The (brief) documentation for scipy.integrate.ode says that two methods (dopri5 and dop853) have stepsize control and dense output. Looking at the examples and the code itself, I can only see a very simple way to get output from an integrator. Namely, it looks like you just step the integrator forward by some fixed dt, get the function value(s) at that time, and repeat.
My problem has pretty variable timescales, so I'd like to just get the values at whatever time steps it needs to evaluate to achieve the required tolerances. That is, early on, things are changing slowly, so the output time steps can be big. But as things get interesting, the output time steps have to be smaller. I don't actually want dense output at equal intervals, I just want the time steps the adaptive function uses.
EDIT: Dense output
A related notion (almost the opposite) is "dense output", whereby the steps taken are as large as the stepper cares to take, but the values of the function are interpolated (usually with accuracy comparable to the accuracy of the stepper) to whatever you want. The fortran underlying scipy.integrate.ode is apparently capable of this, but ode does not have the interface. odeint, on the other hand, is based on a different code, and does evidently do dense output. (You can output every time your right-hand-side is called to see when that happens, and see that it has nothing to do with the output times.)
So I could still take advantage of adaptivity, as long as I could decide on the output time steps I want ahead of time. Unfortunately, for my favorite system, I don't even know what the approximate timescales are as functions of time, until I run the integration. So I'll have to combine the idea of taking one integrator step with this notion of dense output.
EDIT 2: Dense output again
Apparently, scipy 1.0.0 introduced support for dense output through a new interface. In particular, they recommend moving away from scipy.integrate.odeint and towards scipy.integrate.solve_ivp, which as a keyword dense_output. If set to True, the returned object has an attribute sol that you can call with an array of times, which then returns the integrated functions values at those times. That still doesn't solve the problem for this question, but it is useful in many cases.
Since SciPy 0.13.0,
The intermediate results from the dopri family of ODE solvers can
now be accessed by a solout callback function.
import numpy as np
from scipy.integrate import ode
import matplotlib.pyplot as plt
def logistic(t, y, r):
return r * y * (1.0 - y)
r = .01
t0 = 0
y0 = 1e-5
t1 = 5000.0
backend = 'dopri5'
# backend = 'dop853'
solver = ode(logistic).set_integrator(backend)
sol = []
def solout(t, y):
sol.append([t, *y])
solver.set_solout(solout)
solver.set_initial_value(y0, t0).set_f_params(r)
solver.integrate(t1)
sol = np.array(sol)
plt.plot(sol[:,0], sol[:,1], 'b.-')
plt.show()
Result:
The result seems to be slightly different from Tim D's, although they both use the same backend. I suspect this having to do with FSAL property of dopri5. In Tim's approach, I think the result k7 from the seventh stage is discarded, so k1 is calculated afresh.
Note: There's a known bug with set_solout not working if you set it after setting initial values. It was fixed as of SciPy 0.17.0.
I've been looking at this to try to get the same result. It turns out you can use a hack to get the step-by-step results by setting nsteps=1 in the ode instantiation. It will generate a UserWarning at every step (this can be caught and suppressed).
import numpy as np
from scipy.integrate import ode
import matplotlib.pyplot as plt
import warnings
def logistic(t, y, r):
return r * y * (1.0 - y)
r = .01
t0 = 0
y0 = 1e-5
t1 = 5000.0
#backend = 'vode'
backend = 'dopri5'
#backend = 'dop853'
solver = ode(logistic).set_integrator(backend, nsteps=1)
solver.set_initial_value(y0, t0).set_f_params(r)
# suppress Fortran-printed warning
solver._integrator.iwork[2] = -1
sol = []
warnings.filterwarnings("ignore", category=UserWarning)
while solver.t < t1:
solver.integrate(t1, step=True)
sol.append([solver.t, solver.y])
warnings.resetwarnings()
sol = np.array(sol)
plt.plot(sol[:,0], sol[:,1], 'b.-')
plt.show()
result:
The integrate method accepts a boolean argument step that tells the method to return a single internal step. However, it appears that the 'dopri5' and 'dop853' solvers do not support it.
