I need to solve a 2nd order ODE which I have decoupled into two first order ODEs. I've tried solving it using solve_ivp, but it doesn't seem to provide the solution I expect. I have provided the code below.
import numpy as np
import matplotlib.pyplot as plt
from scipy.misc import derivative
from scipy.integrate import solve_ivp
from matplotlib import rc
rc('text', usetex = True)
V0 = 2*10**-10
A = 0.130383
f = 0.129576
phi_i_USR2 = 6.1
phi_Ni_USR2 = 1.2
def V_USR2(phi):
return V0*(np.tanh(phi/np.sqrt(6)) + A*np.sin(1/f*np.tanh(phi/np.sqrt(6))))**2
def V_phi_USR2(phi):
return derivative(V_USR2, phi)
N = np.linspace(0,66,1000)
def USR2(N, X):
phi, g = X
return [g, (g**2/2 - 3)*(V_phi_USR2(phi)/V_USR2(phi) + g)]
X0 = [phi_i_USR2, phi_Ni_USR2]
sol = solve_ivp(USR2, (0,66), X0, method = 'LSODA', t_eval = N)
phi_USR2 = sol.y[0]
phi_N_USR2 = sol.y[1]
N_USR2 = sol.t
plt.plot(phi_USR2, phi_N_USR2)
plt.xlabel("$\phi$")
plt.ylabel("$\phi'$")
plt.title("Phase plot for USR2")
plt.show()
solve_ivp gives me the following plot:
The problem is that there is supposed to be an oscillation near the origin, which is not well captured by solve_ivp. However, for the same equation and initial conditions, Mathematica gives me exactly what I want:
I want the same plot in Python as well. I tried various methods in solve_ivp such as RK45, LSODA, Radau and BDF, but all of them show the same problem (LSODA tries to depict an oscillation but fails, but the other methods don't even move past the point where the oscillation starts). It would be great if someone can shed light on the problem. Thanks in advance.
Related
I'm trying to do a guassian fit for some experimental data but I keep running into error after error. I've followed a few different threads online but either the fit isn't good (it's just a horizontal line) or the code just won't run. I'm following this code from another thread. Below is my code.
I apologize if my code seems a bit messy. There are some bits from other attempts when I tried making it work. Hence the "astropy" import.
import math as m
import matplotlib.pyplot as plt
import numpy as np
from scipy import optimize as opt
import pandas as pd
import statistics as stats
from astropy import modeling
def gaus(x,a,x0,sigma, offset):
return a*m.exp(-(x-x0)**2/(2*sigma**2)) + offset
# Python program to get average of a list
def Average(lst):
return sum(lst) / len(lst)
wavelengths = [391.719, 391.984, 392.248, 392.512, 392.777, 393.041, 393.306, 393.57, 393.835, 394.099, 391.719, 391.455, 391.19, 390.926, 390.661, 390.396]
intensities = [511.85, 1105.85, 1631.85, 1119.85, 213.85, 36.85, 10.85, 6.85, 13.85, 7.85, 511.85, 200.85, 80.85, 53.85, 14.85, 24.85]
n=sum(intensities)
mean = sum(wavelengths*intensities)/n
sigma = m.sqrt(sum(intensities*(wavelengths-mean)**2)/n)
def gaus(x,a,x0,sigma):
return a*m.exp(-(x-x0)**2/(2*sigma**2))
popt,pcov = opt.curve_fit(gaus,wavelengths,intensities,p0=[1,mean,sigma])
print(popt)
plt.scatter(wavelengths, intensities)
plt.title("Helium Spectral Line Peak 1")
plt.xlabel("Wavelength (nm)")
plt.ylabel("Intensity (a.u.)")
plt.show()
Thanks to the kind user, my curve seems to be working more reasonably well. However, one of the points seems to be back connecting to an earlier point? Screenshot below:
There are two problems with your code. The first is that you are performing vector operation on list which gives you the first error in the line mean = sum(wavelengths*intensities)/n. Therefore, you should use np.array instead. The second is that you take math.exp on python list which again throws an error as it takes a real number, so you should use np.exp here instead.
