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SciPy ODEINT not working with certain differential equations

I'm new to Python and trying to create basic trajectory plots of a 2D system. This is what I'm working with right now. It plots forward trajectories for a number of points.
import numpy as np
import matplotlib.pyplot as plt
import scipy as sp
from scipy.integrate import odeint
def dx_dt(x,t):
return [x[0]*2, x[1]]
xmin = ymin = -10
xmax = ymax = 10
plt.figure()
plt.xlim(xmin, xmax)
plt.ylim(ymin, ymax)
ts = np.linspace(0,1,20)
ic = np.linspace(xmin, xmax, 11)
for r in ic:
for s in ic:
x0 = [r, s]
xs = sp.integrate.odeint(dx_dt, x0, ts)
plt.plot(xs[:,0], xs[:,1], "g")
plt.savefig("flowlines.jpg")
The issue that's arising is ODEINT is not working with certain systems, instead saying this half a dozen times
ODEintWarning: Excess work done on this call (perhaps wrong Dfun type). Run with full_output = 1 to get quantitative information.
and subsequently crashing.
I've been playing around with the specific systems of differential equations, and I think I've found the specific problem that's arising. Something like
def dx_dt(x,t):
return [x[1], x[0]**2+x[1]]
works, while
def dx_dt(x,t):
return [x[0], x[1]**2+x[0]]
does not. It seems that it does not like any specific initial conditions where x' =f(x,y) if f(x,y) contains anything involving x other than addition and scalar multiplication. It can do anything it wants with y. Hence, the first one runs just fine, but the second one fails, because y' = y^2+x involves a power of y.
I have no idea how to proceed from here.

You suspicion about the behavior of the solver depending on the algebraic form of the equations is reasonable. It is likely that the systems for which you are experiencing a problem are ones where the solution "blows up" in finite time. That is, one or both components go to infinity as t approaches a finite value. This can easily happen when the right-hand side depends on powers of the variables.
A simple one-dimensional system that exhibits this behavior is
dx/dt = x**2, x(0) = x0
which has the explicit solution
x(t) = x0/(1 - x0*t)
You can verify that it solves the differential equation and initial condition, at least in a neighborhood of t=0. As t → 1/x0 (assuming x0 ≠ 0), the solution diverges to infinity. For example, here's a plot of the solution with x(0) = 0.5:
The red vertical dashed line indicates where the solution blows up.
Equations like this violate the mathematical assumptions on which ODE solvers such as odeint are based. Solvers will generally fail when they encounter such a point, typically because they vainly try to take smaller and smaller steps to bring down the estimated local error as they approach the point where the solution blows up:
In [54]: from scipy.integrate import odeint
In [55]: def sys(x, t):
...: return x**2
...:
In [56]: odeint(sys, 0.5, [0, 2.5])
[...]/lib/python3.9/site-packages/scipy/integrate/odepack.py:247:
ODEintWarning: Excess work done on this call (perhaps wrong Dfun type).
Run with full_output = 1 to get quantitative information.
warnings.warn(warning_msg, ODEintWarning)
Out[56]:
array([[5.00000000e-01],
[4.82437777e+08]])
The solution is the same as the punch line to the old joke:
Patient: Doctor, it hurts when I do this...
Doctor: Then don't do that!

Solving two coupled second order boundary value problems

I have solved a single second order differential equation with two boundary conditions using the module solve_bvp. However, now I am trying to solve the system of two second order differential equations;
U'' + a*B' = 0
B'' + b*U' = 0
with the boundary conditions U(+/-0.5) = +/-0.01 and B(+/-0.5) = 0. I have split this into a system of first ordinary differential equations and I am trying to use solve_bvp to solve them numerically. However, I am just getting arrays full of zeros for my solution. I believe I am implementing the boundary conditions wrong. It is not clear to me how to handle more than two equations from the documentation. My attempt is below
import numpy as np
from scipy.integrate import solve_bvp
import matplotlib.pyplot as plt
%matplotlib inline
from scipy.integrate import solve_bvp
alpha = 1E-8
zeta = 8E-3
C_k = 0.05
sigma = 0.01
def fun(x, y):
return np.vstack((y[1],-((alpha)/(C_k*sigma))*y[2],y[2], -(1/(C_k*zeta))*y[1]))
def bc(ya, yb):
return np.array([ya[0]+0.001, yb[0]-0.001,ya[0]-0, yb[0]-0])
x = np.linspace(-0.5, 0.5, 5000)
y = np.zeros((4, x.size))
print(y)
sol = solve_bvp(fun, bc, x, y)
print(sol)
In my question I have just relabeled a and b, but they're just parameters that I input. I have the analytic solution for this set of equations so I know one exists that is non-trivial. Any help would be greatly appreciated.