The following code shows how you can get the internal steps taken by the solver when the 'vode' solver is used:
import numpy as np
from scipy.integrate import ode
import matplotlib.pyplot as plt
def logistic(t, y, r):
return r * y * (1.0 - y)
r = .01
t0 = 0
y0 = 1e-5
t1 = 5000.0
backend = 'vode'
#backend = 'dopri5'
#backend = 'dop853'
solver = ode(logistic).set_integrator(backend)
solver.set_initial_value(y0, t0).set_f_params(r)
sol = []
while solver.successful() and solver.t < t1:
solver.integrate(t1, step=True)
sol.append([solver.t, solver.y])
sol = np.array(sol)
plt.plot(sol[:,0], sol[:,1], 'b.-')
plt.show()
Result:
FYI, although an answer has been accepted already, I should point out for the historical record that dense output and arbitrary sampling from anywhere along the computed trajectory is natively supported in PyDSTool. This also includes a record of all the adaptively-determined time steps used internally by the solver. This interfaces with both dopri853 and radau5 and auto-generates the C code necessary to interface with them rather than relying on (much slower) python function callbacks for the right-hand side definition. None of these features are natively or efficiently provided in any other python-focused solver, to my knowledge.
Here's another option that should also work with dopri5 and dop853. Basically, the solver will call the logistic() function as often as needed to calculate intermediate values so that's where we store the results:
import numpy as np
from scipy.integrate import ode
import matplotlib.pyplot as plt
sol = []
def logistic(t, y, r):
sol.append([t, y])
return r * y * (1.0 - y)
r = .01
t0 = 0
y0 = 1e-5
t1 = 5000.0
# Maximum number of steps that the integrator is allowed
# to do along the whole interval [t0, t1].
N = 10000
#backend = 'vode'
backend = 'dopri5'
#backend = 'dop853'
solver = ode(logistic).set_integrator(backend, nsteps=N)
solver.set_initial_value(y0, t0).set_f_params(r)
# Single call to solver.integrate()
solver.integrate(t1)
sol = np.array(sol)
plt.plot(sol[:,0], sol[:,1], 'b.-')
plt.show()
Related
I am trying to solve the following system of first order coupled differential equations:
- dR/dt=k2*Y(t)-k1*R(t)*L(t)+k4*X(t)-k3*R(t)*I(t)
- dL/dt=k2*Y(t)-k1*R(t)*L(t)
- dI/dt=k4*X(t)-k3*R(t)*I(t)
- dX/dt=k1*R(t)*L(t)-k2*Y(t)
- dY/dt=k3*R(t)*I(t)-k4*X(t)
The knowed initial conditions are: X(0)=0, Y(0)=0
The knowed constants values are: k1=6500, k2=0.9
This equations defines a kinetick model and I need to solve them to get the Y(t) function to fit my data and find k3 and k4 values. In order to that, I have tried to solve the system simbologically with sympy. There is my code:
import matplotlib.pyplot as plt
import numpy as np
import sympy
from sympy.solvers.ode.systems import dsolve_system
from scipy.integrate import solve_ivp
from scipy.integrate import odeint
k1 = sympy.Symbol('k1', real=True, positive=True)
k2 = sympy.Symbol('k2', real=True, positive=True)
k3 = sympy.Symbol('k3', real=True, positive=True)
k4 = sympy.Symbol('k4', real=True, positive=True)
t = sympy.Symbol('t',real=True, positive=True)
L = sympy.Function('L')
R = sympy.Function('R')
I = sympy.Function('I')
Y = sympy.Function('Y')
X = sympy.Function('X')
f1=k2*Y(t)-k1*R(t)*L(t)+k4*X(t)-k3*R(t)*I(t)
f2=k2*Y(t)-k1*R(t)*L(t)
f3=k4*X(t)-k3*R(t)*I(t)
f4=-f2
f5=-f3
eq1=sympy.Eq(sympy.Derivative(R(t),t),f1)
eq2=sympy.Eq(sympy.Derivative(L(t),t),f2)
eq3=sympy.Eq(sympy.Derivative(I(t),t),f3)
eq4=sympy.Eq(sympy.Derivative(Y(t),t),f4)
eq5=sympy.Eq(sympy.Derivative(X(t),t),f5)
Sys=(eq1,eq2,eq3,eq4,eq5])
solsys=dsolve_system(eqs=Sys,funcs=[X(t),Y(t),R(t),L(t),I(t)], t=t, ics={Y(0):0, X(0):0})
There is the answer:
NotImplementedError:
The system of ODEs passed cannot be solved by dsolve_system.
