Python GEKKO ODE 错误结果

Question

I've been trying to implement the Sedaghat model for simulation of the insulin signaling pathway, as it is presented here , using python's GEKKO . The system modeled is the one without feedback and the model's equations and constants' values can be found in Appendix A. While the first 6 diagrams of my results seem to come out fine, the rest model states (x16-x21) seem a bit problematic (fig. 6,7 of the same paper used for comparison).

I have checked the equations and constant values, tried adding lower and upper bounds to the variables and experimented with the m.options.IMODE and m.options.NODES parameters, but these didn't seem to help.

Any piece of advice would be much appreciated.

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
from gekko import GEKKO

def insulin_pathway_CM(insulin, time_interval): 

    m = GEKKO(remote=False)
    m.time = np.linspace(0, time_interval-1, time_interval)

    # initialization of variables 
    x1 = m.Param(value=insulin)
    x2 = m.Var(9*1e-13)
    x3 = m.Var()
    x4 = m.Var()
    x5 = m.Var()
    x6 = m.Var(1e-13)
    x7 = m.Var()
    x8 = m.Var()
    x9 = m.Var(1e-12)
    x10 = m.Var()
    x11 = m.Var(1e-13)
    x12 = m.Var()
    x13 = m.Var(0.0031)
    x14 = m.Var(0.994)
    x15 = m.Var(0.0029)
    x16 = m.Var(1)
    x17 = m.Var()
    x18 = m.Var(1)
    x19 = m.Var()
    x20 = m.Var(0.96)
    x21 = m.Var(0.04)
  
    # initialization of constants
    km1 = 0.2
    k1 = 6*1e7
    km2 = 100*km1
    k2 = k1
    km3 =  km1
    k3 = 2500
    km4 = 0.003
    k4 = km4/9
    km4a = 2.1*1e-4
    k4a = 2.1*1e-3
    km5 = 1.67*1e-18
    k5 = 10*km5    # simplification made after testing  
                   # x6 seems to be above 1e-13 for all values of interest
    k6 = 0.461
    k7 = 4.16
    km7 = (2.5/7.45)*k7
    km8 = 10
    k8 = km8 * (5/70.775) * 1e12
    km9 = (94/3.1)*1.39
    PI3K = 5*1e-15
    k9basal = (0.31/99.4)*km9
    k9 = m.Var() 
    m.Equation(k9 == (1.39 - k9basal)*(x12/PI3K) + k9basal)
    km10 = 2.77
    k10 = (3.1/2.9)*km10
    km11 = 6.9314718
    k11 = m.Var()
    m.Equation(k11 == (0.1*km11)*(x13-0.31)/2.79)
    km12 = 6.93147
    k12 = m.Var()
    m.Equation(k12 == (0.1*km12)*(x13-0.31)/2.79)
    km13 = 0.167
    k13 = (4/96)*km13
    km14 = 0.001155
    k14 = 96*km14
    
    Effect = m.Var()
    APequil = 100/11
    m.Equation(Effect == (0.2*x17 + 0.8*x19)/APequil)
    k13a = ((4/6) - (4/96))*km13*Effect
    
