# Numerical calculation of Arnold tongue

Hello. I am working on investigation of family of dynamical systems on the torus $$\dot{x}=\cos(x)+b\cos(t)+a$$ $$\dot{t}=1$$ and it's Poincare map $$P:(x,0) \rightarrow (P(x),2\pi=0)$$ I need to find Arnold tongues of map $P$. I tried simple calculation of solution using Runge-Kutta formulas, then iterating and checking rotation number, but it's not working effectively. Arnold tongues was first calculated in 1970s so maybe there is effective algorithm of doing it?

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I'm working on the same family now; Rudolf, do we work in one seminar?;) – Olga Sep 27 '13 at 10:03

There is a good way to compute rotation number of a circle homeomorphism (this was the way Poincaré thinked of it): you calculate the rotation number buy its continued fraction in a direct way.

You start from a point $x$ and $f(x)$: this gives you a decomposition of the circle into points that are on the right side of $x$ (in $]x,f(x)[$) and points which are on its left side (in $]f(x),x[$). You look at $f^2(x)$ and you write $R$ if it is on the right side of $x$, $L$ otherwise. Iteranting $f$ you find a sequence of $R$'s and $L$'s. If you get $LLLLR$, for example, you record 4 (this is the number of $L$'s) and you approximate the rotation number of $f$ by $1/4$.

Renormalizing $f$, you iterate this process finding $\rho=[0,a_1,a_2,\ldots,a_k]$.

I won't be more precise here.

Every detail is very well explained in de Melo & van Strien's One-Dimensional Dynamics, section I.1.

You can find a paper by Bruin (Numerical determination of the continued fraction expansion of the rotation number) in which he compares different methods on Arnold tongues.

EDIT[update]: Recently, I wrote for myself some sage lines implementing the algorithm I described you. This is my second version, now working for rational numbers too : I was originally interested only in irrational rotation numbers (comments are welcome to improve it!).

L=8 #length for cf-expansion, depending on your computer, 8 or 9 suggested for a try run
A=100000 #maximum size of single element of the sequence

def partfrac(x):
return x-floor(x)

##### computing rational approximations given continued fraction expansion
# input b=a continued fraction expansion
# input l=L length of computed expansion
def rational_approximation(b,l):
p=[0,1]
q=[1,b[1]]
for i in range(1,l+1):
p.append(b[i+1]*p[i]+p[i-1])
q.append(b[i+1]*q[i]+q[i-1])
return simplify(p[l+1]/q[l+1])

#computing rotation number of a given circle map f
def rotation(f):
a=[0]
orbit=[]
orbit.append(partfrac(f(0)))
if orbit[0]==0 :
print 'map with a fixed point'
return 0

def shift(x):  #set f(0) as the origin + 1
if partfrac(x)>orbit[0]:
return partfrac(x)-1
return partfrac(x)

def first_return(p,pre_p,y):
x=shift(f(y))
while x<pre_p or x>p:
x=shift(f(x))
return x

a.append(1)
x=orbit[0]

if shift(f(orbit[0]))==0:
print 'map with periodic point of order 2'
return 1/2

if shift(f(orbit[0]))<0:
while shift(f(x))<0:
a[1]=a[1]+1
x = shift(f(x))
if a[1]>A:
print 'approximatively 0'
return 0
if shift(f(x))==0:
print 'periodic point'
a[1]=a[1]+1
return 1/a[1]
orbit.append(shift(x))
z = shift(f(x))
a.append(0)
while z>0:
y = z
z = first_return(shift(orbit[0]),shift(orbit[1]),z)
a[2]=a[2]+1
if a[2]>A:
print 'approximatively rational'
return 1/a[1]
if z==0:
print 'periodic point'
a[2]=a[2]+1
return rational_approximation(a,1)
orbit.append(y)

if shift(f(orbit[0]))>0:
def shift(y):  #set f(0) as the origin
if partfrac(y)>=orbit[0]:
return partfrac(y)-1
return partfrac(y)
orbit.append(orbit[0]-1)
a.append(0)
while shift(f(x))>0:
a[2] = a[2] + 1
x = shift(f(x))
if a[2]>A:
print 'approximatively rational'
return 1/a[1]
if shift(f(x))==0:
print 'periodic point'
a[2]=a[2]+1
return rational_approximation(a,1)
orbit.append(shift(x))
z = shift(f(x))

for i in range(1,L):
a.append(0)
if shift(orbit[i+1])<shift(orbit[i]):
while z>0:
y = z
z = first_return(shift(orbit[i]),shift(orbit[i+1]),z)
a[i+2]=a[i+2]+1
if a[i+2]>A:
print 'approximatively rational'
return rational_approximation(a,i)
if z==0:
print 'periodic point'
a[i+2]=a[i+2]+1
return rational_approximation(a,i+1)
if shift(orbit[i+1])>shift(orbit[i]):
while z<0:
y = z
z = first_return(shift(orbit[i+1]),shift(orbit[i]),z)
a[i+2]=a[i+2]+1
if a[i+2]>A:
print 'approximatively rational'
return rational_approximation(a,i)
if z==0:
print 'periodic point'
a[i+2]=a[i+2]+1
return rational_approximation(a,i+1)
orbit.append(y)

print a
return rational_approximation(a,L)

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@Michele Thanks a lot! Send this algorithm to me please. – Rudolf Dec 14 '12 at 15:30