I have tried with dsolve too, but I get the same.
Is there any other solver I can use or some way of doing this that will allow me to get the function for the fitting? I'm using python 3.8 in Spider with Anaconda in windows64.
Thank you!
# Update
Following
You are saying "experiment". So you have data and want to fit the model to them, find appropriate values for k3 and k4 at least, and perhaps for all coefficients and the initial conditions (the first measured data point might not be the initial condition for the best fit)? See stackoverflow.com/questions/71722061/… for a recent attempt on such a task. –
Lutz Lehmann
23 hours
There is my new code:
t=[0,0.25,0.5,0.75,1.5,2.27,3.05,3.82,4.6,5.37,6.15,6.92,7.7,8.47,13.42,18.42,23.42,28.42,33.42,38.42,43.42,48.42,53.42,58.42,63.42,68.42,83.4,98.4,113.4,128.4,143.4,158.4,173.4,188.4,203.4,218.4,233.4,248.4]
yexp=[832.49,1028.01,1098.12,1190.08,1188.97,1377.09,1407.47,1529.35,1431.72,1556.66,1634.59,1679.09,1692.05,1681.89,1621.88,1716.77,1717.91,1686.7,1753.5,1722.98,1630.14,1724.16,1670.45,1677.16,1614.98,1671.16,1654.03,1661.84,1675.31,1626.76,1638.29,1614.41,1594.31,1584.73,1599.22,1587.85,1567.74,1602.69]
def derivative(S, t, k3, k4):
k1=1798931
k2=0.2629
x, y,r,l,i = S
doty = k1*r*l+k2*y
dotx = k3*r*i-k4*x
dotr = k2*y-k1*r*l+k4*x-k3*r*i
dotl = k2*y-k1*r*l
doti = k4*x-k3*r*i
return np.array([doty, dotx, dotr, dotl, doti])
def solver(XY,t,para):
return odeint(derivative, XY, t, args = para, atol=1e-8, rtol=1e-11)
def integration(XY_arr,*para):
XY0 = para[:5]
para = para[5:]
T = np.arange(len(XY_arr))
res0 = solver(XY0,T, para)
res1 = [ solver(XY0,[t,t+1],para)[-1]
for t,XY in enumerate(XY_arr[:-1]) ]
return np.concatenate([res0,res1]).flatten()
XData =yexp
YData = np.concatenate([ yexp,yexp,yexp,yexp,yexp,yexp[1:],yexp[1:],yexp[1:],yexp[1:],yexp[1:]]).flatten()
p0 =[0,0,100,10,10,1e8,0.01]
params, info = curve_fit(integration,XData,YData,p0=p0, maxfev=5000)
XY0, para = params[:5], params[5:]
print(XY0,tuple(para))
t_plot = np.linspace(0,len(t),500)
x_plot = solver(XY0, t_plot, tuple(para))
But the output are not correct, as are the same as initial condition p0:
[ 0. 0. 100. 10. 10.] (100000000.0, 0.01)
Graph
I understand that the function 'integration' gives packed values of y for each function at each instant of time, but I don't know how to unpack them to make the curve_fitt separately. Maybe I don't quite understand how it works.
Thank you!
As you observed, sympy is not able to solve this system. This might mean that
the procedure to classify ODE in sympy is not complete enough, or
some trick/method is needed above the standard set of methods that is implemented in sympy, or
that there just is no symbolic solution.
