Even for the simple form of (\[Kcuadrado\]), in general we cannot solve (\[Hamil\]) exactly, and one has to resort to numerical methods. In the present analysis, however, we will be interested in solutions for “small” $P$. Since for $P=0$ the solution is $$\phi^{(0)} = 1 + {a \over R}$$ we may solve (\[Hamil\]) iteratively replacing $$\phi = \phi^{(0)}(R) + P^2 \phi^{(2)}(R,\theta) +P^4
\phi^{(4)}(R,\theta) 
+ \cdots$$ in (\[Hamil\]), and imposing $${\partial \phi^{(i)} \over \partial R} +{1 \over 2R} \phi^{(i)} =0 
\;\; \mbox{for} \;\; R= a, \;\; \;\;\; \lim_{R \rightarrow \infty} 
\phi^{(i)} = 0. \;\; \mbox{for} \;\; i\neq 0,$$

The form of (\[Kcuadrado\]) further suggests that we expand $\phi^{(i)}$ in Legendre polynomials $P_{\ell}(\cos\theta)$, so that $\phi^{(2)}$ may be written as $$\label{eqphi}
 \phi = \phi^{(0)}(R) + P^2 \left[\phi^{(2)}_0(R) P_0(\cos \theta)+ 
\phi^{(2)}_2(R) P_2(\cos \theta)\right]  + {\cal{O}}(P^4).$$

Solving for the coefficients taking into account the boundary condition (\[robin\]) we get, $$\begin{aligned}
\phi^{(2)}_0(R) & = & {(R+a)^5-R^5 \over 32 a^2(R+a)^5} +{13 \over 512
a R} \nonumber \\
 \phi^{(2)}_2(R) & = & { 75 R^6+291 a R^5 -650 a^2 R^4 -3400 a^3 R^3
 -4800 a^4 R^2 -2925 a^5 R -669 a^6 \over 400 R^3 (R+a)^5} \nonumber \\
 &  & + {21 a\over 20 R^3} [\ln(2a)-\ln(R+a)] + {121 