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Step by step time integrators in Python

I am solving a first order initial value problem of the form:
dy/dt = f(t,y(t)), y(0)=y0
I would like to obtain y(n+1) from a given numerical scheme, like for example :
using explicit Euler's scheme, we have
y(i) = y(i-1) + f(t-1,y(t-1)) * dt
Example code:
# Test code to evaluate different time integrators for the following equation:
# y' = (1/2) y + 2sin(3t) ; y(0) = -24/37
def dy_dt(y,t):
func = (1/2)*y + 2*np.sin(3*t)
return func
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
tmin = 0
tmax = 50
delt= 1e-2
t = np.arange(tmin,tmax,delt)
total_steps = len(t)
y_explicit=np.zeros(total_steps)
#y_ODEint=np.zeros(total_steps)
y0 = -24/37
y_explicit[0]=y0
#y_ODEint[0]=y0
# exact solution
y_exact = -(24/37)*np.cos(3*t)- (4/37)*np.sin(3*t) + (y0+24/37)*np.exp(0.5*t)
# Solution using ODEint Python
y_ODEint = odeint(dy_dt,y0,t)
for i in range(1,total_steps):
# Explicit scheme
y_explicit[i] = y_explicit[i-1] + (dy_dt(y_explicit[i-1],t[i-1]))*delt
# Update using ODEint
# y_ODEint[i] = odeint(dy_dt,y_ODEint[i-1],[0,delt])[-1]
plt.figure()
plt.plot(t,y_exact)
plt.plot(t,y_explicit)
# plt.plot(t,y_ODEint)
The current issue I am having is that the functions like ODEint in python provide the entire y(t) as opposed to y(i). like in the line "y_ODEint = odeint(dy_dt,y0,t)"
See in the code, how I have coded the explicit scheme, which gives y(i) for every time step. I want to do the same with ODEint, i tried something but didn't work (all commented lines)
I want to obtain y(i) rather than all ys using ODEint. Is that possible ?

Your system is time variant so you cannot translate the time step from (t[i-1], t[i]) to (0, delt).
The step by step integration will is unstable for your differential equation though
Here is what I get
def dy_dt(y,t):
func = (1/2)*y + 2*np.sin(3*t)
return func
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
tmin = 0
tmax = 40
delt= 1e-2
t = np.arange(tmin,tmax,delt)
total_steps = len(t)
y_explicit=np.zeros(total_steps)
#y_ODEint=np.zeros(total_steps)
y0 = -24/37
y_explicit[0]=y0
# exact solution
y_exact = -(24/37)*np.cos(3*t)- (4/37)*np.sin(3*t) + (y0+24/37)*np.exp(0.5*t)
# Solution using ODEint Python
y_ODEint = odeint(dy_dt,y0,t)
# To be filled step by step
y_ODEint_2 = np.zeros_like(y_ODEint)
y_ODEint_2[0] = y0
for i in range(1,len(y_ODEint_2)):
# update your code to run with the correct time interval
y_ODEint_2[i] = odeint(dy_dt,y_ODEint_2[i-1],[tmin+(i-1)*delt,tmin+i*delt])[-1]
plt.figure()
plt.plot(t,y_ODEint, label='single run')
plt.plot(t,y_ODEint_2, label='step-by-step')
plt.plot(t, y_exact, label='exact')
plt.legend()
plt.ylim([-20, 20])
plt.grid()
Important to notice that both methods are unstable, but the step-by-step explodes slightly before than the single odeint call.
With, for example dy_dt(y,t): -(1/2)*y + 2*np.sin(3*t) the integration becomes more stable, for instance, there is no noticeable error after integrating from zero to 200.