The following code solves your problem:
import matplotlib.pyplot as plt
import numpy as np
from scipy import optimize as opt
wavelengths = [391.719, 391.984, 392.248, 392.512, 392.777, 393.041,
393.306, 393.57, 393.835, 394.099, 391.719, 391.455,
391.19, 390.926, 390.661, 390.396]
intensities = [511.85, 1105.85, 1631.85, 1119.85, 213.85, 36.85, 10.85, 6.85,
13.85, 7.85, 511.85, 200.85, 80.85, 53.85, 14.85, 24.85]
wavelengths_new = np.array(wavelengths)
intensities_new = np.array(intensities)
n=sum(intensities)
mean = sum(wavelengths_new*intensities_new)/n
sigma = np.sqrt(sum(intensities_new*(wavelengths_new-mean)**2)/n)
def gaus(x,a,x0,sigma):
return a*np.exp(-(x-x0)**2/(2*sigma**2))
popt,pcov = opt.curve_fit(gaus,wavelengths_new,intensities_new,p0=[1,mean,sigma])
print(popt)
plt.scatter(wavelengths_new, intensities_new, label="data")
plt.plot(wavelengths_new, gaus(wavelengths_new, *popt), label="fit")
plt.title("Helium Spectral Line Peak 1")
plt.xlabel("Wavelength (nm)")
plt.ylabel("Intensity (a.u.)")
plt.show()
I am trying to resolve a special case of boundary value problem which is the boundary layer's equations for natural convection :
Thanks to a contributor of this forum #LutzLehmann, this set of equations is solved by using the function solve_bvp from the scipy library.
Unfortunately, I don't obtain the right result with these particular boundary conditions :
It seems that the initial guess for the solver doesn't work in this case and I wonder what could suit better.
Here is the code :
import numpy as np
from scipy.integrate import solve_bvp
import matplotlib.pyplot as plt
Pr = 5
def odesys(t,u):
F,dF,ddF,θ,dθ = u
return [dF, ddF, θ-0.25/Pr*(2*dF*dF-3*F*ddF), dθ, 0.75*F*dθ]
def bcs(u0,u1): return [u0[0], u0[1], u1[1], u0[3]-1, u1[3]]
x = np.linspace(0,8,25)
u = [x*x, np.exp(-x), 0*x+1, 1-x, 0*x-1]
res = solve_bvp(odesys,bcs,x,u, tol=1e-5)
print(res.message)
plt.subplot(2,1,1)
plt.plot(res.x,res.y[3], color='#801010', label='$\Delta T$')
plt.legend()
plt.grid()
plt.subplot(2,1,2)
plt.plot(res.x,res.y[1], '-', color='C0', label="$F'$")
plt.legend()
plt.grid()
And here are the wrong plots obtained :
Could someone help me by suggesting a better initial case for this problem ?
Thank you for your help,
PS : For the proposition :
u = [0.5*x*x*np.exp(-x), x*np.exp(-x), np.exp(-x), np.exp(-x), -np.exp(-x)]
I got this :
Compared to the literature, I am expecting this result (where g is here theta in the set of equations) :
I need help getting the right plot for tau as a function of c (conversion).
Here's my code:
from __future__ import division, print_function
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from math import *
from scipy.optimize import root
def tau(c,T = 298):
ci = 2
pw = 1000
Cp = 4
k0 = 0.01
e = 1000
Hr = (-3*10**4)
a = np.linspace(0,10000,10000)
t = np.zeros(len(a))
for i in range(len(a)):
t[i] = (ci * k0 * exp(e/298))/(ci - c[i])
plt.plot(t,c)
plt.show()
print(tau(np.linspace(0,1,10000)))
My plot is not correct.
