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From the author's equation 13, 14 We can write by inserting V''(A)=0, Solving for R we get, $$R= \frac{6^{D/4} \sqrt{D}}{\sqrt{-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2}}$$ Now inserting the V into the article equation (11)$$E= \left(\frac{\pi }{2}\right)^{D/2} R^D V,$$ we get, $$E= \left(\frac{\pi }{2}\right)^{D/2} \left(-\left(\frac{2}{3}\right)^{D/2} A^3+2^{\frac{1}{2} (-4-D)} A^4+A^2 \left(1+\frac{D}{2 R^2}\right)\right) R^D$$ Now inserting the value of R, we get, $$E= \left(-\left(\frac{2}{3}\right)^{D/2} A^3+2^{\frac{1}{2} (-4-D)} A^4+A^2 \left(1+2^{-1-\frac{D}{2}} 3^{-D/2} \left(-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2\right)\right)\right) \left(\frac{6^{D/4} \sqrt{D}}{\sqrt{-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2}}\right)^D \left(\frac{\pi }{2}\right)^{D/2}$$

For $D= 3$ we finally get, $$E= \frac{27 6^{3/4} \left(-\frac{2}{3} \sqrt{\frac{2}{3}} A^3+\frac{A^4}{8 \sqrt{2}}+A^2 \left(1+\frac{-12 \sqrt{6}+48 A-9 \sqrt{3} A^2}{12 \sqrt{6}}\right)\right) \pi ^{3/2}}{\left(-12 \sqrt{6}+48 A-9 \sqrt{3} A^2\right)^{3/2}} \tag{1}$$ the graph for equation (1) must satisfy the article graph (FIG 2)

My graph:

Plot[(27 6^(3/4) (-(2/3) Sqrt[2/3] A^3 + A^4/(8 Sqrt[2]) +A^2 (1 + (-12 Sqrt[6] + 48 A - 9 Sqrt[3] A^2)/( 12 Sqrt[6]))) \[Pi]^(3/2))/(-12 Sqrt[6] + 48 A - 9 Sqrt[3] A^2)^(3/2), {A, 0.5, 2.5}]

enter image description here

But the author got,

Output : enter image description here

Am I doing wrong in simulation?

Then The author got like this in Fig 3 enter image description here

`

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Maybe you need to substitute 2 for $D$ in your formula? And add a division somewhere? – copper.hat Jul 2 at 19:24
I have added 2 for D, but that didn't work. – Complex Guy Jul 2 at 19:25
The reason you're not getting any output is because you forgot to replace one of the Ds with a number. However, even if you do this, it doesn't seem to output the answer you want. – Greg Ros Jul 2 at 19:25
Look, this can't be that hard. Try building up the formula slowly. It does look like you are missing a division sign. – copper.hat Jul 2 at 19:27
Ohh, sorry I have eddied my missing, but still it looks different, what division can make this graph alike with the article? – Complex Guy Jul 2 at 19:35
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2 Answers

up vote 1 down vote accepted
+50

Mathematica is more powerful that you give it credit for. You don't have to define any quantities explicitly. It is much more efficient to keep things in symbolic terms and numerically substitute values only when you need them. The below works for $d=3$ and gives results consistent with the paper.

Briefly:

  1. Define the function $V(A)=\frac{A^4}{2^{\frac{d+4}{2}}}-A^3 \left(\frac{2}{3}\right)^{d/2}+A^2 \left(\frac{d}{2 R^2}+1\right);$
  2. Calculate the solution to $V''(A)=0$, take the second $R$ solution
  3. Plug that $R$ into $E(A)=\left(\frac{\pi }{2}\right)^{d/2} V(A) R^d$ and plot
  4. Solve $E'(A)=0$, pick the third solution and call the minimum value $E_\infty$
  5. Solve $\left(\frac{\pi }{2}\right)^{d/2} V(A) R^d=E_\infty$ for $R$ as a function of $A$ (taking the third solution); this is the locus of points that attain the energy $E_\infty$

To get the right answers you have to simply select the "right" solution that Mathematica spits out when solving the various equations. The code for $d=3$ appears below and recreates the paper results.

(*Change this to 2 or 3*)
d = 3;
(*Define the potential*)
V[A_] := (1 + d/(2 R^2)) A^2 - (2/3)^(d/2) A^3 + A^4/2^((d + 4)/2);
(*Find the energy*)
Energy[A_] := 
  Evaluate[(\[Pi]/2)^(d/2) R^d V[A] /. Solve[V''[A] == 0, R][[2]]];
Plot[Energy[A], {A, 0, 2.5}, PlotRange -> {0, 100}]
(*This is the minimum energy, matches article for d=2 and d=3*)
NSolve[D[Energy[A], A] == 0, A][[3]]
Einf = Energy[A] /. NSolve[D[Energy[A], A] == 0, A][[3]]
Plot[R /. NSolve[(\[Pi]/2)^(d/2) R^d V[A] == Einf, R][[3]], {A, 0, 
  2.5}, PlotRange -> {0, 6}]

And the plots are

Energy as function of amplitude R and amplitude points that attain the minimum energy

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Wow, you did it. thanks, you used [[2]]] and [[3]]], are these referring dimension d=2 and d= 3 ? And how do we locate the attractor point with drawing multiple curves? They have said also in three dimensions they didn't predict the attractor point so they guessed, How the guess would be ? – Complex Guy Jul 8 at 16:05
The various solve functions in Mathematica will generally output multiple results because the equations can have multiple solutions. The "[[2]]" and "[[3]]" select the correct one out of the group of solutions. I found these by separating the real answers from the imaginary ones. – rajb245 Jul 8 at 18:43
So, can you please merge these results For Fig:3? – Complex Guy Jul 8 at 18:50
For the multiple curves, the one I present is the "thin solid line" from Figure 4. The thick solid lines are the set of points that satisfy $\omega_\text{gap}=\Gamma_\text{lin}$. What have you tried to do in order to calculate this curve? Where are you stuck? What progress have you made? – rajb245 Jul 8 at 21:49
I got you how you plotted the graph for d=3 in Fig:4 and in the following way I have done for d=2. But my query is, I see in the Fig is thick, thin and dashed curve, how do I trace them out? are these curve same code you given are I will need to formulate more code and then combine? – Complex Guy Jul 8 at 21:59
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up vote 2 down vote

Look at your function in a much more simple way:

$$R(A) = \frac{\sqrt{12}}{\sqrt{24 A - 9 A^2-12}}$$

for $D=2$. The function should have a horizontal tangent when the derivative of the radicand is zero:

$$\frac{d}{dA} (24 A - 9 A^2-12) = 0 \implies A = \frac{4}{3}$$

which seems to agree with your result. Also, $R$ blows up when the denominator goes to zero, or when $A=2$ and $A=2/3$, which also agrees with your plot. So far, I'd say your plot looks good.

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Got you, then which formula they have plotted in the fig:3? – Complex Guy Jul 2 at 20:31
@ComplexGuy: not obvious to me. – Ron Gordon Jul 8 at 15:54

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