how to change variables given event in scipy's solve_ivp - python

TLDR
I'm using scipy.solve_ivp to solve phi''(t) = a(t) - b * (phi'(t))^2 and I want to change the value of b whenever the sign of phi'(t) changes
What I've Tried
More specifically, the function represents an equation of torques on a wing with drag (not really important for the matter but still) and the entire solver is encapsulated in the following Solver class.
This is a runnable code, but the most important functions are sequential_solver, phi_dot_zero_crossing_event and phi_derivatives
# some constants to run the code ...
NUM_SAMPLES = 100 # every time window has this many samples
MASS = 1
WING_LENGTH = 0.07
AERODYNAMIC_CENTER = 0.7 * WING_LENGTH
GYRATION_RADIUS = 0.6 * WING_LENGTH
MoI = MASS * GYRATION_RADIUS ** 2
AIR_DENSITY = 1.2
WING_AREA = 0.5 * WING_LENGTH * (0.5 * WING_LENGTH) * np.pi # 1/2 ellipse with minor radios ~ 1/2 major = length/2
C_D_MAX = 3.4
C_D_0 = 0.4
C_L_MAX = 1.8
class Solver:
def __init__(self) -> None:
self._alpha = 45
def get_c_drag(self) -> float:
"""
calculates the drag coefficient based on the angle of attack
"""
return (C_D_MAX + C_D_0) / 2 - (C_D_MAX - C_D_0) / 2 * np.cos(2 * self._alpha)
def get_c_lift(self) -> float:
"""
calculates the lift coefficient based on the angle of attack
"""
return C_L_MAX * np.sin(2 * self._alpha)
def tau_z(self, t: float) -> np.float:
"""
the moment we apply using the motors
"""
return np.sin(2 * np.pi * t)
def tau_drag(self, phi_dot):
"""
the drag moment
"""
return 0.5 * AIR_DENSITY * WING_AREA * self.get_c_drag() * (GYRATION_RADIUS ** 2) * (phi_dot ** 2)
def phi_derivatives(self, t, y):
"""
A function that defines the ODE that is to be solved: I * phi_ddot = tau_z - tau_drag.
We think of y as a vector y = [phi,phi_dot]. the ode solves dy/dt = f(y,t)
:return:
"""
phi, phi_dot = y[0], y[1]
dy_dt = [phi_dot, (self.tau_z(t) - self.tau_drag(phi_dot)) / MoI] # d[phi,phi_dot]]/dt = [phi_dot,phi_ddot]
return dy_dt
def phi_dot_zero_crossing_event(self, t, y):
return y[1]
def sequential_solver(self, use_windows):
"""
solves the ODE
:return:
"""
phi_0 = 2e-3
phi_dot_0 = 0
prev_t = 0
start_t, end_t, delta_t = 1e-1, 5, 1e-1
inner_delta_t = delta_t / NUM_SAMPLES
time = np.arange(start_t, end_t, inner_delta_t)
sol = solve_ivp(self.phi_derivatives, t_span=(start_t, end_t), y0=[phi_0, phi_dot_0], t_eval=time,
events=self.phi_dot_zero_crossing_event)
phi, phi_dot = sol.y
_, phi_ddot = self.phi_derivatives(time, [phi, phi_dot])
So what's the problem?
my question is how, given phi_dot_zero_crossing_event, I can change the value of self._alpha mid-solving? I currently only track the event of zero crossing, but I don't know how to use this event to set the value of self._alpha (if phi_dot>0 I need self._alpha = +45 and self._alpha=-45 otherwise)

Related

Matplotlib plot's title is missing for unknown reason in Python

Can anyone tell me what is wrong with this code? It is from https://jakevdp.github.io/blog/2012/09/05/quantum-python/ .
Everything in it worked out except the title of the plot.I can't figure it out.
It should look like this
but when the code is run, it polts this
Here is the code given:-
"""
General Numerical Solver for the 1D Time-Dependent Schrodinger's equation.
author: Jake Vanderplas
email: vanderplas#astro.washington.edu
website: http://jakevdp.github.com
license: BSD
Please feel free to use and modify this, but keep the above information. Thanks!
