I have a curve as shown below:
The x coordinates and the y coordinates for this plot are:
path_x= (4.0, 5.638304088577984, 6.785456961280076, 5.638304088577984, 4.0)
path_y =(0.0, 1.147152872702092, 2.7854569612800755, 4.423761049858059, 3.2766081771559668)
And I obtained the above picture by:
x_min =min(path_x)-1
x_max =max(path_x)+1
y_min =min(path_y)-1
y_max =max(path_y)+1
num_pts = len(path_x)
fig = plt.figure(figsize=(8,8))
#fig = plt.figure()
plt.suptitle("Curve and the boundary")
ax = fig.add_subplot(1,1,1)
ax.set_xlim([min(x_min,y_min),max(x_max,y_max)])
ax.set_ylim([min(x_min,y_min),max(x_max,y_max)])
ax.plot(path_x,path_y)
Now my intention is to draw a smooth curve using cubic splines. But looks like for cubic splines you need the x coordinates to be on ascending order. whereas in this case, neither x values nor y values are in the ascending order.
Also this is not a function. That is an x value is mapped with more than one element in the range.
I also went over this post. But I couldn't figure out a proper method to solve my problem.
I really appreciate your help in this regard
As suggested in the comments, you can always parameterize any curve/surface with an arbitrary (and linear!) parameter.
For example, define t as a parameter such that you get x=x(t) and y=y(t). Since t is arbitrary, you can define it such that at t=0, you get your first path_x[0],path_y[0], and at t=1, you get your last pair of coordinates, path_x[-1],path_y[-1].
Here is a code using scipy.interpolate
import numpy
import scipy.interpolate
import matplotlib.pyplot as plt
path_x = numpy.asarray((4.0, 5.638304088577984, 6.785456961280076, 5.638304088577984, 4.0),dtype=float)
path_y = numpy.asarray((0.0, 1.147152872702092, 2.7854569612800755, 4.423761049858059, 3.2766081771559668),dtype=float)
# defining arbitrary parameter to parameterize the curve
path_t = numpy.linspace(0,1,path_x.size)
# this is the position vector with
# x coord (1st row) given by path_x, and
# y coord (2nd row) given by path_y
r = numpy.vstack((path_x.reshape((1,path_x.size)),path_y.reshape((1,path_y.size))))
# creating the spline object
spline = scipy.interpolate.interp1d(path_t,r,kind='cubic')
# defining values of the arbitrary parameter over which
# you want to interpolate x and y
# it MUST be within 0 and 1, since you defined
# the spline between path_t=0 and path_t=1
t = numpy.linspace(numpy.min(path_t),numpy.max(path_t),100)
# interpolating along t
# r[0,:] -> interpolated x coordinates
# r[1,:] -> interpolated y coordinates
r = spline(t)
plt.plot(path_x,path_y,'or')
plt.plot(r[0,:],r[1,:],'-k')
plt.xlabel('x')
plt.ylabel('y')
plt.show()
With output
For non-ascending x splines can be easily computed if you make both x and y functions of another parameter t: x(t), y(t).
In your case you have 5 points so t should be just enumeration of these points, i.e. t = 0, 1, 2, 3, 4 for 5 points.
So if x = [5, 2, 7, 3, 6] then x(t) = x(0) = 5, x(1) = 2, x(2) = 7, x(3) = 3, x(4) = 6. Same for y.
Then compute spline function for both x(t) and y(t). Afterwards compute values of splines in all many intermediate t points. Lastly just use all calculated values x(t) and y(t) as a function y(x).
Once before I implemented cubic spline computation from scratch using Numpy, so I use this code in my example below if you don't mind (it could be useful for you to learn about spline math), replace with your library functions. Also in my code you can see numba lines commented out, if you want you can use these Numba annotations to speed up computation.
You have to look at main() function at the bottom of code, it shows how to compute and use x(t) and y(t).