The last case is the generic one, take a symbolically solvable ODE, add some random term, and almost certainly the resulting ODE is no longer symbolically solvable.
As I understand with the comments, you have an model via ODE system with state space (cX,cY,cR,cL,cI) with equations with 4 parameters k1,k2,k3,k4 and, by the structure of a reaction system R+I <-> X, R+L <-> Y, the sums cR+cX+cY, cL+cY, cI+cX are all constant.
For some other process that is approximately represented by the model, you have time series data t[k],y[k] for the Y component. Also you have partial information on the initial state and the parameter set. If there are sufficiently many data points one could also forget about these, fit for all parameters, and compare how far away the given parameters are to the computed ones.
There are several modules and packages that solve this fitting task in a more or less abstract fashion. I think pyomo and gekko can both be used. More directly one can use the facilities of scipy.odr or scipy.optimize.
Define the forward function that transforms time and parameters
def model(t,u,k1,k2,k3,k4):
X,Y,R,L,I = u
dL = k2*Y - k1*R*L
dI = k4*X - k3*R*I
dR = dL+dI
dX = -dI
dY = -dL
return dX,dY,dR,dL,dI
def solver(t,u0,k):
res = solve_ivp(model, [0, t[-1]], u0, args=tuple(k), t_eval=t,
method="DOP853", atol=1e-7, rtol=1e-10)
return res.y
Prepare some data plus noise
k1o = 6.500; k2o=0.9
T = np.linspace(0,0.05,21)
U = solver(T, [0,0,50,40,25], [k1o, k2o, 5.400, 0.7])
Y = U[1] # equilibrium slightly above 30
Y += np.random.uniform(high=0.05, size=Y.shape)
Prepare the function that splits the combined parameter vector in initial state and coefficients, call the curve fitting function
from scipy.optimize import curve_fit
def partial(t,R,L,I,k3,k4):
print(R,L,I,k3,k4)
U = solver(t,[0,0,R,L,I],[k1o,k2o,k3,k4])
return U[1]
params, info = curve_fit(partial,T,Y, p0=[30,20,10, 0.3,3.000])
R,L,I, k3,k4 = params
print(R,L,I, k3,k4)
It turns out that curve_fit goes into strange regions with large negative values. A likely reason is that the Y component is, in the end, not coupled strongly enough to all the other components, meaning that large changes in some of the parameters have minimal influence on Y, so that minimal noise in Y can lead to large deviations in these parameters. Here this apparently happens (first) to k3.
I was wondering how to speed up the following code in where I compute a probability function which involves numerical integrals and then I compute some confidence margins.
Some possibilities that I have thought about are Numba or vectorization of the code
EDIT:
I have made minor modifications because there was a mistake. I am looking for some modifications that provide major time improvements (I know that there are some minor changes that would provide some minor time improvements, such as repeated functions, but I am not concerned about them)
The code is:
# -*- coding: utf-8 -*-
"""
Created on Tue Jan 26 17:05:46 2021
#author: Ignacio
"""
import numpy as np
from scipy.integrate import simps
def pdf(V,alfa_points):
alfa=np.linspace(0,2*np.pi,alfa_points)
return simps(1/np.sqrt(2*np.pi)/np.sqrt(sigma_R2)*np.exp(-(V*np.cos(alfa)-eR)**2/2/sigma_R2)*1/np.sqrt(2*np.pi)/np.sqrt(sigma_I2)*np.exp(-(V*np.sin(alfa)-eI)**2/2/sigma_I2),alfa)
def find_nearest(array,value):
array=np.asarray(array)
idx = (np.abs(array-value)).argmin()
return array[idx]
N = 20
n=np.linspace(0,N-1,N)
d=1
sigma_An=0.1
sigma_Pn=0.2
An=np.ones(N)
Pn=np.zeros(N)
Vs=np.linspace(0,30,1000)
inc=np.max(Vs)/len(Vs)
th=np.linspace(0,np.pi/2,250)
R=np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
I=np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
fmin=np.zeros(len(th))
fmax=np.zeros(len(th))
for tt in range(len(th)):
eR=np.exp(-sigma_Pn**2/2)*np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[tt])*n*d))
eI=np.exp(-sigma_Pn**2/2)*np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[tt])*n*d))
sigma_R2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)+1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
sigma_I2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)-1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
PDF=np.zeros(len(Vs))
for vv in range(len(Vs)):
PDF[vv]=pdf(Vs[vv],100)
total=simps(PDF,Vs)
values=np.cumsum(PDF)*inc/total
xval_05=find_nearest(values,0.05)
fmin[tt]=Vs[values==xval_05]
xval_95=find_nearest(values,0.95)
fmax[tt]=Vs[values==xval_95]
This version's speedup: 31x
A simple profiling (%%prun) reveals that most of the time is spent in simps.