It is most times really helpful if you state at least once in a comment or by assignment to specifically named variables how you want to compose the state vector.
By the form of the derivative return vector, I would think you intend
U, U', B, B'
which means that U=y[0], U'=y[1] and B=y[2],B'=y[3], so that your derivatives vector should correctly be
return y[1], -((alpha)/(C_k*sigma))*y[3], y[3], -(1/(C_k*zeta))*y[1]
and the boundary conditions
return ya[0]+0.001, yb[0]-0.001, ya[2]-0, yb[2]-0
Especially your boundary condition should throw the algorithm in the first step because of a singular Jacobian, always check the .success field and the .message field of the solution structure.
Note that by default the absolute and relative tolerance of the experimental solve_bvp is 1e-3, and the number of nodes is limited to 500.
Setting the initial node number to 50 (5000 is much too much, the solver refines where necessary), and the tolerance to 1-6, I get the following solution plots that visibly satisfy the boundary conditions.

Is there a way to setup and solve multiple numerical integration in scipy like in mathematica?

I would like to solve the following formulas numerically in Python;
In Mathematica, you can input multiple differential equations and solve it at the same time. Is there a way to do the similar thing with Scipy?
edit: yes, I have looked at scipy.integrate.odeint already, but I'm still not sure how I can solve multiple equations that correlates each other at the same time. Does anyone have suggestion for that?

Eventually I figured out myself and I'm writing down for other people who might be clueless like me;
in odeint, you give three parameters model, y, and t where model is a function that takes in y and t then return dydt, y is the initial value of y, and t is the variable you're trying to take integral over. If you have multiple differential equations that are dependent on each other, you can just pass in all of them to odeint. In my case,
t = np.linspace(0, 20) # range of t
y0 = [No_0, Na_0, Ni_0, Nn_0] # initial condition for each Ns
def model(y, t):
No, Na, Ni, Nn = y
dNodt = -k_oa * No
dNadt = k_oa * No - k_ai * Na
dNidt = k_ai * Na - k_in * Ni
dNndt = k_in * Ni
return [dNodt, dNadt, dNidt, dNndt]
y = odeint(model, y0, t)
You can define multiple differential equations you want to solve within the model you define, and pass in to odeint along with the initial values for all species.

Have a loot at SciPy's Integration package:
https://docs.scipy.org/doc/scipy/reference/integrate.html
specifically at odeint to solve systems of ODE's:
https://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.odeint.html#scipy.integrate.odeint

Need help solving numerical ODE [Python]

I have a physics problem and have never used odeint or any numerical solving for ODE on python and am a little confused. I tried looking at other examples but could not understand and am hoping for a little help. My ODE is:
Where α is a given angle and g is a constant.
r=sqrt(x^2+y^2)
The program will ask for x_0, dx/dt_0 and dy/dt_0.
I'm mostly unsure how to solve ODE's in python. I have seen that I should split my ODE into dr'/dt because odeint will only do a first order ODE's. Could someone help explain how to do this?
I tried using another example to do as much as possible but am stuck:
import numpy as np
import matplotlib.pylab as plt
from scipy.integrate import odeint
pi=np.pi
sin=np.sin
cos=np.cos
sqrt=np.sqrt
alpha=pi/4
g=9.80665
y0=0.0
theta0=0.0
x=[]
y=[]
sina = sin(alpha)**2
second_term = g*sin(alpha)*cos(alpha)
x0 = float(raw_input('What is the initial x in meters?'))
x_vel0 = float(raw_input('What is the initial velocity in the x direction in m/s?'))
y_vel0 = float(raw_input('what is the initial velocity in the y direction in m/s?'))
t_f = float(raw_input('What is the maximum time in seconds?'))
r0 = x0
r = sqrt(float(x)**2 + float(y)**2)
def deriv_z(z,r):
r, rdot=z
return [rdot,r*sina-second_term]
zdot0=x_vel0**2+y_vel0**2
z0 = [r0,zdot0]
times = np.linespace(0, t_f, 1000)
z = odeint(deriv_z,z0,times)

There's a great example I found to help me with my planetary orbit ODE solving.
It uses an adaptive step size solver and plots the orbit nicely. If your solution can handle it, you may also try to use the faster 'lsoda' or 'vode' options instead of 'dopri5' (which is a very solid standard).

On ordinary differential equations (ODE) and optimization, in Python

I want to solve this kind of problem:
dy/dt = 0.01*y*(1-y), find t when y = 0.8 (0<t<3000)
I've tried the ode function in Python, but it can only calculate y when t is given.
So are there any simple ways to solve this problem in Python?
PS: This function is just a simple example. My real problem is so complex that can't be solve analytically. So I want to know how to solve it numerically. And I think this problem is more like an optimization problem:
Objective function y(t) = 0.8, Subject to dy/dt = 0.01*y*(1-y), and 0<t<3000
PPS: My real problem is:
objective function: F(t) = 0.85,
subject to: F(t) = sqrt(x(t)^2+y(t)^2+z(t)^2),
x''(t) = (1/F(t)-1)*250*x(t),
y''(t) = (1/F(t)-1)*250*y(t),
z''(t) = (1/F(t)-1)*250*z(t)-10,
x(0) = 0, y(0) = 0, z(0) = 0.7,
x'(0) = 0.1, y'(0) = 1.5, z'(0) = 0,
0<t<5