Try it online!
import numpy as np, matplotlib.pyplot as plt
# Solves linear system given by Tridiagonal Matrix
# Helper for calculating cubic splines
##numba.njit(cache = True, fastmath = True, inline = 'always')
def tri_diag_solve(A, B, C, F):
n = B.size
assert A.ndim == B.ndim == C.ndim == F.ndim == 1 and (
A.size == B.size == C.size == F.size == n
) #, (A.shape, B.shape, C.shape, F.shape)
Bs, Fs = np.zeros_like(B), np.zeros_like(F)
Bs[0], Fs[0] = B[0], F[0]
for i in range(1, n):
Bs[i] = B[i] - A[i] / Bs[i - 1] * C[i - 1]
Fs[i] = F[i] - A[i] / Bs[i - 1] * Fs[i - 1]
x = np.zeros_like(B)
x[-1] = Fs[-1] / Bs[-1]
for i in range(n - 2, -1, -1):
x[i] = (Fs[i] - C[i] * x[i + 1]) / Bs[i]
return x
# Calculate cubic spline params
##numba.njit(cache = True, fastmath = True, inline = 'always')
def calc_spline_params(x, y):
a = y
h = np.diff(x)
c = np.concatenate((np.zeros((1,), dtype = y.dtype),
np.append(tri_diag_solve(h[:-1], (h[:-1] + h[1:]) * 2, h[1:],
((a[2:] - a[1:-1]) / h[1:] - (a[1:-1] - a[:-2]) / h[:-1]) * 3), 0)))
d = np.diff(c) / (3 * h)
b = (a[1:] - a[:-1]) / h + (2 * c[1:] + c[:-1]) / 3 * h
return a[1:], b, c[1:], d
# Spline value calculating function, given params and "x"
##numba.njit(cache = True, fastmath = True, inline = 'always')
def func_spline(x, ix, x0, a, b, c, d):
dx = x - x0[1:][ix]
return a[ix] + (b[ix] + (c[ix] + d[ix] * dx) * dx) * dx
# Compute piece-wise spline function for "x" out of sorted "x0" points
##numba.njit([f'f{ii}[:](f{ii}[:], f{ii}[:], f{ii}[:], f{ii}[:], f{ii}[:], f{ii}[:])' for ii in (4, 8)],
# cache = True, fastmath = True, inline = 'always')
def piece_wise_spline(x, x0, a, b, c, d):
xsh = x.shape
x = x.ravel()
ix = np.searchsorted(x0[1 : -1], x)
y = func_spline(x, ix, x0, a, b, c, d)
y = y.reshape(xsh)
return y
def main():
x0 = np.array([4.0, 5.638304088577984, 6.785456961280076, 5.638304088577984, 4.0])
y0 = np.array([0.0, 1.147152872702092, 2.7854569612800755, 4.423761049858059, 3.2766081771559668])
t0 = np.arange(len(x0)).astype(np.float64)
plt.plot(x0, y0)
vs = []
for e in (x0, y0):
a, b, c, d = calc_spline_params(t0, e)
x = np.linspace(0, t0[-1], 100)
vs.append(piece_wise_spline(x, t0, a, b, c, d))
plt.plot(vs[0], vs[1])
plt.show()
if __name__ == '__main__':
main()
Output:
I am looking to integrate the difference between my numerical and exact solution to the heat equation though I am not sure what would be the best to way to tackle this. Is there a specific integrator that would allow me to do this ?
I hope to integrate it wrt to $x$.