You are in control of the integration done in pdf(): for example, you can use the trapeze method instead of Simpson with negligible numerical difference if you increase a bit the resolution of alpha. In fact, the higher resolution obtained by a higher sampling of alpha more than makes up for the difference between simps and trapeze (see picture at the bottom as for why). This is by far the highest speedup. We go one bit further by implementing the trapeze method ourselves instead of using scipy, since it is so simple. This alone yields marginal gain, but opens the door for a more drastic optimization (below, about pdf2D.
Also, the remaining simps(PDF, ...) goes faster when it knows that the dx step is constant, so we can just say so instead of passing the whole alpha array.
You can avoid doing the loop to compute PDF and use np.vectorize(pdf) directly on Vs, or better (as in the code below), do a 2-D version of that calculation.
There are some other minor things (such as using an index directly fmin[tt] = Vs[closest(values, 0.05)] instead of finding the index, returning the value, and then using a boolean mask for where values == xval_05), or taking all the constants (including alpha) outside functions and avoid recalculating every time.
This above gives us a 5.2x improvement. There is a number of things I don't understand in your code, e.g. why having An (ones) and Pn (zeros)?
But, importantly, another ~6x speedup comes from the observation that, since we are implementing our own trapeze method by using numpy primitives, we can actually do it in 2D in one go for the whole PDF.
The final speed up of the code below is 31x. I believe that a better understanding of "the big picture" of what you want to do would yield additional, perhaps substantial, speed gains.
Modified code:
import numpy as np
from scipy.integrate import simps
alpha_points = 200 # more points as we'll use trapeze not simps
alpha = np.linspace(0, 2*np.pi, alpha_points)
cosalpha = np.cos(alpha)
sinalpha = np.sin(alpha)
d_alpha = np.mean(np.diff(alpha)) # constant dx
coeff = 1 / np.sqrt(2*np.pi)
Vs=np.linspace(0,30,1000)
d_Vs = np.mean(np.diff(Vs)) # constant dx
inc=np.max(Vs)/len(Vs)
def f2D(Vs, eR, sigma_R2, eI, sigma_I2):
a = coeff / np.sqrt(sigma_R2)
b = coeff / np.sqrt(sigma_I2)
y = a * np.exp(-(np.outer(cosalpha, Vs) - eR)**2 / 2 / sigma_R2) * b * np.exp(-(np.outer(sinalpha, Vs) - eI)**2 / 2 / sigma_I2)
return y
def pdf2D(Vs, eR, sigma_R2, eI, sigma_I2):
y = f2D(Vs, eR, sigma_R2, eI, sigma_I2)
s = y.sum(axis=0) - (y[0] + y[-1]) / 2 # our own impl of trapeze, on 2D y
return s * d_alpha
def closest(a, val):
return np.abs(a - val).argmin()
N = 20
n = np.linspace(0,N-1,N)
d = 1
sigma_An = 0.1
sigma_Pn = 0.2
An=np.ones(N)
Pn=np.zeros(N)
th = np.linspace(0,np.pi/2,250)
R = np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
I = np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
fmin=np.zeros(len(th))
fmax=np.zeros(len(th))
for tt in range(len(th)):
eR=np.exp(-sigma_Pn**2/2)*np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[tt])*n*d))
eI=np.exp(-sigma_Pn**2/2)*np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[tt])*n*d))
sigma_R2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)+1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
sigma_I2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)-1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
PDF=pdf2D(Vs, eR, sigma_R2, eI, sigma_I2)
total = simps(PDF, dx=d_Vs)
values = np.cumsum(PDF) * inc / total
fmin[tt] = Vs[closest(values, 0.05)]
fmax[tt] = Vs[closest(values, 0.95)]
Note: most of the fmin and fmax are np.allclose() compared with the original function, but some of them have a small error: after some digging, it turns out that the implementation here is more precise as that function f() can be pretty abrupt, and more alpha points actually help (and more than compensate the minuscule lack of precision due to using trapeze instead of Simpson).
