions-of-a-complex-variable-fall-2008/lecture-notes/lecture16.pdf . This question is taken from the above notes, in which there is a step I do not understand.
Theorem 20 in the notes:
If $u$ harmonic in $\Omega= \{z: r_1 \leq |z-z_0| \leq r_2\} \subset \mathbb C$, then $\frac{1}{2\pi} \int\limits_0^{2\pi}u(z_0+re^{i\theta}) d\theta = \alpha \log r + \beta$ for all $r_1 \leq r \leq r_2$. The note gives the following justification, by writing the Laplacian in polar coordinate form:
$\triangle = \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r}+ \frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}$.
Then, writing $V(r)= LHS = \frac{1}{2\pi} \int\limits_0^{2\pi}u(z_0+re^{i\theta}) d\theta$, it says $\frac{\partial^2V}{\partial r^2} + \frac{1}{r}\frac{\partial V}{\partial r} = 0$.
The above step is the one I do not get. I think it is trying to say that $V$ is harmonic, and then apply the Laplacian $\triangle$ to $V$, so since $V$ is independent of $\theta$, the last term disappears and we get $\frac{\partial^2V}{\partial r^2} + \frac{1}{r}\frac{\partial V}{\partial r} = 0$. However, I am not sure how to show that $V$ is harmonic.
The proof is showing that the integral evaluates to the $\alpha \log(r)+\beta$. They are showing that $V(r)$ (the integral) satisfies the ODE $$V_{rr} + \frac{1}{r}v_r = 0.$$ If you solve this ODE for $V(r)$, you will get exactly the $\alpha \log(r)+\beta$.
The fact that $V$ is harmonic comes from the fact that $u$ is harmonic. Basically, they are taking the Laplacian operator inside the integral.