I have the following code so far:
import numpy as np
from matplotlib import pyplot
from mpl_toolkits.mplot3d import Axes3D
from scipy import linalg
import matplotlib.pyplot as plt
import math
def initial_data(x):
y = np.zeros_like(x)
for i in range(len(x)):
if (x[i] < 0.25):
y[i] = 0.0
elif (x[i] < 0.5):
y[i] = 4.0 * (x[i] - 0.25)
elif (x[i] < 0.75):
y[i] = 4.0 * (0.75 - x[i])
else:
y[i] = 0.0
return y
def heat_exact(x, t, kmax = 150):
"""Exact solution from separation of variables"""
yexact = np.zeros_like(x)
for k in range(1, kmax):
d = -8.0*
(np.sin(k*np.pi/4.0)-2.0*np.sin(k*np.pi/2.0)+np.sin(3.0*k*np.pi/4.0))/((np.pi*k)**2)
yexact += d*np.exp(-(k*np.pi)**2*t)*np.sin(k*np.pi*x)
return yexact
def ftcs_heat(y, ynew, s):
ynew[1:-1] = (1 - 2.0 * s) * y[1:-1] + s * (y[2:] + y[:-2])
# Trivial boundary conditions
ynew[0] = 0.0
ynew[-1] = 0.0
Nx = 198
h = 1.0 / (Nx + 1.0)
t_end = 0.25
s = 1.0 / 3.0 # s = delta / h^2
delta = s * h**2
Nt = int(t_end / delta)+1
x = np.linspace(0.0, 1.0, Nx+2)
y = initial_data(x)
ynew = np.zeros_like(y)
for n in range(Nt):
ftcs_heat(y, ynew, s)
y = ynew
fig = plt.figure(figsize=(8,6))
ax = fig.add_subplot(111)
x_exact = np.linspace(0.0, 1.0, 200)
ax.plot(x, y, 'kx', label = 'FTCS')
ax.plot(x_exact, heat_exact(x_exact, t_end), 'b-', label='Exact solution')
ax.legend()
plt.show()
diff = (y - heat_exact(x_exact,t_end))**2 # squared difference between numerical and exact solutions
x1 = np.trapz(diff, x=x) #(works!)
import scipy.integrate as integrate
x1 = integrate.RK45(diff,diff[0],0,t_end) #(preferred but does not work)
What I am looking to integrate is the variable diff (the squared difference). Any suggestions are welcomed, thanks.
Edit: I would like to use RK45 method however I am not sure what should my y0 be, I have tried x1 = integrate.RK45(diff,diff[0],0,t_end) but get the following output error:
raise ValueError("`y0` must be 1-dimensional.")
ValueError: `y0` must be 1-dimensional.
By integration you mean you want to find the area between y and heat_exact? Or do you want to know if they are the same within a specific limit? The latter can be found with numpy.isclose. The former you can use several integration functions builtin numpy.
For example:
np.trapz(diff, x=x)
Oh, shouldn't the last line be diff = (y - heat_exact(x_exact,t_end))**2? My integration of this diff gave 8.32E-12, which looks right judging by the plots you gave me.
Check out also scipy.integrate
Currently, I solve the following ODE system of equations using odeint
dx/dt = (-x + u)/2.0
dy/dt = (-y + x)/5.0
initial conditions: x = 0, y = 0
However, I would like to use solve_ivp which seems to be the recommended option for this type of problems, but honestly I don't know how to adapt the code...
Here is the code I'm using with odeint:
import numpy as np
from scipy.integrate import odeint, solve_ivp
import matplotlib.pyplot as plt
def model(z, t, u):
x = z[0]
y = z[1]
dxdt = (-x + u)/2.0
dydt = (-y + x)/5.0
dzdt = [dxdt, dydt]
return dzdt
def main():
# initial condition
z0 = [0, 0]
# number of time points
n = 401
# time points
t = np.linspace(0, 40, n)
# step input
u = np.zeros(n)
# change to 2.0 at time = 5.0
u[51:] = 2.0
# store solution
x = np.empty_like(t)
y = np.empty_like(t)
# record initial conditions
x[0] = z0[0]
y[0] = z0[1]
# solve ODE
for i in range(1, n):
# span for next time step
tspan = [t[i-1], t[i]]
# solve for next step
z = odeint(model, z0, tspan, args=(u[i],))
# store solution for plotting
x[i] = z[1][0]
y[i] = z[1][1]
# next initial condition
z0 = z[1]
# plot results
plt.plot(t,u,'g:',label='u(t)')
plt.plot(t,x,'b-',label='x(t)')
plt.plot(t,y,'r--',label='y(t)')
plt.ylabel('values')
plt.xlabel('time')
plt.legend(loc='best')
plt.show()
main()
It's important that solve_ivp expects f(t, z) as right-hand side of the ODE. If you don't want to change your ode function and also want to pass your parameter u, I recommend to define a wrapper function:
def model(z, t, u):
x = z[0]
y = z[1]
dxdt = (-x + u)/2.0
dydt = (-y + x)/5.0
dzdt = [dxdt, dydt]
return dzdt
def odefun(t, z):
if t < 5:
return model(z, t, 0)
else:
return model(z, t, 2)
Now it's easy to call solve_ivp:
def main():
# initial condition
z0 = [0, 0]
# number of time points
n = 401
# time points
t = np.linspace(0, 40, n)
# step input
u = np.zeros(n)
# change to 2.0 at time = 5.0
u[51:] = 2.0
res = solve_ivp(fun=odefun, t_span=[0, 40], y0=z0, t_eval=t)
x = res.y[0, :]
y = res.y[1, :]
# plot results
plt.plot(t,u,'g:',label='u(t)')
plt.plot(t,x,'b-',label='x(t)')
plt.plot(t,y,'r--',label='y(t)')
plt.ylabel('values')
plt.xlabel('time')
plt.legend(loc='best')
plt.show()
main()
Note that without passing t_eval=t, the solver will automatically choose the time points inside tspan at which the solution will be stored.