For example, at index tt=244, vv=400:
Considering several methods, the one that provides the largest time improvement is the Numba method. The method proposed by Pierre is very interesting and it does not require to install other packages, which is an asset.
However, in the examples that I have computed, the time improvement is not as large as with the numba example, specially when the points in th grows to a few tenths of thousands (which is my actual case). I post here the Numba code just in case someone is interested:
import numpy as np
from numba import njit
#njit
def margins(val_min,val_max):
fmin=np.zeros(len(th))
fmax=np.zeros(len(th))
for tt in range(len(th)):
eR=np.exp(-sigma_Pn**2/2)*np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[tt])*n*d))
eI=np.exp(-sigma_Pn**2/2)*np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[tt])*n*d))
sigma_R2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)+1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
sigma_I2=1/2*np.sum(An*sigma_An**2)+1/2*(1-np.exp(-sigma_Pn**2))*np.sum(An**2)-1/2*np.sum(np.cos(2*(Pn+2*np.pi*np.sin(th[tt])*n*d))*((An**2+sigma_An**2)*np.exp(-2*sigma_Pn**2)-An**2*np.exp(-sigma_Pn**2)))
Vs=np.linspace(0,30,1000)
inc=np.max(Vs)/len(Vs)
integration_points=200
PDF=np.zeros(len(Vs))
for vv in range(len(Vs)):
PDF[vv]=np.trapz(1/np.sqrt(2*np.pi)/np.sqrt(sigma_R2)*np.exp(-(Vs[vv]*np.cos(np.linspace(0,2*np.pi,integration_points))-eR)**2/2/sigma_R2)*1/np.sqrt(2*np.pi)/np.sqrt(sigma_I2)*np.exp(-(Vs[vv]*np.sin(np.linspace(0,2*np.pi,integration_points))-eI)**2/2/sigma_I2),np.linspace(0,2*np.pi,integration_points))
total=np.trapz(PDF,Vs)
values=np.cumsum(PDF)*inc/total
idx = (np.abs(values-val_min)).argmin()
xval_05=values[idx]
fmin[tt]=Vs[np.where(values==xval_05)[0][0]]
idx = (np.abs(values-val_max)).argmin()
xval_95=values[idx]
fmax[tt]=Vs[np.where(values==xval_95)[0][0]]
return fmin,fmax
N = 20
n=np.linspace(0,N-1,N)
d=1
sigma_An=1/2**6
sigma_Pn=2*np.pi/2**6
An=np.ones(N)
Pn=np.zeros(N)
th=np.linspace(0,np.pi/2,250)
R=np.sum(An*np.cos(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
I=np.sum(An*np.sin(Pn+2*np.pi*np.sin(th[:,np.newaxis])*n*d),axis=1)
F=R+1j*I
Fa=np.abs(F)/np.max(np.abs(F))
fmin, fmax = margins(0.05,0.95)
I'm trying to solve a 2D delay differential equation with some parameters. The problem is that I can’t get the right solution (which I know) and I suspect that it comes from the integration step, but I'm not sure and I don't really understand how the JiTCDDE works.
