Suppose that $U$ is a harmonic function on the disk $\Omega=\{r \geq 2\}$ and that $$U(2,\theta)=3\sin(\theta)+1.$$ Without finding the solution, find the maximum and minimum values of $U$ on $\Omega.$
Hmm, so I know that I must use the maximum principle and indeed I am allowed to since $\Omega$ is open and connected. Then, we have that $U$ attains its maximum/minimum on the boundary as it is harmonic. But then how would I explicitly compute the solution at the boundary?
On the boundary $U$ is independent of $r$. Now, nothe that
$$ |3\sin(\theta)+1|\leq3|\sin(\theta)|+1\leq4 $$ So, $4$ is a upper bound. So, the maximum values of $U$ on the boundary is $4$ (with $\theta=\pi/2$).
In the same form you find the minimum.