I am trying to use numpy and scipy to solve the following two equations:
P(z) = sgn(-cos(np.pi*D1) + cos(5*z)) * sgn(-cos(np.pi*D2) + cos(6*z))
1. 0 = 2/2pi ∫ P(z,D1,D2) * cos(5z) dz + z/L
2. 0 = 2/2pi ∫ P(z,D1,D2) * cos(6z) dz - z/L
for D1 and D2 (integral limits are 0 -> 2pi).
My code is:
def equations(p, z):
D1, D2 = p
period = 2*np.pi
P1 = lambda zz, D1, D2: \
np.sign(-np.cos(np.pi*D1) + np.cos(6.*zz)) * \
np.sign(-np.cos(np.pi*D2) + np.cos(5.*zz)) * \
np.cos(6.*zz)
P2 = lambda zz, D1, D2: \
np.sign(-np.cos(np.pi*D1) + np.cos(6.*zz)) * \
np.sign(-np.cos(np.pi*D2) + np.cos(5.*zz)) * \
np.cos(5.*zz)
eq1 = 2./period * integrate.quad(P1, 0., period, args=(D1,D2), epsabs=0.01)[0] + z
eq2 = 2./period * integrate.quad(P2, 0., period, args=(D1,D2), epsabs=0.01)[0] - z
return (eq1, eq2)
z = np.arange(0., 1000., 0.01)
N = int(len(z))
D1 = np.empty([N])
D2 = np.empty([N])
for i in range(N):
D1[i], D2[i] = fsolve(equations, x0=(0.5, 0.5), args=z[i])
print D1, D2
Unfortunately, it does not seem to converge. I don't know much about numerical methods and was hoping someone could give me a hand.
Thank you.
P.S. I'm also trying the following which should be equivalent:
import numpy as np
from scipy.optimize import fsolve
from scipy import integrate
from scipy import signal
def equations(p, z):
D1, D2 = p
period = 2.*np.pi
K12 = 1./L * z
K32 = -1./L * z + 1.
P1 = lambda zz, D1, D2: \
signal.square(6.*zz, duty=D1) * \
signal.square(5.*zz, duty=D2) * \
np.cos(6.*zz)
P2 = lambda zz, D1, D2: \
signal.square(6.*zz, duty=D1) * \
signal.square(5.*zz, duty=D2) * \
np.cos(5.*zz)
eq1 = 2./period * integrate.quad(P1, 0., period, args=(D1,D2))[0] + K12
eq2 = 2./period * integrate.quad(P2, 0., period, args=(D1,D2))[0] - K32
return (eq1, eq2)
h = 0.01
L = 10.
z = np.arange(0., L, h)
N = int(len(z))
D1 = np.empty([N])
D2 = np.empty([N])
for i in range(N):
D1[i], D2[i] = fsolve(equations, x0=(0.5, 0.5), args=z[i])
print
print z[i]
print ("%0.8f,%0.8f" % (D1[i], D2[i]))
print
PSS:
I implemented what you wrote (I think I understand it!), very nicely done. Thank you. Unfortunately, I really don't have much skill in this field and don't really know how to make a suitable guess, so I just guess 0.5 (I also added a small amount of noise to the initial guess to try and improve it). The result I'm getting have numerical errors it seems, and I'm not sure why, I was hoping you could point me in the right direction. So essentially, I did an FFT sweep (did an FFT for each dutycycle variation and looked at the frequency component at 5, which is shown below in the graph) and found that the linear part (z/L) is slightly jagged.