This differential equation can be solved analytically quite easily:
dy/dt = 0.01 * y * (1-y)
rearrange to gather y and t terms on opposite sides
100 dt = 1/(y * (1-y)) dy
The lhs integrates trivially to 100 * t, rhs is slightly more complicated. We can always write a product of two quotients as a sum of the two quotients * some constants:
1/(y * (1-y)) = A/y + B/(1-y)
The values for A and B can be worked out by putting the rhs on the same denominator and comparing constant and first order y terms on both sides. In this case it is simple, A=B=1. Thus we have to integrate
1/y + 1/(1-y) dy
The first term integrates to ln(y), the second term can be integrated with a change of variables u = 1-y to -ln(1-y). Our integrated equation therefor looks like:
100 * t + C = ln(y) - ln(1-y)
not forgetting the constant of integration (it is convenient to write it on the lhs here). We can combine the two logarithm terms:
100 * t + C = ln( y / (1-y) )
In order to solve t for an exact value of y, we first need to work out the value of C. We do this using the initial conditions. It is clear that if y starts at 1, dy/dt = 0 and the value of y never changes. Thus plug in the values for y and t at the beginning
100 * 0 + C = ln( y(0) / (1 - y(0) )
This will give a value for C (assuming y is not 0 or 1) and then use y=0.8 to get a value for t. Note that because of the logarithm and the factor 100 multiplying t y will reach 0.8 within a relatively short range of t values, unless the initial value of y is incredibly small. It is of course also straightforward to rearrange the equation above to express y in terms of t, then you can plot the function as well.
Edit: Numerical integration
For a more complexed ODE which cannot be solved analytically, you will have to try numerically. Initially we only know the value of the function at zero time y(0) (we have to know at least that in order to uniquely define the trajectory of the function), and how to evaluate the gradient. The idea of numerical integration is that we can use our knowledge of the gradient (which tells us how the function is changing) to work out what the value of the function will be in the vicinity of our starting point. The simplest way to do this is Euler integration:
y(dt) = y(0) + dy/dt * dt
Euler integration assumes that the gradient is constant between t=0 and t=dt. Once y(dt) is known, the gradient can be calculated there also and in turn used to calculate y(2 * dt) and so on, gradually building up the complete trajectory of the function. If you are looking for a particular target value, just wait until the trajectory goes past that value, then interpolate between the last two positions to get the precise t.
The problem with Euler integration (and with all other numerical integration methods) is that its results are only accurate when its assumptions are valid. Because the gradient is not constant between pairs of time points, a certain amount of error will arise for each integration step, which over time will build up until the answer is completely inaccurate. In order to improve the quality of the integration, it is necessary to use more sophisticated approximations to the gradient. Check out for example the Runge-Kutta methods, which are a family of integrators which remove progressive orders of error term at the cost of increased computation time. If your function is differentiable, knowing the second or even third derivatives can also be used to reduce the integration error.
Fortunately of course, somebody else has done the hard work here, and you don't have to worry too much about solving problems like numerical stability or have an in depth understanding of all the details (although understanding roughly what is going on helps a lot). Check out http://docs.scipy.org/doc/scipy/reference/generated/scipy.integrate.ode.html#scipy.integrate.ode for an example of an integrator class which you should be able to use straightaway. For instance
from scipy.integrate import ode
def deriv(t, y):
return 0.01 * y * (1 - y)
my_integrator = ode(deriv)
my_integrator.set_initial_value(0.5)
t = 0.1 # start with a small value of time
while t < 3000:
y = my_integrator.integrate(t)
if y > 0.8:
print "y(%f) = %f" % (t, y)
break
t += 0.1
This code will print out the first t value when y passes 0.8 (or nothing if it never reaches 0.8). If you want a more accurate value of t, keep the y of the previous t as well and interpolate between them.

As an addition to Krastanov`s answer:
Aside of PyDSTool there are other packages, like Pysundials and Assimulo which provide bindings to the solver IDA from Sundials. This solver has root finding capabilites.

Use scipy.integrate.odeint to handle your integration, and analyse the results afterward.
import numpy as np
from scipy.integrate import odeint
ts = np.arange(0,3000,1) # time series - start, stop, step
def rhs(y,t):
return 0.01*y*(1-y)
y0 = np.array([1]) # initial value
ys = odeint(rhs,y0,ts)
Then analyse the numpy array ys to find your answer (dimensions of array ts matches ys). (This may not work first time because I am constructing from memory).
This might involve using the scipy interpolate function for the ys array, such that you get a result at time t.
EDIT: I see that you wish to solve a spring in 3D. This should be fine with the above method; Odeint on the scipy website has examples for systems such as coupled springs that can be solved for, and these could be extended.

What you are asking for is a ODE integrator with root finding capabilities. They exist and the low-level code for such integrators is supplied with scipy, but they have not yet been wrapped in python bindings.
For more information see this mailing list post that provides a few alternatives: http://mail.scipy.org/pipermail/scipy-user/2010-March/024890.html
You can use the following example implementation which uses backtracking (hence it is not optimal as it is a bolt-on addition to an integrator that does not have root finding on its own): https://github.com/scipy/scipy/pull/4904/files