# Box-ball soliton system?

1. Sep 27, 2007

### parsifal

Well, I don't have a clue on where to put this, but I'll go with this because solitons are a physical phenomenom, too. But I'll guess this is a wrong place for this anyway, so I apologize in advance.

Anyway, could somebody help me understand the box-ball soliton system (soliton, specifically, because I'm considering the form that doesn't disperse).

I've found very little information on box-ball soliton system. I've found this but I don't completely get it.

It seems to me that the last line with superscripts differs from the first colored line.

And it is said that: "Find the first maximal string of only 0s or only 1s no longer than the one to its immediate right, and denote its length m. Label each element of the string and the leftmost m elements of the adjacent string with the number m. Now remove these 2m elements from consideration, yielding a new system with fewer 1s. Repeat this process until all 1s (and as many 0s) have been labeled."

There is a line: 1 1 1 1 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 ...
So the first "11110000" is reduced from the line, resulting in: 0 1 1 0 1 0 0 0 1 0 0 ...

But why then, isn't the first zero taken into account?
1. "find the first maximal string of only 0s or only 1s"
2. "no longer that the one to its immediate right"

So the first 0 is a string of lenght 1 and the following string of 1s is of lenght 2.

Well.. I just don't get it.

How should the line 0 1 1 1 1 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
be labeled and how do the solitons move when colliding with each other?

It isn't probably anything complicated, but as said, I just don't get it.

2. Apr 11, 2011

### Evan Wilson

I have no clue as to the meaning of the text you quoted, but the type of box-ball system I know follows two simple time evolution rules:

1) Starting from the left, choose the first '1' not already moved and move it to the right across all the consecutive '1's to its left until you get to the first '0',

2) Replace the initial '1' with a '0' and the final '0' with a '1'. Once you have moved a '1' it is no longer available to be moved, but still will counts when moving any subsequent '1's.

Once you have done the above for all the '1's in the initial state, you have computed one time step of the box-ball system, and can repeat the procedure for as long as you want. For the example you gave

111100000110100010000111000000000000

becomes
t=1
000011110001011001000000111000000000

after one time step. Thereafter we get
t=2
000000001110100110110000000111000000
t=3
000000000001011001001111000000111000
t=4
000000000000100110100000111100000111

So we see the following behavior:
1) Longer solitons move faster than shorter ones,
2) After collision, solitons retain their same shape, but phase shifted from where they would
be if no exchange had taken place.

Hope this helps!

Last edited: Apr 11, 2011
3. Apr 11, 2011

### Evan Wilson

About the phrase: "Find the first maximal string of only 0s or only 1s no longer than the one to its immediate right, and denote its length m." Here, "maximal" means a string of '1's (respectively '0's) surrounded by '0's (respectively '1's). We select such a string of '0's (respectively '1's) if and only if the following (maximal) string of '1's (respectively '0's) has <=m elements. So in the example you gave, after removing the string 11110000 from consideration, we are left with 01101.... The initial '0' string is followed by '11', which clearly is longer than the string preceding it, so we skip it to the next '0' string followed by '1'. This can now be removed from consideration and we proceed recursively until all the '1' strings have been labeled (since there are infinitely many '0's we don't want to label all those).

Point being: a typical soliton "looks like" a string 00001111 or 11110000 with as many '0's as '1's. There may be many overlapping solitons, however. After collision, each soliton will be in the same position as it would have been without collision, but it may change from 00001111 to 11110000 or vice-versa.

Last edited: Apr 11, 2011