0
$\begingroup$

I hope you can help with a problem I am having. I want to solve this equation

$$\left( n^2 \Theta^2 + 2\alpha n\Theta -1\right)R_n = \frac{1}{R_0^2}\left( 2 \sum_{m=1}^\infty R_{n+m}R_m + \sum_{m=1}^n R_{n-m}R_m\right)$$,

for different value of $\Theta$ and $\alpha$, here 0.1 and 1. This is done by guessing a trial function $R_n \propto \exp(-n)$ and doing successive iteration. Thereby calculating the sums for a fixed value. Here I choose $m \leq 256$ divided by iteration number. So first one is 256 next is 128 etc. In the end I want to sum over all the number R_n to get a single value. My idea so far is to do the following code, by brute force

    f[n_] = Exp[-n]
    E2 = (Sum[2*f[n + m]*f[m], {m, 256}] + Sum[f[n - m]*f[m], {m, 256}])/(n^2*0.1^2 + 2*n*0.1*1-1)*1/(f[0]^2)
    D1[n_] = E2; 

and then doing it all again

      E3 = (Sum[2*D1[n + m]*D1[m], {m, 256/2}] + Sum[D1[n - m]*D1[m], {m, 256/2}])/(n^2*0.1^2 + 2*n*0.1*1 - 1)*1/(D1[0]^2);
      D2[n_] = E3;

and again

      E4 = (Sum[2*D2[n + m]*D2[m], {m, 256/4}] + Sum[D2[n - m]*D2[m], {m, 256/4}])/(n^2*0.1^2 + 2*n*0.1*1 - 1)*1/(D2[0]^2);
      D3[n_] = E4;

The problem now is that the evaluation of E5 etc. take a long time. Am I doing it the right clever way? I want in the end a list of R_n values. Hope I have provided enough information.

I am trying to replicate the calculation in this paper https://journals.aps.org/prb/abstract/10.1103/PhysRevB.36.1931, from equation 7.1, where they for $\alpha=1$ and $\theta=0.1$ find that

$$ s=2\pi/\Theta \left( 1+1/8 \sum_{n,m=-\infty}^{\infty} R_{n+m} R_n R_m \right)=22.2$$

$\endgroup$
11
  • $\begingroup$ yes it should be and is in my code, will change in question $\endgroup$ Commented Apr 1, 2018 at 21:23
  • $\begingroup$ I am confused. Negative indices of R are allowed? $\endgroup$ Commented Apr 1, 2018 at 21:35
  • $\begingroup$ in the article they state that $R_{n}=R_{-n}$, so that negative indices are just the positive ones $\endgroup$ Commented Apr 1, 2018 at 21:37
  • $\begingroup$ Ah. Okay. And we can expect that $R_n \to 0$ for $n \to \infty$? $\endgroup$ Commented Apr 1, 2018 at 21:38
  • $\begingroup$ yes, exactly so we are only interested in a fixed value of $R_n$ $\endgroup$ Commented Apr 1, 2018 at 21:40

1 Answer 1

1
$\begingroup$

I am not completely sure but I think, you are looking for the roots of the function F below. For given R, F[R] is the difference of both side of the equation

$$\left( n^2 \Theta^2 + 2\alpha n\Theta -1\right)R_n = \frac{1}{R_0^2}\left( 2 \sum_{m=1}^\infty R_{n+m}R_m + \sum_{m=1}^n R_{n-m}R_m\right)$$

Caution: R[[1]] corresponds to $R_0$, R[[2]] to $R_1$, R[[3]] to $R_2$ and so on.

nn = 100;
R0 = Exp[-Range[0., nn - 1]];
Θ = 0.1;
α = 1.;

F = R \[Function] 
   Join[{(-1) R[[1]] - 2. R[[1 ;;]].R[[1 ;; nn]] / R[[1]]^2}, 
    Table[(n^2 Θ^2 + 2 α n Θ - 1) R[[n + 1]] - (2. R[[n + 1 ;;]].R[[1 ;; nn - n]] + R[[2 ;; n + 1]].R[[n ;; 1 ;; -1]])/R[[1]]^2, {n, 1, nn - 1}]
    ];

We can employ FindRoot to search for roots as follows

R = Array[r, nn];
FindRoot[Evaluate[F[R]], Evaluate@Transpose[{R, Evaluate@R0}],
 MaxIterations -> 1000]

In a nutshell, FindRoot applies Newton's method

$$R^{(k+1)} = R^{(k)} - DF(R^{(k)})^{-1} \, F(R^{(k)})$$

Unfortunately, $R^{(k)}$ seems to converge towards the constant vector $0$ for $k \to \infty$.

Edit

If I got it correctly, the fixed point iteration they use in the article is generated by the following function:

G = R \[Function] 
   Join[{ 2. R[[1 ;;]].R[[1 ;; nn]]/(-1 R[[1]]^2)}, 
    Table[(2. R[[n + 1 ;;]].R[[1 ;; nn - n]] + 
        R[[2 ;; n + 1]].R[[n ;; 1 ;; -1]])/((n^2 Θ^2 + 
          2 α n Θ - 1) R[[1]]^2), {n, 1, 
      nn - 1}]];

This is how you can optain the list of the first 1000 iterates:

sequence = NestList[G, R0, 1000];

And indeed, the iterates seem to converge to a fixed point (the differences converge to the zero vector):

ListPlot[Max /@ Abs[Differences[sequence[[1 ;; 100]]]], PlotRange -> All]

enter image description here

I am surprised.

$\endgroup$
4
  • $\begingroup$ Would you be able to explain how and what the F function represents. If it converges towards zero, then it could be due to $\frac{1}{F_0}$. $\endgroup$ Commented Apr 2, 2018 at 0:26
  • $\begingroup$ I am not entirely sure, but the F function looks a bit weird in the table part. it seems like you also divide with $1/F_0^2$ on the bit to the left which you subtract. The bits with $((n^2 Θ^2 + 2 α n Θ - 1))$. $\endgroup$ Commented Apr 2, 2018 at 9:13
  • $\begingroup$ Ah, right. I corrected it. Still, FindRoot doesn't find a nontrivial solution with the given initial conditions... $\endgroup$ Commented Apr 2, 2018 at 10:41
  • $\begingroup$ okay, I will have a go at it and accept your answer. I think the method is right, it is just confusing me because the article says it is a easy iteration . $\endgroup$ Commented Apr 2, 2018 at 11:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.