This is the DDE:
This is my model:
def model(p, q, r, alpha, T, tau, tmax, ci):
f = [1/T * (p*y(0)+alpha*y(1, t-tau)), 1/T * (r*y(0)+q*y(1))]
DDE = jitcdde(f)
DDE.constant_past(ci)
DDE.step_on_discontinuities()
data = []
for time in np.arange(DDE.t, DDE.t+tmax, 0.09):
data.append( DDE.integrate(time)[1])
return data
Where I'm only interested in the y(1) solution
And the parameters:
T=32 #escala temporal
p=-2.4/T
q=-1.12/T
r=1.5/T
alpha=.6/T
tau=T*2.4 #delay
tmax=400
ci = np.array([4080, 0])
This is the plot I have with that model and parameters:
And this is (the blue line) the correct solution (someone give me the plot not the data)
The following code works for me and produces a result similar to your control:
import numpy as np
from jitcdde import y,t,jitcdde
T = 1
p = -2.4/T
q = -1.12/T
r = 1.5/T
alpha = .6/T
tau = T*2.4
tmax = 10
ci = np.array([4080, 0])
f = [
1/T * (p*y(0)+alpha*y(1, t-tau)),
1/T * (r*y(0)+q*y(1))
]
DDE = jitcdde(f)
DDE.constant_past(ci)
DDE.adjust_diff()
times = np.linspace( DDE.t, DDE.t+tmax, 1000 )
values = [DDE.integrate(time)[1] for time in times]
from matplotlib.pyplot import subplots
fig,axes = subplots()
axes.plot(times,values)
fig.show()
Note the following:
I set T=1 (and adjusted tmax accordingly). I presume that there still is a mistake here.
I used adjust_diff instead of step_on_discontinuities. The problem with your model is that it has a heavy discontinuity in the derivative at t=0. (A discontinuity is normal, but none of this kind). This causes problems with the adaptive step size control at the very beginning of the integration. Such a discontinuity suggests that there is something wrong with either your model or your initial past. The latter doesn’t matter if you only care about long-term behaviour, but this doesn’t seem to be the case here. I added a passage to the documentation about this kind of issue.
I am trying to solve a simple example with the dopri5 integrator in scipy.integrate.ode. As the documentation states
This is an explicit runge-kutta method of order (4)5 due to Dormand & Prince (with stepsize control and dense output).
this should work. So here is my example:
import numpy as np
from scipy.integrate import ode
import matplotlib.pyplot as plt
def MassSpring_with_force(t, state):
""" Simple 1DOF dynamics model: m ddx(t) + k x(t) = f(t)"""
# unpack the state vector
x = state[0]
xd = state[1]
# these are our constants
k = 2.5 # Newtons per metre
m = 1.5 # Kilograms
# force
f = force(t)
# compute acceleration xdd
xdd = ( ( -k*x + f) / m )
# return the two state derivatives
return [xd, xdd]
def force(t):
""" Excitation force """
f0 = 1 # force amplitude [N]
freq = 20 # frequency[Hz]
omega = 2 * np.pi *freq # angular frequency [rad/s]
return f0 * np.sin(omega*t)
# Time range
t_start = 0
t_final = 1
# Main program
state_ode_f = ode(MassSpring_with_force)
state_ode_f.set_integrator('dopri5', rtol=1e-6, nsteps=500,
first_step=1e-6, max_step=1e-3)
state2 = [0.0, 0.0] # initial conditions
state_ode_f.set_initial_value(state2, 0)
sol = np.array([[t_start, state2[0], state2[1]]], dtype=float)
print("Time\t\t Timestep\t dx\t\t ddx\t\t state_ode_f.successful()")
while state_ode_f.t < (t_final):
state_ode_f.integrate(t_final, step=True)
sol = np.append(sol, [[state_ode_f.t, state_ode_f.y[0], state_ode_f.y[1]]], axis=0)
print("{0:0.8f}\t {1:0.4e} \t{2:10.3e}\t {3:0.3e}\t {4}".format(
state_ode_f.t, sol[-1, 0]- sol[-2, 0], state_ode_f.y[0], state_ode_f.y[1], state_ode_f.successful()))
The result I get is:
Time Timestep dx ddx state_ode_f.successful()
0.49763822 4.9764e-01 2.475e-03 -8.258e-04 False
0.99863822 5.0100e-01 3.955e-03 -3.754e-03 False
1.00000000 1.3618e-03 3.950e-03 -3.840e-03 False
with a warning:
c:\python34\lib\site-packages\scipy\integrate_ode.py:1018: UserWarning: dopri5: larger nmax is needed
self.messages.get(idid, 'Unexpected idid=%s' % idid))
The result is incorect. If I run the same code with vode integrator, I get the expected result.