How to include known parameter that changes over time in solve_bvp

I am trying to use scipy's solve_bvp in python to solve differential equations that depend on a known parameter that changes over time. I have this parameter saved in a numpy array. However, when I try to use this array in the derivatives function, I get the following error ValueError: operands could not be broadcast together with shapes (10,) (11,).
Below is a simplified version of my code. I want the variable d2 to take certain values at different times according to an array, d2_set_values. The differential equations for some of the 12 variables then depend on d2. I hope it's clear from this code what I'm trying to achieve.
import numpy as np
from scipy.integrate import solve_bvp
t = np.linspace(0, 10, 11)
# Known parameter that changes over time
d2_set_values = np.zeros(t.size)
d2_set_values[:4] = 0.1
d2_set_values[4:8] = 0.2
d2_set_values[8:] = 0.1
# Initialise y vector
y = np.zeros((12, t.size))
# ODEs
def fun(x, y):
S1, I1, R1, S2, I2, R2, lamS1, lamI1, lamR1, lamS2, lamI2, lamR2 = y
d1 = 0.5*(I1 + 0.1*I2)*(lamS1 - lamI1)
d2 = d2_set_values
dS1dt = -0.5*S1*(1-d1)*(I1 + 0.1*I2)
dS2dt = -0.5*S2*(1-d2)*(I2 + 0.1*I1)
dI1dt = 0.5*S1*(1-d1)*(I1 + 0.1*I2) - 0.2*I1
dI2dt = 0.5*S2*(1-d2)*(I2 + 0.1*I1) - 0.2*I2
dR1dt = 0.2*I1
dR2dt = 0.2*I2
dlamS1dt = 0.5*(1-d1)*S1*lamS1
dlamS2dt = 0.5*(1-d2)*S2*lamS2
dlamI1dt = 0.5*(1-d1)*I1*lamI1
dlamI2dt = 0.5*(1-d2)*I2*lamI2
dlamR1dt = lamR1
dlamR2dt = lamR2
return np.vstack((dS1dt, dI1dt, dR1dt, dS2dt, dI2dt, dR2dt, dlamS1dt, dlamI1dt, dlamR1dt, dlamS2dt, dlamI2dt, dlamR2dt))
# Boundary conditions
def bc(ya, yb):
return np.array([ya[0]-0.99, ya[1]-0.01, ya[2]-0., ya[3]-1.0, ya[4]-0., ya[5]-0.,
yb[6]-0., yb[7]-1., yb[8]-0., yb[9]-0, yb[10]-0, yb[11]-0])
# Run the solver
sol = solve_bvp(fun, bc, t, y)
I have even tried reducing the size of d2_set_values by one, but that doesn't solve the issue.
Any help I can get would be much appreciated!

Why `numpy.fft.irfft` is so imprecise?

I understand that most FFT/IFFT routines have an error floor. I was expecting NumPy's FFT to have an error floor in the same orders as FFTW (say 1e-15), but the following experiment shows errors in the order of 1e-5.
Consider calculating the IDFT of a box. It is well-known that the result is the sinc-like Dirichlet kernel. But that is not what I get from numpy.fft.irfft. In fact even the first sample that should simply equal the width of the box divided by the number of FFT points is off by an amount around 4e-5 as the following example shows:
import numpy as np
import matplotlib.pyplot as plt
from scipy.special import diric
N = 40960
K = 513
X = np.ones(K, dtype=np.complex)
x = np.fft.irfft(X, N)
print("x[0] = %g: expected %g - error = %g" % (x[0], (2*K+1)/N, x[0]-(2*K+1)/N))
# expected IDFT of a box is Dirichlet function (see
# https://en.wikipedia.org/wiki/Discrete_Fourier_transform#Some_discrete_Fourier_transform_pairs)
y = diric(2*np.pi*np.arange(N)/N, 2*K+1) * (2*K+1) / N
plt.figure()
plt.plot(x[:1024] - y[:1024])
plt.title('error')
plt.show(block=True)
It looks like the error is of sinusoidal form:
Has anybody experience same issue? Am I misunderstanding something about the NumPy's FFT pack or it is just not accurate?
Update
Here is the equivalent of part of the script in Octave:
N = 40960;
K = 513;
X = zeros(1, N);
X(1:K) = 1;
X(N-K:N) = 1;
x = ifft(X);
fprintf("x[0] = %g, expected = %g - error = %g\n", x(1), (2*K+1)/N, x(1)-(2*K+1)/N);
The error on x[0] is practically zero in Octave. (I did not check other samples because I am not aware of equivalent of diric function in Octave.)