"""
import numpy as np
from matplotlib import pyplot as pl
from matplotlib import animation
from scipy.fftpack import fft,ifft
class Schrodinger(object):
"""
Class which implements a numerical solution of the time-dependent
Schrodinger equation for an arbitrary potential
"""
def __init__(self, x, psi_x0, V_x,
k0 = None, hbar=1, m=1, t0=0.0):
"""
Parameters
----------
x : array_like, float
length-N array of evenly spaced spatial coordinates
psi_x0 : array_like, complex
length-N array of the initial wave function at time t0
V_x : array_like, float
length-N array giving the potential at each x
k0 : float
the minimum value of k. Note that, because of the workings of the
fast fourier transform, the momentum wave-number will be defined
in the range
k0 < k < 2*pi / dx
where dx = x[1]-x[0]. If you expect nonzero momentum outside this
range, you must modify the inputs accordingly. If not specified,
k0 will be calculated such that the range is [-k0,k0]
hbar : float
value of planck's constant (default = 1)
m : float
particle mass (default = 1)
t0 : float
initial tile (default = 0)
"""
# Validation of array inputs
self.x, psi_x0, self.V_x = map(np.asarray, (x, psi_x0, V_x))
N = self.x.size
assert self.x.shape == (N,)
assert psi_x0.shape == (N,)
assert self.V_x.shape == (N,)
# Set internal parameters
self.hbar = hbar
self.m = m
self.t = t0
self.dt_ = None
self.N = len(x)
self.dx = self.x[1] - self.x[0]
self.dk = 2 * np.pi / (self.N * self.dx)
# set momentum scale
if k0 == None:
self.k0 = -0.5 * self.N * self.dk
else:
self.k0 = k0
self.k = self.k0 + self.dk * np.arange(self.N)
self.psi_x = psi_x0
self.compute_k_from_x()
# variables which hold steps in evolution of the
self.x_evolve_half = None
self.x_evolve = None
self.k_evolve = None
# attributes used for dynamic plotting
self.psi_x_line = None
self.psi_k_line = None
self.V_x_line = None
def _set_psi_x(self, psi_x):
self.psi_mod_x = (psi_x * np.exp(-1j * self.k[0] * self.x)
* self.dx / np.sqrt(2 * np.pi))
def _get_psi_x(self):
return (self.psi_mod_x * np.exp(1j * self.k[0] * self.x)
* np.sqrt(2 * np.pi) / self.dx)
def _set_psi_k(self, psi_k):
self.psi_mod_k = psi_k * np.exp(1j * self.x[0]
* self.dk * np.arange(self.N))
def _get_psi_k(self):
return self.psi_mod_k * np.exp(-1j * self.x[0] *
self.dk * np.arange(self.N))
def _get_dt(self):
return self.dt_
def _set_dt(self, dt):
if dt != self.dt_:
self.dt_ = dt
self.x_evolve_half = np.exp(-0.5 * 1j * self.V_x
/ self.hbar * dt )
self.x_evolve = self.x_evolve_half * self.x_evolve_half
self.k_evolve = np.exp(-0.5 * 1j * self.hbar /
self.m * (self.k * self.k) * dt)
psi_x = property(_get_psi_x, _set_psi_x)
psi_k = property(_get_psi_k, _set_psi_k)
dt = property(_get_dt, _set_dt)
def compute_k_from_x(self):
self.psi_mod_k = fft(self.psi_mod_x)
def compute_x_from_k(self):
self.psi_mod_x = ifft(self.psi_mod_k)
def time_step(self, dt, Nsteps = 1):
"""
Perform a series of time-steps via the time-dependent
Schrodinger Equation.
Parameters
----------
dt : float
the small time interval over which to integrate
Nsteps : float, optional
the number of intervals to compute. The total change
in time at the end of this method will be dt * Nsteps.
