I am trying to solve below problem of the book Partial Differential Equations(Third edition) written by jurgen just, problem 3.9. Can any one give an idea? Thanks in advance.
Let $\Bbb R^2:=\{(x_1,x_2)\}$, $\Omega$:=$B^{\circ}(0,R_2)$$\setminus$$B(0,R_1)$ with $R_2$$>$$R_1$$>$0. The function $\phi:= a + b \log(|x|)$ is harmonic in $\Omega$ for all a,b. Let u$\in$$C^2$($\Omega$)$\cap$$C^0(\bar{\Omega})$ be subharmonic, i.e., $\hspace{7.2cm}\Delta$u$\geq$0, $\hspace{0.6cm}$ $x$$\in$$\Omega$.
Show that
$\hspace{6cm}$M(r)$\leq$${M(R_1)\log({R_2\over r}) + M(R_2)\log({r\over R_1})}\over{\log({R_2\over R_1})}$
with $$M(r):=\max_{\partial B(0,r)} u(x)$$ and $R_1$$\leq$r$\leq$$R_2$.
Since you're given a hint in the form of function $\phi$, it's natural to relate it to the stated inequality. To this end, define $$ \phi(x) = {{M(R_1)\log({R_2\over |x|}) + M(R_2)\log({|x|\over R_1})}\over{\log({R_2\over R_1})}}$$ which is a harmonic function. Observe that $\phi(x)=M(R_k)$ when $|x|=R_k$, for $k=1,2$. Thus, $u - \phi\le 0$ on the boundary. By the maximum principle, $u-\phi\le 0$ in the domain, which was to be proved.
Remark. This inequality, often known as the 3-circle lemma, expresses the fact that $M(r)$ is a convex function of $\log r$. Convex functions are distinguished by the property "graph lies below secant". Similarly, subharmonic functions "lie below harmonic". In this setting, $\phi$ is a harmonic function that plays the role of the secant.