Develop Reference

Double Direct Integration

I am trying to solve the set of coupled boundary value problems such that;
U'' +aB'+ b*(cosh(lambda z))^{-2}tanh(lambda*z) = 0,
B'' + c*U' = 0,
T'' = (gamma^{-1} - 1)*(d*(U')^2 + e*(B')^2)
subject to the boundary conditions U(+/- 1/2) = +/-U_0*tanh(lambda/2), B(+/- 1/2) = 0 and T(-1/2) = 1, T(1/2) = 4. I have decomposed this set of equations into a set of first order differential equations, and used the derivative array such that [U, U', B, B', T, T']. But bvp solve is returning the error that I have a single Jacobian. When I remove the last two equations, I get a solution for U and B and that works fine. However, I am unsure why adding the other two equations results in this issue.
import numpy as np
from scipy.integrate import solve_bvp
import matplotlib.pyplot as plt
%matplotlib inline
alpha = 1E-7
zeta = 8E-3
C_k = 0.01
sigma = 0.005
Q = 30
U_0 = 0.1
gamma = 5/3
theta = 3
def fun(x, y):
return y[1], -2*U_0*Q**2*(1/np.cosh(Q*x))**2*np.tanh(Q*x)-((alpha)/(C_k*sigma))*y[3], y[3],\
-(1/(C_k*zeta))*y[1], y[4], (1/gamma - 1)*(sigma*(y[1])**2 + zeta*alpha*(y[3])**2)
def bc(ya, yb):
return ya[0]+U_0*np.tanh(Q*0.5), yb[0]-U_0*np.tanh(Q*0.5), ya[2]-0, yb[2]-0, ya[4] - 1, yb[4] - 4
x = np.linspace(-0.5, 0.5, 500)
y = np.zeros((6, x.size))
sol = solve_bvp(fun, bc, x, y)
print(sol)
However, the error that I am getting is that 'setting an array with sequence'. The first function and boundary conditions solves two coupled equations, then I use these results to evaluate the equation I have given. I have tried writing all of my equations in one function, however this seems to be returning trivial solutions i.e an array full of zeros.
Any help would be appreciated.

When the expressions become larger it is often more helpful to keep the computations human readable instead of compact.
def fun(x, y):
U, dU, B, dB, T, dT = y;
d2U = -2*U_0*Q**2*(1/np.cosh(Q*x))**2*np.tanh(Q*x)-((alpha)/(C_k*sigma))*dB;
d2B = -(1/(C_k*zeta))*dU;
d2T = (1/gamma - 1)*(sigma*dU**2 + zeta*alpha*dB**2);
return dU, d2U, dB, d2B, dT, d2T
This avoids missing an index error as there are no indices in this computation, all has names close to the original formulas.
Then the solution components (using initialization with only 5 points, resulting in a refinement with 65 points) plots as

Solving set of Boundary Value Problems

I am trying to solve a set of boundary value problems given by 4 differential equations. I am using bvp_solver in python, and I am getting errors which state 'invalid value encountered in division'. I am assuming this means I am dividing by NaN or 0 at some point, but I am unsure where.
import numpy as np
from scipy.integrate import solve_bvp
import matplotlib.pyplot as plt
%matplotlib inline
alpha = 1
zeta = 1
C_k = 1
sigma = 1
Q = 30
U_0 = 0.1
gamma = 5/3
theta = 3
m = 1.5
def fun(x, y):
U, dU, B, dB, T, dT, M, dM = y;
d2U = -2*U_0*Q**2*(1/np.cosh(Q*x))**2*np.tanh(Q*x)-((alpha)/(C_k*sigma))*dB;
d2B = -(1/(C_k*zeta))*dU;
d2T = (1/gamma - 1)*(sigma*dU**2 + zeta*alpha*dB**2);
d2M = -(dM/T)*dT + (dM/T)*theta*(m+1) - (alpha/T)*B*dB
return dU, d2U, dB, d2B, dT, d2T, dM, d2M
def bc(ya, yb):
return ya[0]+U_0*np.tanh(Q*0.5), yb[0]-U_0*np.tanh(Q*0.5), ya[2]-0, yb[2]-0, ya[4] - 1, yb[4] - 4, ya[6], yb[6] - 1
x = np.linspace(-0.5, 0.5, 500)
y = np.zeros((8, x.size))
sol = solve_bvp(fun, bc, x, y)
If I remove the last two equations for M and dM, then the solution works fine. I have had trouble in the past understanding the return arrays of bvp_solver, but I am confident I understand that now. But I continue to get errors each time I add more equations. Any help is greatly appreciated.