default is N = 1
"""
self.dt = dt
if Nsteps > 0:
self.psi_mod_x *= self.x_evolve_half
for i in xrange(Nsteps - 1):
self.compute_k_from_x()
self.psi_mod_k *= self.k_evolve
self.compute_x_from_k()
self.psi_mod_x *= self.x_evolve
self.compute_k_from_x()
self.psi_mod_k *= self.k_evolve
self.compute_x_from_k()
self.psi_mod_x *= self.x_evolve_half
self.compute_k_from_x()
self.t += dt * Nsteps
######################################################################
# Helper functions for gaussian wave-packets
def gauss_x(x, a, x0, k0):
"""
a gaussian wave packet of width a, centered at x0, with momentum k0
"""
return ((a * np.sqrt(np.pi)) ** (-0.5)
* np.exp(-0.5 * ((x - x0) * 1. / a) ** 2 + 1j * x * k0))
def gauss_k(k,a,x0,k0):
"""
analytical fourier transform of gauss_x(x), above
"""
return ((a / np.sqrt(np.pi))**0.5
* np.exp(-0.5 * (a * (k - k0)) ** 2 - 1j * (k - k0) * x0))
######################################################################
# Utility functions for running the animation
def theta(x):
"""
theta function :
returns 0 if x<=0, and 1 if x>0
"""
x = np.asarray(x)
y = np.zeros(x.shape)
y[x > 0] = 1.0
return y
def square_barrier(x, width, height):
return height * (theta(x) - theta(x - width))
######################################################################
# Create the animation
# specify time steps and duration
dt = 0.01
N_steps = 50
t_max = 120
frames = int(t_max / float(N_steps * dt))
# specify constants
hbar = 1.0 # planck's constant
m = 1.9 # particle mass
# specify range in x coordinate
N = 2 ** 11
dx = 0.1
x = dx * (np.arange(N) - 0.5 * N)
# specify potential
V0 = 1.5
L = hbar / np.sqrt(2 * m * V0)
a = 3 * L
x0 = -60 * L
V_x = square_barrier(x, a, V0)
V_x[x < -98] = 1E6
V_x[x > 98] = 1E6
# specify initial momentum and quantities derived from it
p0 = np.sqrt(2 * m * 0.2 * V0)
dp2 = p0 * p0 * 1./80
d = hbar / np.sqrt(2 * dp2)
k0 = p0 / hbar
v0 = p0 / m
psi_x0 = gauss_x(x, d, x0, k0)
# define the Schrodinger object which performs the calculations
S = Schrodinger(x=x,
psi_x0=psi_x0,
V_x=V_x,
hbar=hbar,
m=m,
k0=-28)
######################################################################
# Set up plot
fig = pl.figure()
# plotting limits
xlim = (-100, 100)
klim = (-5, 5)
# top axes show the x-space data
ymin = 0
ymax = V0
ax1 = fig.add_subplot(211, xlim=xlim,
ylim=(ymin - 0.2 * (ymax - ymin),
ymax + 0.2 * (ymax - ymin)))
psi_x_line, = ax1.plot([], [], c='r', label=r'$|\psi(x)|$')
V_x_line, = ax1.plot([], [], c='k', label=r'$V(x)$')
center_line = ax1.axvline(0, c='k', ls=':',
label = r"$x_0 + v_0t$")
title = ax1.set_title("")
ax1.legend(prop=dict(size=12))
ax1.set_xlabel('$x$')
ax1.set_ylabel(r'$|\psi(x)|$')
# bottom axes show the k-space data
ymin = abs(S.psi_k).min()
ymax = abs(S.psi_k).max()
ax2 = fig.add_subplot(212, xlim=klim,
ylim=(ymin - 0.2 * (ymax - ymin),
ymax + 0.2 * (ymax - ymin)))
psi_k_line, = ax2.plot([], [], c='r', label=r'$|\psi(k)|$')
p0_line1 = ax2.axvline(-p0 / hbar, c='k', ls=':', label=r'$\pm p_0$')
p0_line2 = ax2.axvline(p0 / hbar, c='k', ls=':')
mV_line = ax2.axvline(np.sqrt(2 * V0) / hbar, c='k', ls='--',
label=r'$\sqrt{2mV_0}$')
ax2.legend(prop=dict(size=12))
ax2.set_xlabel('$k$')
ax2.set_ylabel(r'$|\psi(k)|$')
V_x_line.set_data(S.x, S.V_x)
######################################################################
# Animate plot
def init():
psi_x_line.set_data([], [])
V_x_line.set_data([], [])
center_line.set_data([], [])
psi_k_line.set_data([], [])
title.set_text("")
return (psi_x_line, V_x_line, center_line, psi_k_line, title)
def animate(i):
S.time_step(dt, N_steps)
psi_x_line.set_data(S.x, 4 * abs(S.psi_x))
V_x_line.set_data(S.x, S.V_x)
center_line.set_data(2 * [x0 + S.t * p0 / m], [0, 1])
psi_k_line.set_data(S.k, abs(S.psi_k))
title.set_text("t = %.2f" % S.t)
return (psi_x_line, V_x_line, center_line, psi_k_line, title)
