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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

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The values of $E_{\infty}$ were probably calculated numerically. They explain in the paper that they rewrote the equation for the energy to contain only $A$. So what will have done is input a range of realistic values for $A$ and plot the energy in that range. They found a minimum and calculated the value of $E$ in that minimum (that's not too hard).

The reason for writing $E_{\infty}$ will be because there is apparently an attractor point in configuration space to which the oscillon tends. This means that after a long time the oscillon will come infinitesimally close to that point. It will 'settle down' there. The energy corresponding to that point is therefore a good approximation for the energy after a long time ($t\rightarrow\infty$). That's most likely why this energy is subscripted with an infinity symbol.

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Can you elaborate that how did they get $E_\infty$? –  Raisa Jul 1 '13 at 8:26
    
Well, so they got an equation for $E$ in terms of $A$ alone, making it possible for them to calculate any $E$ corresponding to some $A$ (at a fixed $d$). They most likely used an algorithm to scan a physically relevant range of $A$'s, trying to determine for which $A$ the energy was minimal. (have a look at their figure 2, they basically searched for the minimum of that curve numerically) This minimal energy then is $E_{\infty}$. –  Wouter Jul 1 '13 at 8:48
    
Is the figure representing equation(11), if so then what would be the value or radius and the potential? Can you represents that numerically please(IF possible)? –  Raisa Jul 1 '13 at 9:55
    
The figure is indeed representing equation (11), but the radius and potential were substituted. Equation (14) was used to get an expression for $R$ by setting that equation equal to $0$. Then the obtained expression for $R$ was substituted into equation (11) and equation (13). Lastly, this adjusted equation (13) was used for the potential in equation (11). So by this procedure they got an expression $E(A)$ for the energy, only depending on $A$. This explanation is roughly present in the paper as well, under equation (14). –  Wouter Jul 1 '13 at 14:08
    
Got it, but are they using $V(A_{max})$ in equation because V''=0 ? If this is true then why they wrote( below the figure 2) $V(A_{max})$=$V(A_{min})$? –  Raisa Jul 1 '13 at 15:53

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