A Lagrangian is given by, $$L= \left(\frac{\pi}{2}\right)^2 R^d \left[\frac{1}{2}\dot A^2 - V(A_{max})\right]$$ $$E=\left(\frac{\pi}{2}\right)^2R^d V(A_{max}) $$ where V (A) now includes nonlinear terms and E is the energy which is found by taking the appropriate Legendre transform of the Lagrangian and evaluating it at the upper turning point of an oscillation, $A_{max}$. Now using the potential $V= \phi^2-\phi^3+\frac{\phi^4}{4}$, and $\phi=A(t)e^\frac{-r^2}{R^2}$we can write, $$V(A)= (1+\frac{d}{2R^2})A^2-\left(\frac{2}{3}\right)^\frac{d}{2} A^3+ \frac{A^4}{2^\frac{d+4}{2}}$$ $$V''(A)= (2+\frac{d}{R^2})-6\left(\frac{2}{3}\right)^\frac{d}{2} A+ 3\frac{A^2}{2^\frac{d}{2}}$$
For $d=2$, they got $E_{\infty}=4.44$ and $d=3$ they found the value $E_{\infty}=39.69$, but how? Why do we write here $E_{\infty}$? For more information please check equations 13 and 14 in the link