# call the animator. blit=True means only re-draw the parts that have changed.
anim = animation.FuncAnimation(fig, animate, init_func=init,
frames=frames, interval=30, blit=True)
# uncomment the following line to save the video in mp4 format. This
# requires either mencoder or ffmpeg to be installed on your system
#anim.save('schrodinger_barrier.mp4', fps=15, extra_args=['-vcodec', 'libx264'])
pl.show()
For your ease I mention the lines of this code which I do suspect for the error are line number 238,247,255,286,287,296,297
Thanks in advance
The problem is resolved when blit=False, though it may slow down your animation.
Just quoting from a previous answer:
"Possible solutions are:
Put the title inside the axes.
Don't use blitting"
See: How to update plot title with matplotlib using animation?
You also need ffmpeg installed. There are other answers on stackoverflow that help you through that installation. But for this script, here are my recommended new lines you need to add, assuming you're using Windows:
pl.rcParams['animation.ffmpeg_path'] = r"PUT_YOUR_FULL_PATH_TO_FFMPEG_HERE\ffmpeg.exe"
Writer = animation.writers['ffmpeg']
Then adjust the anim.save() line to:
anim.save('schrodinger_barrier.mp4', writer=Writer(fps=15, codec='libx264'))

Electrostatic Force - Simulate Trajectory of Test Particle using Runge Kutta - Force always Repels

In the center of a 2D-Plane a positive static charge Q is placed with position r_prime. This charge creates a static electrical Field E.
Now i want to place a test particle with charge Q and position vector r in this static E-Field and compute its trajectory using the 4th order Runge-Kutta method.
For the initial conditions
Q = 1, r_prime=(0,0)
q = -1, r = (9, 0), v = (0,0)
one would expect, that the negative charged test particle should move towards the positive charge in the center. Instead i get the following result for the time evolution of the test particles x component:
[9.0,
9.0,
8.999876557604697,
8.99839964155741,
8.992891977287334,
8.979313669171093,
8.95243555913327,
8.906134626441052,
8.83385018027209,
8.729257993736123,
8.587258805422984,
8.405449606446608,
8.186368339303788,
7.940995661159361,
7.694260386250479,
7.493501689700884,
7.420415546859942,
7.604287312065716,
8.226733652039988,
9.498656905483394,
11.60015461031076,
14.621662121713964,
18.56593806599109,
....
The results of the first iteration steps show the correct behavior, but then the particle is strangely repelled to infinity. There must be a major flaw in my implementation of the Runge-Kutta Method, but I checked it several times and I cant find any...
Could someone please take a quick look over my implementation and see if they can find a bug.
