quote:

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Originally posted by aplotnikov

It's a simple quadratic equation.

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quote:

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Originally posted by AKazak

Tried applying Equation Solver app to solve the above equation, but for about 1/3 of all points, it returns non-realistic negative values of x.
Is there a way to set some conditions on the searched solution?

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quote:

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Originally posted by Sam Fang

Hi AK,

You can check Initial Values for Dependent Variables in the dialog, set a positive value for initial x, and the final solution of x may be positive values.

If the problem still exists, you can tell us other parameter values, we can try it.
e.g. do you want to solve x=f(sp, sm, phi), and A, B are constants?

quote:

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Originally posted by AKazak

Tried applying Equation Solver app to solve the above equation, but for about 1/3 of all points, it returns non-realistic negative values of x.
Is there a way to set some conditions on the searched solution?

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Sam
OriginLab Technical Services

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pip install scipy

from scipy.optimize import fsolve
from math import sin
import numpy as np
import originpro as op


def equation(x, *Params):
    sp, sm, D, Phi = Params
    return Phi*(sp-x)/(x+D*(sp-x))+(1-Phi)*(sm-x)/(x+D*(sm-x))
    #sp, sm, A, B, Phi = Params
    #return Phi*(sp-X)/(sp+A*X)-(1-Phi)*(sm-X)/(sm+B*X) 
    #return [Phi*(sp-X)/(sp+A*X)-(1-Phi)*(sm-X)/(sm+B*X) for x in X]


Phis = np.linspace(0, 1, 101, endpoint=True)
Xs = []
Params = (0.1, 0.001, 0.3333, )
for Phi in Phis:
    CurrentParam = Params + (Phi, )
    Xs.append(fsolve(equation, 0, args=CurrentParam))
wks = op.new_sheet()
wks.from_list(0, Phis, lname='Phi')
wks.from_list(1, Xs, lname='Root')
graph = op.new_graph(template='line')
plot = graph[0].add_plot(wks, coly=1, colx=0, type='line')
plot.color = '#167BB2'
graph[0].yscale = 2 
graph[0].set_xlim(begin=0, end=1, step=0.1)
graph[0].set_ylim(begin=6e-4, end=0.2, step=1)
graph[0].label('legend').remove()
graph.set_int('aa', 1)

Phi*(0.1-x)*(x+1.0/3*(0.001-x)) + (1-Phi)*(0.001-x)*(x+1.0/3*(0.1-x)) = 0

quote:

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Originally posted by AKazak
Dear Sam,

Please test the following example project:
https://my.originlab.com/ftp/forum_and_kbase/Images/CM Test.opj

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AK

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quote:

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Originally posted by Sam Fang

Hi AK,

Thanks for your project file.

I defined the equation as folows:

Phi*(0.1-x)*(x+1.0/3*(0.001-x)) + (1-Phi)*(0.001-x)*(x+1.0/3*(0.1-x)) = 0

And set 1 to the initial value for x. The result looks like as below:
Can you check whether the result makes sense?

Thanks.

Sam
OriginLab Technical Services

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quote:

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Originally posted by AKazak

The results make sense.
However, I noticed that you replaced the equation parameters with the corresonding constants:
original equation: Phi*(0.1-x)*(x+1.0/3*(0.001-x)) + (1-Phi)*(0.001-x)*(x+1.0/3*(0.1-x)) = 0
modified equation: Phi*(Cp-x)/(x+D*(Cp-x))+(1-Phi)*(Cm-x)/(x+D*(Cm-x))=0

What is the problem with using equation parameters instead of constants?

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AK

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quote:

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Originally posted by Sam Fang

Hi AK,

You'd better use constants instead of constant independent variables because too many independent variables may cause the problem to be complex.

We will support to define constant parameters in a list. I added it to our bug tracking database. It will be convenient and clear for your case.

Sam
OriginLab Technical Services

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quote:

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Originally posted by lkb0221

Please try the following Python which I made with the awsome 2021 beta:

If not yet, install scipy from script window

pip install scipy

Python code:

from scipy.optimize import fsolve
from math import sin
import numpy as np
import originpro as op


def equation(x, *Params):
    sp, sm, D, Phi = Params
    return Phi*(sp-x)/(x+D*(sp-x))+(1-Phi)*(sm-x)/(x+D*(sm-x))
    #sp, sm, A, B, Phi = Params
    #return Phi*(sp-X)/(sp+A*X)-(1-Phi)*(sm-X)/(sm+B*X) 
    #return [Phi*(sp-X)/(sp+A*X)-(1-Phi)*(sm-X)/(sm+B*X) for x in X]


Phis = np.linspace(0, 1, 101, endpoint=True)
Xs = []
Params = (0.1, 0.001, 0.3333, )
for Phi in Phis:
    CurrentParam = Params + (Phi, )
    Xs.append(fsolve(equation, 0, args=CurrentParam))
wks = op.new_sheet()
wks.from_list(0, Phis, lname='Phi')
wks.from_list(1, Xs, lname='Root')
graph = op.new_graph(template='line')
plot = graph[0].add_plot(wks, coly=1, colx=0, type='line')
plot.color = '#167BB2'
graph[0].yscale = 2 
graph[0].set_xlim(begin=0, end=1, step=0.1)
graph[0].set_ylim(begin=6e-4, end=0.2, step=1)
graph[0].label('legend').remove()
graph.set_int('aa', 1)

Which will produce this graph:


FYI, for anyone being interested in this new version, you can request here:
https://www.originlab.com/doc/python/originpro

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quote:

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Originally posted by AKazak
Dear Sam,

Got it, thank you.
Can I set each independent variables to point to specific single column-label-row cell?

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AK

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quote:

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Originally posted by Sam Fang

quote:

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Originally posted by AKazak
Dear Sam,

Got it, thank you.
Can I set each independent variables to point to specific single column-label-row cell?

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AK

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Hi AK,

Yes. We will support to define constants by column label row cell in Constants list box.

Sam
OriginLab Technical Services

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quote:

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Originally posted by AKazak
Glad to read this.
Is this the near future?

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AK

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