Thanks to MarkDickinson, I realized that my math was wrong. The correct comparison would be carried out by:
import numpy as np
import matplotlib.pyplot as plt
from scipy.special import diric
N = 40960
K = 513
X = np.ones(K+1, dtype=np.complex)
x = np.fft.irfft(X, N)
print("x[0] = %g: expected %g - error = %g" % (x[0], (2*K+1)/N, x[0]-(2*K+1)/N))
# expected IDFT of a box is Dirichlet function (see
# https://en.wikipedia.org/wiki/Discrete_Fourier_transform#Some_discrete_Fourier_transform_pairs)
y = diric(2*np.pi*np.arange(N)/N, 2*K+1) * (2*K+1) / N
plt.figure()
plt.plot(x[:1024] - y[:1024])
plt.title('error')
plt.show(block=True)
that shows irfft is accurate. Here is the error plot:
Numpy is correct, my math was incorrect. I am sorry for posting this misleading question. I don't know what is the standard procedure in these cases. Should I delete my question or leave it here with this answer? I just don't want it to be undermining NumPy or challanging its accuracy (as this was clearly a false alarm).

Jacobi Method & Basic Matrix Math using NUMPY

I'm getting an import error for "norm". What am I not doing correct??
I'm open to constructive feedback on improving the code, however I have to keep the parameters as they are!
Thanks!!!
Code is below:
import numpy as np
from numpy import norm, inalg, array, zeros, diag, diagflat, dot, linalg
"""Test Case Data"""
A = np.matrix([[4,-1,-1],[-2,6,1],[-1,1,7]])
b = np.matrix([[3],[9],[-6]])
x = np.matrix([[0],[0],[0]])
"""Main Function"""
def jacobi(A, b, x, Tolerance, Iterations):
V = np.diag(A)
D = np.diag(V)
R = D-A
D_I = D.I
D = np.asmatrix(D)
Counter_1 = 1
tol_gauge = 100
while Counter_1 <= Iterations:
# I considered using the "dot" function in NUMPY but I was wary of mixed results
iterative_approach_form = D_I * ((R*x)+b)
tol_gauge = np.linalg.norm(iterative_approach_form-x)
x = iterative_approach_form
if initial_tol <= Tolerance:
return("The Solution x = {},y={}, z={} ".format(x[0], x[1], x[2]))
return("The Solution was found in %s interation(s)" %(Counter_1))
else:
pass
Counter_1 +=1
return("The Solution was not found in {} iteration(s)".format(Iterations))

You need to specify which numpy module you are importing from. The following works if you want to use a function only by its name:
from numpy import linalg
from numpy.linalg import norm
from numpy import zeros, array, diag, diagflat, dot
Looking at you code however, you don't need the second import line, because in the rest of the code the numpy functions are specified according to the accepted norm. For example, norm is already present in your code as np.linalg.norm.
There are three more issues with your code: 1) initial_tol is not assigned a value; 2) tol_gauge is assigned but not used in the code; 3) the last return statement is not indented properly (perhaps only here) and the same is very likely for the block in your while loop.

fitting an ODE with python leastsq gives a cast error when initial conditions is passed as parameter