"""
Computes the trajectory of a test particle with Charge q with position vector r = R[:2] in a
electrostatic field that is generated by charge Q with fixed position r_prime
"""
import numpy as np
import matplotlib.pyplot as plt
def distance(r, r_prime, n):
"""
computes the euclidean distance to the power n between position x and x_prime
"""
return np.linalg.norm(r - r_prime)**n
def f(R, r_prime, q, Q):
"""
The equations of motion for the particle is given by:
d^2/dt^2 r(t) = F = constants * q * Q * (r - r_prime)/||r - r_prime||^3
To apply the Runge-Kutta-Method we transform the above (constants are set to 1)
two dimensional second order ODE into four one dimensional ODEs:
d/dt r_x = v_x
d/dt r_y = v_y
d/dt v_x = q * Q * (r_x - r_prime_x)/||r - r_prime||^3
d/dt v_y = q * Q * (r_y - r_prime_y)/||r - r_prime||^3 '''
"""
r_x, r_y = R[0], R[1]
v_x, v_y = R[2], R[3]
dist = 1 / distance(np.array([r_x, r_y]), r_prime, 3)
# Right Hand Side of the 1D Odes
f1 = v_x
f2 = v_y
f3 = q * Q * dist * (r_x - r_prime[0])
f4 = q * Q * dist * (r_y - r_prime[1])
return np.array([f1, f2, f3, f4], float)
# Constants for the Simulation
a = 0.0 # t_0
b = 10.0 # t_end
N = 100 # number of iterations
h = (b-a) / N # time step
tpoints = np.arange(a,b+h,h)
# Create lists to store the computed values
xpoints, ypoints = [], []
vxpoints, vypoints = [], []
# Initial Conditions
Q, r_prime = 1, np.array([0,0], float) # charge and position of particle that creates the static E-Field
q, R = -1, np.array([9,0,0,0], float) # charge and its initial position + velocity r=R[:2], v=[2:]
for dt in tpoints:
xpoints.append(R[0])
ypoints.append(R[1])
vxpoints.append(R[2])
vypoints.append(R[3])
# Runge-Kutta-4th Order Method
k1 = dt * f(R, r_prime, q, Q)
k2 = dt * f(R + 0.5 * k1, r_prime, q, Q)
k3 = dt * f(R + 0.5 * k2, r_prime, q, Q)
k4 = dt * f(R + k3, r_prime, q, Q)
R += (k1 + 2*k2 * 2*k3 + k4)
plt.plot(tpoints, xpoints) # should converge to 0

Stop Integrating when Output Reaches 0 in scipy.integrate.odeint

I've written a code which looks at projectile motion of an object with drag. I'm using odeint from scipy to do the forward Euler method. The integration runs until a time limit is reached. I would like to stop integrating either when this limit is reached or when the output for ry = 0 (i.e. the projectile has landed).
def f(y, t):
# takes vector y(t) and time t and returns function y_dot = F(t,y) as a vector
# y(t) = [vx(t), vy(t), rx(t), ry(t)]
# magnitude of velocity calculated
v = np.sqrt(y[0] ** 2 + y[1] **2)
# define new drag constant k
k = cd * rho * v * A / (2 * m)
return [-k * y[0], -k * y[1] - g, y[0], y[1]]
def solve_f(v0, ang, h0, tstep, tmax=1000):
# uses the forward Euler time integration method to solve the ODE using f function
# create vector for time steps based on args
t = np.linspace(0, tmax, tmax / tstep)
# convert angle from degrees to radians
ang = ang * np.pi / 180
return odeint(f, [v0 * np.cos(ang), v0 * np.sin(ang), 0, h0], t)
solution = solve_f(v0, ang, h0, tstep)
I've tried several loops and similar to try to stop integrating when ry = 0. And found this question below but have not been able to implement something similar with odeint. solution[:,3] is the output column for ry. Is there a simple way to do this with odeint?
Does scipy.integrate.ode.set_solout work?
Checkout scipy.integrate.ode here. It is more flexible than odeint and helps with what you want to do.
A simple example using a vertical shot, integrated until it touches ground:
from scipy.integrate import ode, odeint
import scipy.constants as SPC
def f(t, y):
return [y[1], -SPC.g]
v0 = 10
y0 = 0
r = ode(f)
r.set_initial_value([y0, v0], 0)
dt = 0.1
while r.successful() and r.y[0] >= 0:
print('time: {:.3f}, y: {:.3f}, vy: {:.3f}'.format(r.t + dt, *r.integrate(r.t + dt)))
Each time you call r.integrate, r will store current time and y value. You can pass them to a list if you want to store them.
Let's solve this as the boundary value problem that it is. We have the conditions x(0)=0, y(0)=h0, vx(0)=0, vy(0)=0 and y(T)=0. To have a fixed integration interval, set t=T*s, which means that dx/ds=T*dx/dt=T*vx etc.