PSSS:
Thank you for that, I've noted some of the techniques you've suggested. I tried replicated your second graph as it seems very useful. To do this, I kept D1 (D2) fixed and swept D2 (D1), and I did this for various z values. fmin did not always find the correct minimum (it was dependent on the initial guess) so I swept the initial guess of fmin until I found the correct answer. I get a similar answer to you. (I think it's correct?)
Also, I would just like to say that you might like to give me your contact details, as this solution as a step in finding the solution to a problem I have (I'm a student doing research), and I will most certainly acknowledge you in any papers in which this code is used.
#!/usr/bin/env python
import numpy as np
from scipy.optimize import fsolve
from scipy import integrate
from scipy import optimize
from scipy import signal
######################################################
######################################################
altsigns = np.ones(50)
altsigns[1::2] = -1
def get_breaks(x, y, a, b):
sa = np.arange(0, 2*a, 2)
sb = np.arange(0, 2*b, 2)
zx = (( x + sa) % (2*a))*np.pi/a
zx2 = ((-x + sa) % (2*a))*np.pi/a
zy = (( y + sb) % (2*b))*np.pi/b
zy2 = ((-y + sb) % (2*b))*np.pi/b
zi = np.r_[np.sort(np.hstack((zx, zx2, zy, zy2))), 2*np.pi]
if zi[0]:
zi = np.r_[0, zi]
return zi
def integrals(x, y, a, b):
zi = get_breaks(x % 1., y % 1., a, b)
sins = np.vstack((np.sin(b*zi), np.sin(a*zi)))
return (altsigns[:zi.size-1]*(sins[:,1:] - sins[:,:-1])).sum(1) / np.array((b, a))
def equation1(p, z, d2):
D2 = d2
D1 = p
I1, _ = integrals(D1, D2, deltaK1, deltaK2)
eq1 = 1. / np.pi * I1 + z
return abs(eq1)
def equation2(p, z, d1):
D1 = d1
D2 = p
_, I2 = integrals(D1, D2, deltaK1, deltaK2)
eq2 = 1. / np.pi * I2 - z + 1
return abs(eq2)
######################################################
######################################################
z = [0.2, 0.4, 0.6, 0.8, 1.0]#np.arange(0., 1., 0.1)
step = 0.05
deltaK1 = 5.
deltaK2 = 6.
f = open('data.dat', 'w')
D = np.arange(0.0, 1.0, step)
D1eq1 = np.empty([len(D)])
D2eq2 = np.empty([len(D)])
D1eq1Err = np.empty([len(D)])
D2eq2Err = np.empty([len(D)])
for n in z:
for i in range(len(D)):
# Fix D2 and solve for D1.
for guessD1 in np.arange(0.,1.,0.1):
D2 = D
tempD1 = optimize.fmin(equation1, guessD1, args=(n, D2[i]), disp=False, xtol=1e-8, ftol=1e-8, full_output=True)
if tempD1[1] < 1.e-6:
D1eq1Err[i] = tempD1[1]
D1eq1[i] = tempD1[0][0]
break
else:
D1eq1Err[i] = -1.
D1eq1[i] = -1.
# Fix D1 and solve for D2.
for guessD2 in np.arange(0.,1.,0.1):
D1 = D
tempD2 = optimize.fmin(equation2, guessD2, args=(n, D1[i]), disp=False, xtol=1e-8, ftol=1e-8, full_output=True)
if tempD2[1] < 1.e-6:
D2eq2Err[i] = tempD2[1]
D2eq2[i] = tempD2[0][0]
break
else:
D2eq2Err[i] = -2.