I have a set of data that I am trying to fit to an ODE model using scipy's leastsq function. My ODE has parameters beta and gamma, so that it looks for example like this:
# dS/dt = -betaSI
# dI/dt = betaSI - gammaI
# dR/dt = gammaI
# with y0 = y(t=0) = (S(0),I(0),R(0))
The idea is to find beta and gamma so that the numerical integration of my system of ODE's best approximates the data. I am able to do this just fine using leastsq if I know all the points in my initial condition y0.
Now, I am trying to do the same thing but to pass now one of the entries of y0 as an extra parameter. Here is where the Python and me stop communicating...
I did a function so that now the first entry of the parameters that I pass to leastsq is the initial condition of my variable R.
I get the following message:
*Traceback (most recent call last):
File "/Users/Laura/Dropbox/SHIV/shivmodels/test.py", line 73, in <module>
p1,success = optimize.leastsq(errfunc, initguess, args=(simpleSIR,[y0[0]],[Tx],[mydata]))
File "/Library/Frameworks/Python.framework/Versions/7.2/lib/python2.7/site-packages/scipy/optimize/minpack.py", line 283, in leastsq
gtol, maxfev, epsfcn, factor, diag)
TypeError: array cannot be safely cast to required type*
Here is my code. It is a little more involved that what it needs to be for this example because in reality I want to fit another ode with 7 parameters and want to fit to several data sets at once. But I wanted to post here something simpler... Any help will be very very much appreciated! Thank you very much!
import numpy as np
from matplotlib import pyplot as plt
from scipy import optimize
from scipy.integrate import odeint
#define the time span for the ODE integration:
Tx = np.arange(0,50,1)
num_points = len(Tx)
#define a simple ODE to fit:
def simpleSIR(y,t,params):
dydt0 = -params[0]*y[0]*y[1]
dydt1 = params[0]*y[0]*y[1] - params[1]*y[1]
dydt2 = params[1]*y[1]
dydt = [dydt0,dydt1,dydt2]
return dydt
#generate noisy data:
y0 = [1000.,1.,0.]
beta = 12*0.06/1000.0
gamma = 0.25
myparam = [beta,gamma]
sir = odeint(simpleSIR, y0, Tx, (myparam,))
mydata0 = sir[:,0] + 0.05*(-1)**(np.random.randint(num_points,size=num_points))*sir[:,0]
mydata1 = sir[:,1] + 0.05*(-1)**(np.random.randint(num_points,size=num_points))*sir[:,1]
mydata2 = sir[:,2] + 0.05*(-1)**(np.random.randint(num_points,size=num_points))*sir[:,2]
mydata = np.array([mydata0,mydata1,mydata2]).transpose()
#define a function that will run the ode and fit it, the reason I am doing this
#is because I will use several ODE's to see which one fits the data the best.
def fitfunc(myfun,y0,Tx,params):
myfit = odeint(myfun, y0, Tx, args=(params,))
return myfit
#define a function that will measure the error between the fit and the real data:
def errfunc(params,myfun,y0,Tx,y):
"""
INPUTS:
params are the parameters for the ODE
myfun is the function to be integrated by odeint
y0 vector of initial conditions, so that y(t0) = y0
Tx is the vector over which integration occurs, since I have several data sets and each
one has its own vector of time points, Tx is a list of arrays.
y is the data, it is a list of arrays since I want to fit to multiple data sets at once
"""
res = []
for i in range(len(y)):
V0 = params[0][i]
myparams = params[1:]
initCond = np.zeros([3,])
initCond[:2] = y0[i]
initCond[2] = V0
myfit = fitfunc(myfun,initCond,Tx[i],myparams)
res.append(myfit[:,0] - y[i][:,0])
res.append(myfit[:,1] - y[i][:,1])
res.append(myfit[1:,2] - y[i][1:,2])
#end for
all_residuals = np.hstack(res).ravel()
return all_residuals
#end errfunc
#example of the problem:
V0 = [0]
params = [V0,beta,gamma]
y0 = [1000,1]
#this is just to test that my errfunc does work well.
errfunc(params,simpleSIR,[y0],[Tx],[mydata])
initguess = [V0,0.5,0.5]
p1,success = optimize.leastsq(errfunc, initguess, args=(simpleSIR,[y0[0]],[Tx],[mydata]))

The problem is with the variable initguess. The function optimize.leastsq has the following call signature:
http://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.leastsq.html
It's second argument, x0, has to be an array. Your list
initguess = [v0,0.5,0.5]
won't be converted to an array because v0 is a list instead of an int or float. So you get an error when you try to convert initguess from a list to an array in the leastsq function.
I would adjust the variable params from
def errfunc(params,myfun,y0,Tx,y):
so that it is a 1-D array. Make the first few entries the values of v0 then append beta and gamma to that.