    SHIP = 1
    PTEN = 1
    PTP = 1
    IRp = 8.97*1e-13

    # equations
    m.Equation(x2.dt() == km1*x3 + km3*PTP*x5 - k1*x1*x2 + km4*x6 - k4*x2)
    m.Equation(x3.dt() == k1*x1*x2 - km1*x3 - k3*x3)
    m.Equation(x4.dt() == k2*x1*x5 -km2*x4 + km4a*x7 -k4a*x4)
    m.Equation(x5.dt() == k3*x3 + km2*x4 - k2*x1*x5 - km3*PTP*x5 + km4a*x8 - k4a*x5)
    m.Equation(x6.dt() == k5 - km5*x6 + k6*PTP*(x7 + x8) + k4*x2 - km4*x6)
    m.Equation(x7.dt() == k4a*x4 - km4a*x7 - k6*PTP*x7)
    m.Equation(x8.dt() == k4a*x5 - km4a*x8 - k6*PTP*x8)
    m.Equation(x9.dt() == km7*PTP*x10 - k7*x9*(x4+x5)/IRp)
    m.Equation(x10.dt() == k7*x9*(x4+x5)/IRp + km8*x12 - (km7*PTP + k8*x11)*x10)
    m.Equation(x11.dt() == km8*x12 - k8*x10*x11)
    m.Equation(x12.dt() == k8*x10*x11 - km8*x12)
    m.Equation(x13.dt() == k9*x14 + k10*x15 - (km9*PTEN + km10*SHIP)*x13)
    m.Equation(x14.dt() == km9*PTEN*x13 - k9*x14) 
    m.Equation(x15.dt() == km10*SHIP*x13 - k10*x15)
    m.Equation(x16.dt() == km11*x17 - k11*x16) 
    m.Equation(x17.dt() == k11*x16 - km11*x17)  
    m.Equation(x18.dt() == km12*x19 - k12*x18)  
    m.Equation(x19.dt() == k12*x18 - km12*x19)
    m.Equation(x20.dt() == km13*x21 - (k13 + k13a)*x20 + k14 - km14*x20)  
    m.Equation(x21.dt() == (k13 + k13a)*x20 - km13*x21) 
  
    m.options.IMODE = 7
    m.options.OTOL  = 1e-8
    m.options.RTOL  = 1e-8
    m.options.NODES = 3
    m.solve(disp=False)
  
    # plotting
    x45 = []
    for i in range(len(x4.value)):
        x45.append(x4[i]+x5[i])
  
    fig, axs = plt.subplots(4, 2, figsize=(20, 20))
    axs[0, 0].set_title('Free surface receptors (x2)')
    axs[0, 0].plot(m.time, x2)
    axs[0, 1].set_title('Once and twice bound phosphorylated receptors (x4:orange, x5:green)')
    axs[0, 1].plot(m.time, x4, 'tab:orange')
    axs[0, 1].plot(m.time, x5, 'tab:green')
    axs[1, 0].set_title('Surface phosphorylated receptors (x4 + x5)')
    axs[1, 0].plot(m.time, x45, 'tab:red')
    axs[1, 1].set_title('Unphosphorylated and tyrosine phosphorylated IRS-1 (x9:orange, x10:green)')
    axs[1, 1].plot(m.time, x9, 'tab:orange')
    axs[1, 1].plot(m.time, x10, 'tab:green')
    axs[2, 0].set_title('Activated PI 3-kinase (x12)')
    axs[2, 0].plot(m.time, x12, 'tab:orange')
    axs[2, 1].set_title('PI(3,4,5)P3 and PI(3,4,5)P2 (x13:orange, x15:green)')
    axs[2, 1].plot(m.time, x13, 'tab:orange')
    axs[2, 1].plot(m.time, x15, 'tab:green')
    axs[3, 0].set_title('Activated PKC-ζ (x19)')
    axs[3, 0].plot(m.time, x19, 'tab:red')
    axs[3, 1].set_title('Percentage of cell surface GLUT4 (x21)')
    axs[3, 1].plot(m.time, x21, 'tab:cyan')
    return

# input as in the paper
time_interval = 60
insulin = np.zeros(60)
insulin[:15] = 1e-7

insulin_pathway_CM(insulin, time_interval)

Answer 1

Here are a few suggestions:

• Replace some of the states with Intermediate type. This gives same solution but calculates faster.

    #k9 = m.Var() 
    #m.Equation(k9 == (1.39 - k9basal)*(x12/PI3K) + k9basal)
    k9 = m.Intermediate(1.39 - k9basal)*(x12/PI3K) + k9basal)

• Add a few small time steps at the beginning in case of early fast dynamics that are not tracking well with time step of 1. This didn't have an effect, but it is a good thing to try.

    m = GEKKO(remote=False)
    t = np.linspace(0, time_interval-1, time_interval)
    m.time = np.insert(t,1,[1e-5,1e-4,1e-3,1e-2,0.1])
    insulin = np.insert(insulin,1,[insulin[0]]*5)

• Investigate any non-intuitive solutions such x17 and x19 as negative numbers.