def fall_ode(t,u,p):
vx, vy, rx, ry = u
T = p[0]
# magnitude of velocity calculated
v = np.hypot(vx, vy)
# define new drag constant k
k = cd * rho * v * A / (2 * m)
return np.array([-k * vx, -k * vy - g, vx, vy])*T
def solve_fall(v0, ang, h0):
# convert angle from degrees to radians
ang = ang * np.pi / 180
vx0, vy0 = v0*np.cos(ang), v0*np.sin(ang)
def fall_bc(y0, y1, p): return [ y0[0]-vx0, y0[1]-vy0, y0[2], y0[3]-h0, y1[3] ]
t_init = [0,1]
u_init = [[0,0],[0,0],[0,0], [h0,0]]
p_init = [1]
res = solve_bvp(fall_ode, fall_bc, t_init, u_init, p_init)
print res.message
if res.success:
print "time to ground: ", res.p[0]
# res.sol is a dense output, evaluation interpolates the computed values
return res.sol
sol = solve_fall(300, 30, 20)
s = np.linspace(0,1,501)
u = sol(s)
vx, vy, rx, ry = u
plt.plot(rx, ry)

Explicit Euler method doesn't behave how I expect

I have implemented the following explicit euler method in python:
def explicit_euler(df, x0, h, N):
"""Solves an ODE IVP using the Explicit Euler method.
Keyword arguments:
df - The derivative of the system you wish to solve.
x0 - The initial value of the system you wish to solve.
h - The step size.
N - The number off steps.
"""
x = np.zeros(N)
x[0] = x0
for i in range(0, N-1):
x[i+1] = x[i] + h * df(x[i])
return x
Following the article on wikipedia I can plot the function and verify that I get the same plot: . I believe that here the method I have written is working correctly.
Next I tried to use it to solve the last system given on this page and instead of the plot shown there I obtain this:
I am not sure why my plot doesn't match the one shown on the webpage. The explicit euler method seems to work fine when I use it to solve systems where the slope doesn't change, but for an oscillating function it never seems to mimic it at all. Not even showing the expected error gain as indicated on the linked webpage. I am not sure what is wrong with the method I have implemented.
Here is the code used for plotting and the derivative:
def g(t):
return -0.5 * np.exp(t * 0.5) * np.sin(5 * t) + 5 * np.exp(t * 0.5)
* np.cos(5 * t)
h = 0.001
x0 = 0
tn = 4
N = int(tn / h)
x = ee.explicit_euler(f, x0, h, N)
t = np.arange(0, tn, h)
fig = plt.figure()
plt.plot(t, x, label="Explicit Euler")
plt.plot(t, (np.exp(0.5 * t) * np.sin(5 * t)), label="Analytical
solution")
#plt.plot(t, np.exp(0.5 * t), label="Analytical solution")
plt.xlabel('Timesteps t')
plt.ylabel('x(t)=e^(0.5*t) * sin(5*t)')
plt.legend()
plt.grid()
plt.show()
Edit:
As requested here is the current equation I am applying the method to:
y'-y=-0.5*e^(t/2)*sin(5t)+5e^(t/2)*cos(5t)
Where y(0)=0.
I would like to make clear however that this behaviour doesn't occur just for this equation but all equations where the slope has a change in sign, or oscillating behaviour.
Edit 2:
Ok thanks. Yes the code below does indeed work. But I have one further question. In the simple example I had for the exponential function, I had defined a method:
def f(x):
return x
for the system f'(x)=x. This gave the output of my first graph which looks correct. I then defined another function:
def k(x):
return cos(x)
for the system f'(x)=cos(x), this does not give expected output. But when I change the function definition to
def k(t, x):
return cos(t)
I get the expected output. If I change my function
def f(t, x):
return t
I get an incorrect output. Am I always actually evaluating the function at a time step and is it just by chance for the system x'=x that at each time step the value is just the value of x?
I had understood that the Euler method used the value of the previously calculated value in order to get the next value. But if I run code for my function k(x)=cos(x), I get output pictured below, which must be incorrect. This now uses the updated code you provided.
def k(t, x):
return np.cos(x)
h = 0.1 # Step size
x0 = (0, 0) # Initial point of iteration
tn = 10 # Time step to iterate to
N = int(tn / h) # Number of steps
x = ee.explicit_euler(k, x0, h, N)
t = np.arange(0, tn, h)
The problem is that you have incorrectly raised the function g, you want to solve the equation:
From where we observe that:
y' = y -0.5*e^(t/2)*sin(5t)+5e^(t/2)*cos(5t)
Then we define the function f(t, y) = y -0.5*e^(t/2)*sin(5t)+5e^(t/2)*cos(5t) as:
def f(t, y):
return y -0.5 * np.exp(t * 0.5) * np.sin(5 * t) + 5 * np.exp(t * 0.5) * np.cos(5 * t)
The initial point of iteration is f0=(t(0), y(0)):
f0 = (0, 0)
Then from Euler's equations:
def explicit_euler(df, x0, h, N):
"""Solves an ODE IVP using the Explicit Euler method.
