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Solve a strongly nonlinear equation for x in Python

I am trying to solve the following equation for dB (for simplicity, I wrote dB as x in the title of the question):

image of equation

All other terms of the equation are known. I tried using SymPy for a symbolic solution for dB, but I kept getting time errors. I also tried using fminboundfrom scipy.optimize, but the answer for db was wrong (see below Python code using approach fminbound).

Does anyone know how to solve the equation for dB using Python?

import numpy as np

from scipy.optimize import fminbound

#------------------------------------------------------------------------------
# parameters

umf = 0.063         # minimum fluidization velocity, m/s
dbed = 0.055        # bed diameter, m
z0 = 0              # position bubbles are generated, m
z = 0.117           # bed vertical position, m
g = 9.81            # gravity, m/s^2

#------------------------------------------------------------------------------
# calculations

m = 3                       # multiplier for Umf
u = m*umf                   # gas superficial velocity, m/s

abed = (np.pi*dbed**2)/4.0  # bed cross-sectional area, m^2

# calculate parameters used in equation

dbmax = 2.59*(g**-0.2)*(abed*(u-umf))**0.4
dbmin = 3.77*(u-umf)**2/g

c1 = 2.56*10**-2*((dbed / g)**0.5/umf)

c2 = (c1**2 + (4*dbmax)/dbed)**0.5

c3 = 0.25*dbed*(c1 + c2)**2

dbeq = 0.25*dbed*(-c1 + (c1**2 + 4*(dbmax/dbed))**0.5 )**2

# general form of equation ... (term1)^power1 * (term2)^power2 = term3

power1 = 1 - c1/c2

power2 = 1 + c1/c2

term3 = np.exp(-0.3*(z - z0)/dbed)

def dB(d):
    term1 = (np.sqrt(d) - np.sqrt(dbeq)) / (np.sqrt(dbmin) - np.sqrt(dbeq))
    term2 = (np.sqrt(d) + np.sqrt(c3)) / (np.sqrt(dbmin) + np.sqrt(c3))
    return term1**power1 * term2**power2 - term3

# solve main equation for dB

dbub = fminbound(dB, 0.01, dbed)

print 'dbub = ', dbub
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2 answers

Here are four one-time root methods:

from scipy.optimize import brentq, brenth, ridder, bisect
for rootMth in [brentq, brenth, ridder, bisect]:
    dbub = rootMth(dB, 0.01, dbed)
    print 'dbub = ', dbub, '; sanity check (is it a root?):', dB(dbub)

Also the newton-raphson (secant / haley) method:

from scipy.optimize import newton
dbub = newton(dB, dbed)
print 'dbub = ', dbub, '; sanity check (is it a root?):', dB(dbub)

The scipy documentation recommends brentq if you have a bracketing interval.

0

, :

In [9]:

import numpy as np

import scipy.optimize as so

In [10]:

def f(x):

    return ((x-0.32)**0.8+(x+1.45)**1.1-np.exp(0.8))**2

In [11]:

so.fmin(f, x0=5)

Optimization terminated successfully.
         Current function value: 0.000000
         Iterations: 20
         Function evaluations: 40

Out[11]:

array([ 0.45172119])

In [12]:

f(0.45172119)

Out[12]:

4.7663411535618792e-13

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