Equations for x16 to x21 do not appear to influence other parts of the model and the only input is k11 that is also dependent on x13 , x14 , and x15 .

    km11 = 6.9314718
    k11 = m.Intermediate((0.1*km11)*(x13-0.31)/2.79)
    m.Equation(x13.dt() == k9*x14 + k10*x15 - (km9*PTEN + km10*SHIP)*x13)
    m.Equation(x14.dt() == km9*PTEN*x13 - k9*x14) 
    m.Equation(x15.dt() == km10*SHIP*x13 - k10*x15)

    m.Equation(x16.dt() == km11*x17 - k11*x16) 
    m.Equation(x17.dt() == k11*x16 - km11*x17)  
    m.Equation(x18.dt() == km12*x19 - k12*x18)  
    m.Equation(x19.dt() == k12*x18 - km12*x19)
    m.Equation(x20.dt() == km13*x21 - (k13 + k13a)*x20 + k14 - km14*x20)  
    m.Equation(x21.dt() == (k13 + k13a)*x20 - km13*x21) 

There may be a model error in those equations, possibly even in the original paper (it sometimes happens). Focusing on pairs x16 and x17 , it appears that x17 is just the change in x16 . This is similar for x18 and x19 . At first glance, I couldn't see any obvious differences in the equations between the Python source code and the paper.

Here is the complete script with some changes with Intermediates, 2 plots added, and the additional time points at the beginning.

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
from gekko import GEKKO

def insulin_pathway_CM(insulin, time_interval): 

    m = GEKKO(remote=False)
    t = np.linspace(0, time_interval-1, time_interval)
    m.time = np.insert(t,1,[1e-5,1e-4,1e-3,1e-2,0.1])
    print(m.time)
    insulin = np.insert(insulin,1,[insulin[0]]*5)

    # initialization of variables 
    x1 = m.Param(value=insulin)
    x2 = m.Var(9*1e-13)
    x3 = m.Var()
    x4 = m.Var()
    x5 = m.Var()
    x6 = m.Var(1e-13)
    x7 = m.Var()
    x8 = m.Var()
    x9 = m.Var(1e-12)
    x10 = m.Var()
    x11 = m.Var(1e-13)
    x12 = m.Var()
    x13 = m.Var(0.0031)
    x14 = m.Var(0.994)
    x15 = m.Var(0.0029)
    x16 = m.Var(1)
    x17 = m.Var()
    x18 = m.Var(1)
    x19 = m.Var()
    x20 = m.Var(0.96)
    x21 = m.Var(0.04)
  
    # initialization of constants
    km1 = 0.2
    k1 = 6*1e7
    km2 = 100*km1
    k2 = k1
    km3 =  km1
    k3 = 2500
    km4 = 0.003
    k4 = km4/9
    km4a = 2.1*1e-4
    k4a = 2.1*1e-3
    km5 = 1.67*1e-18
    k5 = 10*km5    # simplification made after testing  
                   # x6 seems to be above 1e-13 for all values of interest
    k6 = 0.461
    k7 = 4.16
    km7 = (2.5/7.45)*k7
    km8 = 10
    k8 = km8 * (5/70.775) * 1e12
    km9 = (94/3.1)*1.39
    PI3K = 5*1e-15
    k9basal = (0.31/99.4)*km9
    k9 = m.Intermediate((1.39 - k9basal)*(x12/PI3K) + k9basal)
    km10 = 2.77
    k10 = (3.1/2.9)*km10
    km11 = 6.9314718
    k11 = m.Intermediate((0.1*km11)*(x13-0.31)/2.79)
    km12 = 6.93147
    k12 = m.Intermediate((0.1*km12)*(x13-0.31)/2.79)
    km13 = 0.167
    k13 = (4/96)*km13
    km14 = 0.001155
    k14 = 96*km14
    
    APequil = 100/11
    Effect = m.Intermediate((0.2*x17 + 0.8*x19)/APequil)
    k13a = ((4/6) - (4/96))*km13*Effect
    