Keyword arguments:
df - The derivative of the system you wish to solve.
x0 - The initial value of the system you wish to solve.
h - The step size.
N - The number off steps.
"""
x = np.zeros(N)
t, x[0] = x0
for i in range(0, N-1):
x[i+1] = x[i] + h * df(t ,x[i])
t += h
return x
Complete Code:
def explicit_euler(df, x0, h, N):
"""Solves an ODE IVP using the Explicit Euler method.
Keyword arguments:
df - The derivative of the system you wish to solve.
x0 - The initial value of the system you wish to solve.
h - The step size.
N - The number off steps.
"""
x = np.zeros(N)
t, x[0] = x0
for i in range(0, N-1):
x[i+1] = x[i] + h * df(t ,x[i])
t += h
return x
def df(t, y):
return -0.5 * np.exp(t * 0.5) * np.sin(5 * t) + 5 * np.exp(t * 0.5) * np.cos(5 * t) + y
h = 0.001
f0 = (0, 0)
tn = 4
N = int(tn / h)
x = explicit_euler(df, f0, h, N)
t = np.arange(0, tn, h)
fig = plt.figure()
plt.plot(t, x, label="Explicit Euler")
plt.plot(t, (np.exp(0.5 * t) * np.sin(5 * t)), label="Analytical solution")
#plt.plot(t, np.exp(0.5 * t), label="Analytical solution")
plt.xlabel('Timesteps t')
plt.ylabel('x(t)=e^(0.5*t) * sin(5*t)')
plt.legend()
plt.grid()
plt.show()
Screenshot:
Dump y' and what is on the right side is what you should place in the df function.
We will modify the variables to maintain the same standard for the variables, and will y be the dependent variable, and t the independent variable.
Equation 2: In this case the equation f'(x)=cos(x) will be rewritten to:
y'=cos(t)
Then:
def df(t, y):
return np.cos(t)
In conclusion, if we have an equation of the following form:
y' = f(t, y)
Then:
def df(t, y):
return f(t, y)

Approximation by sin waves using DFT on python. What's wrong?

I'm writing the prorgram on python that can approximate time series by sin waves.
The program uses DFT to find sin waves, after that it chooses sin waves with biggest amplitudes.
Here's my code:
__author__ = 'FATVVS'
import math
# Wave - (amplitude,frequency,phase)
# This class was created to sort sin waves:
# - by anplitude( set freq_sort=False)
# - by frequency (set freq_sort=True)
class Wave:
#flag for choosing sort mode:
# False-sort by amplitude
# True-by frequency
freq_sort = False
def __init__(self, amp, freq, phase):
self.freq = freq #frequency
self.amp = amp #amplitude
self.phase = phase
def __lt__(self, other):
if self.freq_sort:
return self.freq < other.freq
else:
return self.amp < other.amp
def __gt__(self, other):
if self.freq_sort:
return self.freq > other.freq
else:
return self.amp > other.amp
def __le__(self, other):
if self.freq_sort:
return self.freq <= other.freq
else:
return self.amp <= other.amp
def __ge__(self, other):
if self.freq_sort:
return self.freq >= other.freq
else:
return self.amp >= other.amp
def __str__(self):
s = "(amp=" + str(self.amp) + ",frq=" + str(self.freq) + ",phase=" + str(self.phase) + ")"
return s
def __repr__(self):
return self.__str__()
#Discrete Fourier Transform
def dft(series: list):
n = len(series)
m = int(n / 2)
real = [0 for _ in range(n)]
imag = [0 for _ in range(n)]
amplitude = []
phase = []
angle_const = 2 * math.pi / n
for w in range(m):
a = w * angle_const
for t in range(n):
real[w] += series[t] * math.cos(a * t)
imag[w] += series[t] * math.sin(a * t)
amplitude.append(math.sqrt(real[w] * real[w] + imag[w] * imag[w]) / n)
phase.append(math.atan(imag[w] / real[w]))
return amplitude, phase
#extract waves from time series
# series - time series
# num - number of waves
def get_waves(series: list, num):
amp, phase = dft(series)
m = len(amp)
waves = []
for i in range(m):
waves.append(Wave(amp[i], 2 * math.pi * i / m, phase[i]))
waves.sort()
waves.reverse()
waves = waves[0:num]#extract best waves