Of course this will fail in the first step. You initialize everything to zero, and then in the derivatives function, you divide by T, which is zero from the initialization.
Find a more realistic initialization of T, for instance
x = np.linspace(-0.5, 0.5, 15)
y = np.zeros((8, x.size))
y[4] = 2.5+3*x
y[5] = 3+0*x
or
desingularize the division, which usually is done similar to
d2M = (-dM*dT + dM*theta*(m+1) - alpha*B*dB) * T/(1e-12+T*T)
It makes always sense to print after sol = solve_bvp(...) the error message print(sol.message). Now that there are more than a few components, I changed the construction of the output tableau to the systematic
%matplotlib inline
plt.figure(figsize=(10,2.5*4))
for k in range(4):
v,c = ['U','B','T','M'][k],['-+b','-*r','-xg','-om'][k];
plt.subplot(4,2,2*k+1); plt.plot(sol.x,sol.y[2*k ],c, ms=2); plt.grid(); plt.legend(["$%c$"%v]);
plt.subplot(4,2,2*k+2); plt.plot(sol.x,sol.y[2*k+1],c, ms=2); plt.grid(); plt.legend(["$%c'$"%v]);
plt.tight_layout(); plt.savefig("/tmp/bvp3.png"); plt.show(); plt.close()

add PID controller to odeint integration

I have an integral to which I want to control the input by adding a P-controller to.
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
xs = []
yd = []
def dY(y, x):
a = 0.001
yin = 1
C = 0.01
N = 1
dC = C/N
b1 = 0
y_diff = -np.copy(y)
y_diff[0] += yin
y_diff[1:] += y[:-1]
xs.append(x)
yd.append(y_diff)
return (a/dC)*y_diff+b1*dC
x = np.linspace(0,20,1000)
y0 = np.zeros(4)
res = odeint(dY, y0, x)
plt.plot(x,res, '-')
plt.gca().set_prop_cycle(plt.rcParams['axes.prop_cycle'])
plt.plot(np.array(xs),np.array(yd), '-.')
plt.show()
y_in is a variable that I want to control by a P/PI-controller in order to reach a certain setpoint. I have looked at it in different ways but I honestly have no idea where to start with implementing the controller.
I am familiar with the concept of PID controller and all related equations however this is a total new thing for me to implement in a case where ODEINT is used.
Hopefully someone could help me out.

Pendulum-like odeint integration

I'm trying to solve a pendulum-like differential equation that goes like d2(phi)/dt = -(g/R) *sin(phi) (it's the skateboard problem in Taylor's Classical Mechanics). I'm new to scipy and odeint and the like, so I'm doing this to prepare for more sophisticated numerical solutions in the future.
I've used code from here to try and navigate the coding, but all I'm coming up with is a flat line for phi(t). I think it stems from the fact that I'm trying to split a second order differential equation into two first orders, where one doesn't depend on the other (because d(phi)/dt to make an appearance); but I'm not sure how to go about fixing it.
Anyone know what's wrong with it?
# integrate skateboard problem, plot result
from numpy import *
from scipy.integrate import odeint
import matplotlib.pyplot as plt
def skate(y, t, params):
phi, omega = y
g, R = params
derivs = [omega, -(g/R)*np.sin(phi)]
return derivs
# Parameters
g = 9.81
R = 5
params = [g, R]
#Initial values
phi0 = 20
omega0 = 0
y0 = [phi0, omega0]
t = linspace(0, 20, 5000)
solution = odeint(skate, y0, t, args=(params,))
plt.plot(t, solution[:,0])
plt.xlabel('time [s]')
plt.ylabel('angle [rad]')
plt.show()

I suspect a bug here: -(g/R)*np.sin(phi). Perhaps you forget to define the alias for numpy lib import (for example: import numpy as np). Instead you just did (from numpy import *). Try this:
def skate(y, t, params):
phi, omega = y
g, R = params
derivs = [omega, -(g/R)*sin(phi)]
return derivs