Edit
A similar issue is described here:
Using adaptive step sizes with scipy.integrate.ode
The suggested solution recommends setting nsteps=1, which solves the ODE correctly and with step-size control. However the integrator returns state_ode_f.successful() as False.
No, there is nothing wrong. You are telling the integrator to perform an integration step to t_final and it performs that step. Internal steps of the integrator are not reported.
The sensible thing to do is to give the desired sampling points as input of the algorithm, set for example dt=0.1 and use
state_ode_f.integrate( min(state_ode_f.t+dt, t_final) )
There is no single-step method in dopri5, only vode has it defined, see the source code https://github.com/scipy/scipy/blob/v0.14.0/scipy/integrate/_ode.py#L376, this could account for the observed differences.
As you found in Using adaptive step sizes with scipy.integrate.ode, one can force single-step behavior by setting the iteration bound nsteps=1. This will produce a warning every time, so one has to suppress these specific warnings to see a sensible result.
You should not use a parameter (which is a constant for the integration interval) for a time-dependent force. Use inside MassSpring_with_force the evaluation f=force(t). Possibly you could pass the function handle of force as parameter.
So I have the function
f(x) = I_0(exp(Q*x/nKT)
Where Q, K and T are constants, for the sake of clarity I'll add the values
Q = 1.6x10^(-19)
K = 1.38x10^(-23)
T = 77.6
and n and I_0 are the two constraints that I'm trying to minimize.
my xdata is a list of 50 datapoints and as is my ydata. So as of yet this is my code:
from __future__ import division
import scipy.optimize as optimize
import numpy
xdata = numpy.array([1.07,1.07994,1.08752,1.09355,
1.09929,1.10536,1.10819,1.11321,
1.11692,1.12099,1.12435,1.12814,
1.13181,1.13594,1.1382,1.14147,
1.14443,1.14752,1.15023,1.15231,
1.15514,1.15763,1.15985,1.16291,1.16482])
ydata = [0.00205,
0.004136,0.006252,0.008252,0.010401,
0.012907,0.014162,0.016498,0.018328,
0.020426,0.022234,0.024363,0.026509,
0.029024,0.030457,0.032593,0.034576,
0.036725,0.038703,0.040223,0.042352,
0.044289,0.046043,0.048549,0.050146]
#data and ydata is experimental data, xdata is voltage and ydata is current
def f(x,I0,N):
# I0 = 7.85E-07
# N = 3.185413895
Q = 1.66E-19
K = 1.38065E-23
T = 77.3692
return I0*(numpy.e**((Q*x)/(N*K*T))-1)
result = optimize.curve_fit(f, xdata,ydata) #trying to minize I0 and N
But the answer doesn't give suitably optimized constraints
Any help would be hugely appreciated I realize there may be something obvious I am missing, I just can't see what it is!
I have tried this, but for some reason if you throw out those constants so function becomes
def f(x,I0,N):
return I0*(numpy.exp(x/N)-1)
you get something reasonable.
1.86901114e-13, 4.41838309e-02
Its true, that when we get rid off constants its better. Define function as:
def f(x,A,B):
return A*(np.e**(B*x)-1)
and fit it by curve_fit, you'll be able to get A that is explicitly I0 (A=I0) and B (you can obtain N simply by N=Q/(BKT) ). I managed to get pretty good fit.
I think if there is too much constants, algorithm gets confused some way.