print("the program found the next %s sin waves:"%(num))
print(waves)#print best waves
return waves
#approximation by sin waves
#series - time series
#num- number of sin waves
def sin_waves_appr(series: list, num):
n = len(series)
freq = get_waves(series, num)
m = len(freq)
model = []
for i in range(n):
summ = 0
for j in range(m): #sum by sin waves
summ += freq[j].amp * math.sin(freq[j].freq * i + freq[j].phase)
model.append(summ)
return model
if __name__ == '__main__':
import matplotlib.pyplot as plt
N = 500 # length of time series
num = 2 # number of sin wawes, that we want to find
#y - generate time series
y = [2 * math.sin(0.05 * t + 0.5) + 0.5 * math.sin(0.2 * t + 1.5) for t in range(N)]
model = sin_waves_appr(y, num) #generate approximation model
## ------------------plotting-----------------
plt.figure(1)
# plotting of time series and his approximation model
plt.subplot(211)
h_signal, = plt.plot(y, label='source timeseries')
h_model, = plt.plot(model, label='model', linestyle='--')
plt.legend(handles=[h_signal, h_model])
plt.grid()
# plotting of spectre
amp, _ = dft(y)
xaxis = [2*math.pi*i / N for i in range(len(amp))]
plt.subplot(212)
h_freq, = plt.plot(xaxis, amp, label='spectre')
plt.legend(handles=[h_freq])
plt.grid()
plt.show()
But I've got a strange result:
In the program I've created a time series from two sin waves:
y = [2 * math.sin(0.05 * t + 0.5) + 0.5 * math.sin(0.2 * t + 1.5) for t in range(N)]
And my program found wrong parameters of the sin waves:
the program found the next 2 sin waves:
[(amp=0.9998029885151699,frq=0.10053096491487339,phase=1.1411803525843616), (amp=0.24800925225626422,frq=0.40212385965949354,phase=0.346757128184013)]
I suppuse, that my problem is wrong scaling of wave parameters, but I'm not sure.
There're two places, where the program does scaling. The first place is creating of waves:
for i in range(m):
waves.append(Wave(amp[i], 2 * math.pi * i / m, phase[i]))
And the second place is sclaling of the x-axis:
xaxis = [2*math.pi*i / N for i in range(len(amp))]
But my suppose may be wrong. I've tried to change scaling many times, and it haven't solved my problem.
What may be wrong with the code?
So, these lines I believe are wrong:
for t in range(n):
real[w] += series[t] * math.cos(a * t)
imag[w] += series[t] * math.sin(a * t)
amplitude.append(math.sqrt(real[w] * real[w] + imag[w] * imag[w]) / n)
phase.append(math.atan(imag[w] / real[w]))
I believe it should be dividing by m instead of n, since you are only working with computing half the points. That will fix the amplitude problem. Also, the computation of imag[w] is missing a negative sign. Taking into account the atan2 fix, it would look like:
for t in range(n):
real[w] += series[t] * math.cos(a * t)
imag[w] += -1 * series[t] * math.sin(a * t)
amplitude.append(math.sqrt(real[w] * real[w] + imag[w] * imag[w]) / m)
phase.append(math.atan2(imag[w], real[w]))
The next one is here:
for i in range(m):
waves.append(Wave(amp[i], 2 * math.pi * i / m, phase[i]))
The divide by m is not right. amp has only half the points it should, so using the length of amp isn't right here. It should be:
for i in range(m):
waves.append(Wave(amp[i], 2 * math.pi * i / (m * 2), phase[i]))
Finally, your model reconstruction has a problem:
for j in range(m): #sum by sin waves
summ += freq[j].amp * math.sin(freq[j].freq * i + freq[j].phase)
It should use cosine instead (sine introduces a phase shift):
for j in range(m): #sum by cos waves
summ += freq[j].amp * math.cos(freq[j].freq * i + freq[j].phase)
When I fix all of that, the model and the DFT both make sense:

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