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I want to solve for the variables s,L for different values of t. t is a part of my second equation and its values changes, I tried to solve for s,L for different t values then append the values to an empty list so that i could have different values of s, L for diiferent t values. But what i was getting is just an empty list.PLease help me with this enter image description here

from  scipy.optimize import fsolve
import numpy as np
import math as m
q0=0.0011
thetas,thetai,thetar=0.43,0.1,0.05
ks=0.0022#m/hr
psib=-0.15# m
lamda=1
eta=2+3*lamda
ki=8.676*10**(-8)
si=0.13157
t=np.array([3,18,24])

S=0.02/24
delta=-0.1001
b=[]
n=[]
for i in range(3):
    def equations(p):
        s, L = p
        f1=(ks*s**(3+(2/lamda))-(psib/(1-eta))*(((ki*si**(-1/lamda))-(ks*s**(3+(1/lamda))))/L)-q0)
        f2=(L*(s*(thetas-thetar))+S*t[i]*0.5*(m.exp(-delta*psib*(-1+s**(-1/lamda))))-(q0-ki)*t[i])
        return(f1,f2)

        s,L=fsolve(equations,([0.19,0.001]))
        b.append(s)
        n.append(L)
print(b)
print(n)
Mahesh
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  • For a problem like this: relax, take a step back from the code. A blackboard and chalk, or a notebook and pen, will be much more helpful than a screen and keyboard. – Stef Oct 03 '22 at 12:50
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    In other words: could you please add the actual equations to your post? They are likely more relevant to the question than your code. – Stef Oct 03 '22 at 12:50
  • I added the actual equations, solving by using pen and paper is a tedious task – Mahesh Oct 03 '22 at 15:47
  • That empty list solution might be correct. Perhaps there is no solution in the range you've given. These are non-linear in (s, L). Do you understand what fsolve function is doing to solve them? – duffymo Oct 03 '22 at 16:02
  • There is a solution, i can solve for one value of t at a time, – Mahesh Oct 03 '22 at 16:04

1 Answers1

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There are several ways to evaluate this system with an adjustable parameter. You could plug each value in before solving, which would make it compatible with additional solvers if fsolve didn't give you the desired results, or you could utilize the args parameter within fsolve. If I set up a dummy system such that I try to find x,y,z for some initial guess, and step through a parameter, I can append a preallocated solution array with the results

import numpy as np
from scipy.optimize import fsolve

a = np.linspace(0,10,21)

def equations(variables, a):
    x,y,z = variables
    eq1 = x+y+z*a
    eq2 = x-y-z
    eq3 = x*y*x*a   
    return tuple([eq1, eq2, eq3])

solutions = np.zeros((21,3))
for idx, i in enumerate(a):
    solutions[idx] = fsolve(equations, [-1,0,1], args=(i))

print(solutions)

which gives

[[ 5.00000000e-01 -5.00000000e-01  1.00000000e+00]
 [ 9.86864911e-17 -2.96059473e-16  3.94745964e-16]
 [ 1.62191889e-39 -1.28197512e-16  1.28197512e-16]
 [-2.15414908e-17 -1.07707454e-16  8.61659633e-17]
 [ 2.19853562e-28  6.59560686e-28 -4.39707124e-28]
 [-1.20530409e-28 -2.81237621e-28  1.60707212e-28]
 [-3.34744837e-17 -6.69489674e-17  3.34744837e-17]
 [ 6.53567253e-17  1.17642106e-16 -5.22853803e-17]
 [-3.14018492e-17 -5.23364153e-17  2.09345661e-17]
 [-5.99115518e-17 -9.41467242e-17  3.42351724e-17]
 [ 5.18199815e-29  7.77299722e-29 -2.59099907e-29]
 [-2.70691440e-17 -3.90998747e-17  1.20307307e-17]
 [-2.57288510e-17 -3.60203914e-17  1.02915404e-17]
 [-2.44785120e-17 -3.33797891e-17  8.90127708e-18]
 [-1.27252940e-28 -1.69670587e-28  4.24176466e-29]
 [ 2.24744956e-56  2.93897250e-56 -6.91522941e-57]
 [-2.12580678e-17 -2.73318015e-17  6.07373366e-18]
 [-2.03436865e-17 -2.57686696e-17  5.42498307e-18]
 [-3.89960988e-17 -4.87451235e-17  9.74902470e-18]
 [-1.87148635e-17 -2.31183608e-17  4.40349730e-18]
 [-7.19531738e-17 -8.79427680e-17  1.59895942e-17]]
t.o.
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