Numerical ODE solving in Python

How do I numerically solve an ODE in Python?
Consider
\ddot{u}(\phi) = -u + \sqrt{u}
with the following conditions
u(0) = 1.49907
and
\dot{u}(0) = 0
with the constraint
0 <= \phi <= 7\pi.
Then finally, I want to produce a parametric plot where the x and y coordinates are generated as a function of u.
The problem is, I need to run odeint twice since this is a second order differential equation.
I tried having it run again after the first time but it comes back with a Jacobian error. There must be a way to run it twice all at once.
Here is the error:
odepack.error: The function and its Jacobian must be callable functions
which the code below generates. The line in question is the sol = odeint.
import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt
from numpy import linspace
def f(u, t):
return -u + np.sqrt(u)
times = linspace(0.0001, 7 * np.pi, 1000)
y0 = 1.49907
yprime0 = 0
yvals = odeint(f, yprime0, times)
sol = odeint(yvals, y0, times)
x = 1 / sol * np.cos(times)
y = 1 / sol * np.sin(times)
plot(x,y)
plt.show()
Edit
I am trying to construct the plot on page 9
Classical Mechanics Taylor
Here is the plot with Mathematica
In[27]:= sol =
NDSolve[{y''[t] == -y[t] + Sqrt[y[t]], y[0] == 1/.66707928,
y'[0] == 0}, y, {t, 0, 10*\[Pi]}];
In[28]:= ysol = y[t] /. sol[[1]];
In[30]:= ParametricPlot[{1/ysol*Cos[t], 1/ysol*Sin[t]}, {t, 0,
7 \[Pi]}, PlotRange -> {{-2, 2}, {-2.5, 2.5}}]

import scipy.integrate as integrate
import matplotlib.pyplot as plt
import numpy as np
pi = np.pi
sqrt = np.sqrt
cos = np.cos
sin = np.sin
def deriv_z(z, phi):
u, udot = z
return [udot, -u + sqrt(u)]
phi = np.linspace(0, 7.0*pi, 2000)
zinit = [1.49907, 0]
z = integrate.odeint(deriv_z, zinit, phi)
u, udot = z.T
# plt.plot(phi, u)
fig, ax = plt.subplots()
ax.plot(1/u*cos(phi), 1/u*sin(phi))
ax.set_aspect('equal')
plt.grid(True)
plt.show()

The code from your other question is really close to what you want. Two changes are needed:
You were solving a different ODE (because you changed two signs inside function deriv)
The y component of your desired plot comes from the solution values, not from the values of the first derivative of the solution, so you need to replace u[:,0] (function values) for u[:, 1] (derivatives).
This is the end result:
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
def deriv(u, t):
return np.array([u[1], -u[0] + np.sqrt(u[0])])
time = np.arange(0.01, 7 * np.pi, 0.0001)
uinit = np.array([1.49907, 0])
u = odeint(deriv, uinit, time)
x = 1 / u[:, 0] * np.cos(time)
y = 1 / u[:, 0] * np.sin(time)
plt.plot(x, y)
plt.show()
However, I suggest that you use the code from unutbu's answer because it's self documenting (u, udot = z) and uses np.linspace instead of np.arange. Then, run this to get your desired figure:
x = 1 / u * np.cos(phi)
y = 1 / u * np.sin(phi)
plt.plot(x, y)
plt.show()

You can use scipy.integrate.ode. To solve dy/dt = f(t,y), with initial condition y(t0)=y0, at time=t1 with 4th order Runge-Kutta you could do something like this:
from scipy.integrate import ode
solver = ode(f).set_integrator('dopri5')
solver.set_initial_value(y0, t0)
dt = 0.1
while t < t1:
y = solver.integrate(t+dt)
t += dt
Edit: You have to get your derivative to first order to use numerical integration. This you can achieve by setting e.g. z1=u and z2=du/dt, after which you have dz1/dt = z2 and dz2/dt = d^2u/dt^2. Substitute these into your original equation, and simply iterate over the vector dZ/dt, which is first order.
Edit 2: Here's an example code for the whole thing:
import numpy as np
import matplotlib.pyplot as plt
from numpy import sqrt, pi, sin, cos
from scipy.integrate import ode
# use z = [z1, z2] = [u, u']
# and then f = z' = [u', u''] = [z2, -z1+sqrt(z1)]
def f(phi, z):
return [z[1], -z[0]+sqrt(z[0])]
# initialize the 4th order Runge-Kutta solver
solver = ode(f).set_integrator('dopri5')
# initial value
z0 = [1.49907, 0.]
solver.set_initial_value(z0)
values = 1000
phi = np.linspace(0.0001, 7.*pi, values)
u = np.zeros(values)
for ii in range(values):
u[ii] = solver.integrate(phi[ii])[0] #z[0]=u
x = 1. / u * cos(phi)
y = 1. / u * sin(phi)
plt.figure()
plt.plot(x,y)
plt.grid()
plt.show()

scipy.integrate() does ODE integration. Is that what you are looking for?