D2eq2[i] = -2.
for i in range(len(D)):
f.write('%0.8f,%0.8f,%0.8f,%0.8f,%0.8f\n' %(D[i], D1eq1[i], D2eq2[i], D1eq1Err[i], D2eq2Err[i]))
f.write('\n\n')
f.close()
This is a very ill-posed problem. Let's recap what you are trying to do:
You want to solve 100000 optimization problems
Each optimization problem is 2 dimensional, so you need O(10000) function evaluations (estimating O(100) function evaluations for a 1D optimization problem)
Each function evaluation depends on the evaluation of two numerical integrals
The integrands contain jumps, i.e. they are 0-times contiguously differentiable
The integrands are composed of periodic functions, so they have multiple minima and maxima
So you are off to a very hard time. In addition, even in the most optimistic estimate in which all factors in the integrand that are < 1 are replaced by 1, the integrals can only take values between -2*pi and 2*pi. Much less than that in reality. So you can already see that you only have a chance of a solution for
I1 - z = 0
I2 + z = 0
for very small numbers of z. So there is no point in trying up to z = 1000.
I am almost certain that this is not the problem you need to solve. (I cannot imagine a context in which such a problem would appear. It seems like a weird twist on Fourier coefficient computation...) But in case you insist, your best bet is to work on the inner loop first.
As you noted, the numerical evaluation of the integrals is subject to large errors. This is due to the jumps introduced by the sgn() function. Functions such as scipy.integrate.quad() tend to use higher order algorithms which assume that the integrands are smooth. If they are not, they perform very badly. You either need to hand-pick an algorithm that can deal with jumps or, much better in this case, do the integrals by hand:
The following algorithm calculates the jump points of the sgn() function and then evaluates the analytic integrals on all pieces:
altsigns = np.ones(50)
altsigns[1::2] = -1
def get_breaks(x, y, a, b):
sa = np.arange(0, 2*a, 2)
sb = np.arange(0, 2*b, 2)
zx = (( x + sa) % (2*a))*np.pi/a
zx2 = ((-x + sa) % (2*a))*np.pi/a
zy = (( y + sb) % (2*b))*np.pi/b
zy2 = ((-y + sb) % (2*b))*np.pi/b
zi = np.r_[np.sort(np.hstack((zx, zx2, zy, zy2))), 2*pi]
if zi[0]:
zi = np.r_[0, zi]
return zi
def integrals(x, y, a, b):
zi = get_breaks(x % 1., y % 1., a, b)
sins = np.vstack((np.sin(b*zi), np.sin(a*zi)))
return (altsigns[:zi.size-1]*(sins[:,1:] - sins[:,:-1])).sum(1) / np.array((b, a))
This gets rid of the problem of the numerical integration. It is very accurate and fast. However, even the integrals will not be perfectly contiguous for all parameters, so in order to solve your optimization problem, you are better off using an algorithm that doesn't rely on the existence of any derivatives. The only choice in scipy is scipy.optimize.fmin(), which you can use like:
def equations2(p, z):
x, y = p
I1, I2 = integrals(x, y, 6., 5.)
fact = 1. / pi
eq1 = fact * I1 + z
eq2 = fact * I2 - z
return eq1, eq2
def norm2(p, z):
eq1, eq2 = equations2(p, z)
return eq1**2 + eq2**2 # this has the minimum when eq1 == eq2 == 0
z = 0.25
res = fmin(norm2, (0.25, 0.25), args=(z,), xtol=1e-8, ftol=1e-8)
print res
# -> [ 0.3972 0.5988]
print equations2(res, z)
# -> (-2.7285737558280232e-09, -2.4748670890417657e-09)
You are still left with the problem of finding suitable starting values for all z, which is still a tricky business. Good Luck!
Edit
To check if you still have numerical errors, plug the result of the optimization back in the equations and see if they are satisfied to the required accuracy, which is what I did above. Note that I used (0.25, 0.25) as a starting value, since starting at (0.5, 0.5) didn't lead to convergence. This is normal for optimizations problems with local minima (such as yours). There is no better way to deal with this other than trying multiple starting values, rejecting non-converged results. In the case above, if equations2(res, z) returns anything higher than, say, (1e-6, 1e-6), I would reject the result and try again with a different starting value. A very useful technique for successive optimization problems is to use the result of the previous problem as the starting value for the next problem.