How do I modify my equation to get the following kind of plot?
It's difficult to say why your plot of conversion is incorrect without knowing the original problem statement. Can you please provide more information? Your plot of conversion does increase with residence time (tau), which it should do. It looks like you've defined some constants that are unused, such as the enthalpy of reaction (Hr), pw, and the temperature. At any rate, it is apparent that your question is not about python or plotting, but rather about how to correctly solve a chemical engineering reactor design problem.
Here is my first steps within the NumPy world.
As a matter of fact the target is plotting below 2-D function as a 3-D mesh:
N = \frac{n}{2\sigma\sqrt{\pi}}\exp^{-\frac{n^{2}x^{2}}{4\sigma^{2}}}
That could been done as a piece a cake in Matlab with below snippet:
[x,n] = meshgrid(0:0.1:20, 1:1:100);
mu = 0;
sigma = sqrt(2)./n;
f = normcdf(x,mu,sigma);
mesh(x,n,f);
But the bloody result is ugly enough to drive me trying Python capabilities to generate scientific plots.
I searched something and found that the primary steps to hit above mark in Pyhton might be acquired by below snippet:
from matplotlib.patches import Polygon
import numpy as np
from scipy.integrate import quad
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
sigma = 1
def integrand(x,n):
return (n/(2*sigma*np.sqrt(np.pi)))*np.exp(-(n**2*x**2)/(4*sigma**2))
t = np.linespace(0, 20, 0.01)
n = np.linespace(1, 100, 1)
lower_bound = -100000000000000000000 #-inf
upper_bound = t
tt, nn = np.meshgrid(t,n)
real_integral = quad(integrand(tt,nn), lower_bound, upper_bound)
Axes3D.plot_trisurf(real_integral, tt,nn)
Edit: With due attention to more investigations on Greg's advices, above code is the most updated snippet.
Here is the generated exception:
RuntimeError: infinity comparisons don't work for you
It is seemingly referring to the quad call...
Would you please helping me to handle this integrating-plotting problem?!...
Best
Just a few hints to get you in the right direction.
numpy.meshgrid can do the same as MatLABs function:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.meshgrid.html
When you have x and n you can do math just like in matlab:
sigma = numpy.sqrt(2)/n
(in python multiplication/division is default index by index - no dot needed)
scipy has a lot more advanced functions, see for example How to calculate cumulative normal distribution in Python for a 1D case.
For plotting you can use matplotlibs pcolormesh:
import matplotlib.pyplot as plt
plt.pcolormesh(x,n,real_integral)
Hope this helps until someone can give you a more detailed answer.
I am using python's fsolve to find roots of a set of equations using the script below:
from scipy.optimize import fsolve
from scipy.interpolate import spline
import math
import numpy as np
#from mpldatacursor import datacursor
%pylab
from matplotlib import pyplot as plt
import matplotlib.dates as mdates
%matplotlib inline
def equations(p):
x,y,z = p
f1 = (1 - 2*math.cos(math.radians(x)) + 2*math.cos(math.radians(y)) - 2*math.cos(math.radians(z)) + 0.8)
f2 = (1 - 2*math.cos(math.radians(5*x)) + 2*math.cos(math.radians(5*y)) - 2*math.cos(math.radians(5*z)))
f3 = (1 - 2*math.cos(math.radians(7*x)) + 2*math.cos(math.radians(7*y)) - 2*math.cos(math.radians(7*z)))
return (f1,f2,f3)
x,y,z = fsolve(equations,(0,0,0))
#print equations((x,y,z))
This works perfect with one small issue. This generates only one set of solution i.e., one value for each x, y, z to satisfy those equations, but since these are trigonometric there should be atleast 2 solutions. How can I use fsolve (or anything else) to find two solutions for x, y, z for a given starting point.
Example: for x,y,z = fsolve(equations,(0,0,0)), is it possible to generate two values for x,y,z instead of one solution?
Thanks for the help.