    SHIP = 1
    PTEN = 1
    PTP = 1
    IRp = 8.97*1e-13

    # equations
    m.Equation(x2.dt() == km1*x3 + km3*PTP*x5 - k1*x1*x2 + km4*x6 - k4*x2)
    m.Equation(x3.dt() == k1*x1*x2 - km1*x3 - k3*x3)
    m.Equation(x4.dt() == k2*x1*x5 -km2*x4 + km4a*x7 -k4a*x4)
    m.Equation(x5.dt() == k3*x3 + km2*x4 - k2*x1*x5 - km3*PTP*x5 + km4a*x8 - k4a*x5)
    m.Equation(x6.dt() == k5 - km5*x6 + k6*PTP*(x7 + x8) + k4*x2 - km4*x6)
    m.Equation(x7.dt() == k4a*x4 - km4a*x7 - k6*PTP*x7)
    m.Equation(x8.dt() == k4a*x5 - km4a*x8 - k6*PTP*x8)
    m.Equation(x9.dt() == km7*PTP*x10 - k7*x9*(x4+x5)/IRp)
    m.Equation(x10.dt() == k7*x9*(x4+x5)/IRp + km8*x12 - (km7*PTP + k8*x11)*x10)
    m.Equation(x11.dt() == km8*x12 - k8*x10*x11)
    m.Equation(x12.dt() == k8*x10*x11 - km8*x12)
    m.Equation(x13.dt() == k9*x14 + k10*x15 - (km9*PTEN + km10*SHIP)*x13)
    m.Equation(x14.dt() == km9*PTEN*x13 - k9*x14) 
    m.Equation(x15.dt() == km10*SHIP*x13 - k10*x15)
    m.Equation(x16.dt() == km11*x17 - k11*x16) 
    m.Equation(x17.dt() == k11*x16 - km11*x17)  
    m.Equation(x18.dt() == km12*x19 - k12*x18)  
    m.Equation(x19.dt() == k12*x18 - km12*x19)
    m.Equation(x20.dt() == km13*x21 - (k13 + k13a)*x20 + k14 - km14*x20)  
    m.Equation(x21.dt() == (k13 + k13a)*x20 - km13*x21) 
  
    m.options.IMODE = 7
    m.options.OTOL  = 1e-10
    m.options.RTOL  = 1e-10
    m.options.NODES = 3
    m.solve(disp=False)
  
    # plotting
    x45 = []
    for i in range(len(x4.value)):
        x45.append(x4[i]+x5[i])
  
    fig, axs = plt.subplots(4, 2, figsize=(20, 20))
    axs[0, 0].set_title('Free surface receptors (x2)')
    axs[0, 0].plot(m.time, x2)
    axs[0, 1].set_title('Once and twice bound phosphorylated receptors (x4:orange, x5:green)')
    axs[0, 1].plot(m.time, x4, 'tab:orange')
    axs[0, 1].plot(m.time, x5, 'tab:green')
    axs[1, 0].set_title('Surface phosphorylated receptors (x4 + x5)')
    axs[1, 0].plot(m.time, x45, 'tab:red')
    axs[1, 1].set_title('Unphosphorylated and tyrosine phosphorylated IRS-1 (x9:orange, x10:green)')
    axs[1, 1].plot(m.time, x9, 'tab:orange')
    axs[1, 1].plot(m.time, x10, 'tab:green')
    axs[2, 0].set_title('Activated PI 3-kinase (x12)')
    axs[2, 0].plot(m.time, x12, 'tab:orange')
    axs[2, 1].set_title('PI(3,4,5)P3 and PI(3,4,5)P2 (x13:orange, x15:green)')
    axs[2, 1].plot(m.time, x13, 'tab:orange')
    axs[2, 1].plot(m.time, x15, 'tab:green')
    axs[3, 0].set_title('Activated PKC-ζ (x19)')
    axs[3, 0].plot(m.time, x19, 'tab:red')
    axs[3, 1].set_title('Percentage of cell surface GLUT4 (x21)')
    axs[3, 1].plot(m.time, x21, 'tab:cyan')
    
    plt.figure()
    for ni in range(16,22):
        nx = 'x'+str(ni); x = eval(nx)
        plt.subplot(3,2,ni-15)
        plt.plot(m.time,x.value,label=nx); plt.legend()

    plt.figure()
    plt.subplot(2,1,1)
    plt.plot(m.time,k11.value,label='k11')
    plt.subplot(2,1,2)
    plt.plot(m.time,x13.value,label='x13')

    return

# input as in the paper
time_interval = 60
insulin = np.zeros(60)
insulin[:15] = 1e-7

insulin_pathway_CM(insulin, time_interval)

plt.show()