Note however that you have no guarantee of a smooth solution for D1(z) and D2(z). Just a tiny change in D1 could push one break point off the integration interval, resulting in a big change of the value of the integral. The algorithm may well adjust by using D2, leading to jumps in D1(z) and D2(z). Note also that you can take any result modulo 1, due to the symmetries of cos(pi*D1).
The bottom line: There shouldn't be any remaining numerical inaccuracies if you use the analytical formula for the integrals. If the residuals are less than the accuracy you specified, this is your solution. If they are not, you need to find better starting values. If you can't, a solution may not exist. If the solutions are not contiguous as a function of z, that is also expected, since your integrals are not contiguous. Good luck!
Edit 2
It appears your equations have two solutions in the interval z in [0, ~0.46], and no solutions for z > 0.46, see the first figure below. To prove this, see the good old graphical solution in the second figure below. The contours represent solutions of Eq. 1 (vertical) and Eq. 2 (horizontal), for different z. You can see that the contours cross twice for z < 0.46 (two solutions) and not at all for z > 0.46 (no solution that simultaneously satisfies both equations). If this is not what you expected, you need to write down different equations (which was my suspicion in the first place...)
Here is the final code I was using:
import numpy as np
from numpy import sin, cos, sign, pi, arange, sort, concatenate
from scipy.optimize import fmin
a = 6.0
b = 5.0
def P(z, x, y):
return sign((cos(a*z) - cos(pi*x)) * (cos(b*z) - cos(pi*y)))
def P1(z, x, y):
return P(z, x, y) * cos(b*z)
def P2(z, x, y):
return P(z, x, y) * cos(a*z)
altsigns = np.ones(50)
altsigns[1::2] = -1
twopi = 2*pi
pi_a = pi/a
da = 2*pi_a
pi_b = pi/b
db = 2*pi_b
lim = np.array([0., twopi])
def get_breaks(x, y):
zx = arange(x*pi_a, twopi, da)
zx2 = arange((2-x)*pi_a, twopi, da)
zy = arange(y*pi_b, twopi, db)
zy2 = arange((2-y)*pi_b, twopi, db)
zi = sort(concatenate((lim, zx, zx2, zy, zy2)))
return zi
ba = np.array((b, a))[:,None]
fact = np.array((1. / b, 1. / a))
def integrals(x, y):
zi = get_breaks(x % 1., y % 1.)
sins = sin(ba*zi)
return fact * (altsigns[:zi.size-1]*(sins[:,1:] - sins[:,:-1])).sum(1)
def equations2(p, z):
x, y = p
I1, I2 = integrals(x, y)
fact = 1. / pi
eq1 = fact * I1 + z
eq2 = fact * I2 - z
return eq1, eq2
def norm2(p, z):
eq1, eq2 = equations2(p, z)
return eq1**2 + eq2**2
def eval_integrals(Nx=100, Ny=101):
x = np.arange(Nx) / float(Nx)
y = np.arange(Ny) / float(Ny)
I = np.zeros((Nx, Ny, 2))
for i in xrange(Nx):
xi = x[i]
Ii = I[i]
for j in xrange(Ny):
Ii[j] = integrals(xi, y[j])
return x, y, I
def solve(z, start=(0.25, 0.25)):
N = len(z)
res = np.zeros((N, 2))
res.fill(np.nan)
for i in xrange(N):
if i < 100:
prev = start
prev = fmin(norm2, prev, args=(z[i],), xtol=1e-8, ftol=1e-8)
if norm2(prev, z[i]) < 1e-7:
res[i] = prev
else:
break
return res
#x, y, I = eval_integrals(Nx=1000, Ny=1001)
#zlvl = np.arange(0.2, 1.2, 0.2)
#contour(x, y, -I[:,:,0].T/pi, zlvl)
#contour(x, y, I[:,:,1].T/pi, zlvl)
N = 1000
z = np.linspace(0., 1., N)
res = np.zeros((N, 2, 2))
res[:,0,:] = solve(z, (0.25, 0.25))
res[:,1,:] = solve(z, (0.05, 0.95))