Solving a System of ODE for Steady State

In summary, the conversation discusses the task of finding the steady states in an ODE system and calculating the values of y1, y2, and y3 at the steady state. The given information includes the values of various parameters and the initial condition for y0. However, after analyzing the equations and setting them equal to zero to find the steady state, it is determined that the conditions for a steady state are impossible to achieve with the given data. This is in contrast to the results shown in a paper that the system is based on, leaving uncertainty about the accuracy of the equations or the possibility of finding a steady state.
  • #1
Bewilder
2
0
I am trying to find the steady states in the ODE system. Assuming y0 = 2.5 * 10^5, I want to calculate y1, y2, y3 at the steady state. I do not understand how this would be possible, because only y0 is given and the following:

d0 = 0.003,
d1 = 0.008,
d2 = 0.05,
d3 = 1,
ry = 0.008,
ay = 1.6/100,
by = 10/750,
cy = 100,
u = 4 * 10^−8,
y0 = 2.5 * 10^5.

This is my ODE system:
EY0x6.png


My task says: Find algebraically the steady state for the equations. Set y0 = 2.5 * 10^5 and calculate y1, y2, y3 based on the derived equations at the steady state.

Is it even possible to find the exact value of y1, y2, and y3 with the given information?

I have tried it in R, but it's impossible to get an answer.. How can I do this manually to get the solution for y1, y2, y3 at the steady state?

Code:
model <- function(t,x,params){
  y0 <- x[1]
  y1 <- x[2]
  y2 <- x[3]
  y3 <- x[4]
  ry <- params[1]
  mu <- params[2]
  d0 <- params[3]
  ay <- params[4]
  d1 <- params[5]
  by <- params[6]
  d2 <- params[7]
  cy <- params[8]
  d3 <- params[9]  m <- rep(0,4)
  m[1] = ((ry*(1-mu)) - d0) * y0
  m[2] = (ay * y0) - (d1 * y1)
  m[3] = (by * y1) - (d2 * y2)
  m[4] = (cy * y2) - (d3 * y3)
 
  return(m)

}

x <- ode23(model, y0 = c(y0=250000, y1=y_1, y2=y_2, y3=y_3), t0=0,tf=400, params = c(0.008,4*10^-8,0.003,1.6/100,0.008,10/750,0.05,100,1))
 
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  • #2
Bewilder said:
I am trying to find the steady states in the ODE system. Assuming y0 = 2.5 * 10^5, I want to calculate y1, y2, y3 at the steady state. I do not understand how this would be possible, because only y0 is given and the following:

d0 = 0.003,
d1 = 0.008,
d2 = 0.05,
d3 = 1,
ry = 0.008,
ay = 1.6/100,
by = 10/750,
cy = 100,
u = 4 * 10^−8,
y0 = 2.5 * 10^5.

This is my ODE system:
EY0x6.png


My task says: Find algebraically the steady state for the equations. Set y0 = 2.5 * 10^5 and calculate y1, y2, y3 based on the derived equations at the steady state.

Is it even possible to find the exact value of y1, y2, and y3 with the given information?

I have tried it in R, but it's impossible to get an answer.. How can I do this manually to get the solution for y1, y2, y3 at the steady state?

Code:
model <- function(t,x,params){
  y0 <- x[1]
  y1 <- x[2]
  y2 <- x[3]
  y3 <- x[4]
  ry <- params[1]
  mu <- params[2]
  d0 <- params[3]
  ay <- params[4]
  d1 <- params[5]
  by <- params[6]
  d2 <- params[7]
  cy <- params[8]
  d3 <- params[9]  m <- rep(0,4)
  m[1] = ((ry*(1-mu)) - d0) * y0
  m[2] = (ay * y0) - (d1 * y1)
  m[3] = (by * y1) - (d2 * y2)
  m[4] = (cy * y2) - (d3 * y3)

  return(m)

}

x <- ode23(model, y0 = c(y0=250000, y1=y_1, y2=y_2, y3=y_3), t0=0,tf=400, params = c(0.008,4*10^-8,0.003,1.6/100,0.008,10/750,0.05,100,1))

If by steady-state you mean that ##\dot{y}_i = 0 ## for ##i = 0, \ldots ,3## then your sought-for conditions are impossible. The only way to have ##y_0 \neq 0## but ##\dot{y}_0 = 0## would be to have the coefficient ##r_y(1-u) - d_0 =0##, and that is not true for the given data.
 
  • #3
Ray Vickson said:
If by steady-state you mean that ##\dot{y}_i = 0 ## for ##i = 0, \ldots ,3## then your sought-for conditions are impossible. The only way to have ##y_0 \neq 0## but ##\dot{y}_0 = 0## would be to have the coefficient ##r_y(1-u) - d_0 =0##, and that is not true for the given data.

That's what has confused me. I am setting the equations equal to zero but it does not work, but it should since this system comes from a paper and they have solved it there.

http://ped.fas.harvard.edu/files/ped/files/nature05b_0.pdf?m=1425933222

This is the paper and the equations are in page 6 of the PDF. When the steady states are found then the simulation is run for 400 days and the results are plotted in page 3, Figure 4.a. But when I plot my simulation for 400 days (after setting equations equal to zero to find the steady state) what I get as plots is completely different from those shown in the paper..
 

FAQ: Solving a System of ODE for Steady State

What is a System of ODE for Steady State?

A System of ODE (Ordinary Differential Equations) for Steady State refers to a set of equations that describe the behavior of a physical system over time when it has reached a state of equilibrium or stability. This means that the values of the variables in the system do not change over time and remain constant.

What is the importance of solving a System of ODE for Steady State?

Solving a System of ODE for Steady State is important because it allows us to understand the behavior of a physical system at its equilibrium state. This information can be used to make predictions, design experiments, and optimize processes.

What are the methods used for solving a System of ODE for Steady State?

The most commonly used methods for solving a System of ODE for Steady State are analytical methods, such as separation of variables and substitution, and numerical methods, such as Euler's method and Runge-Kutta method. These methods help to find the exact or approximate solutions to the system of equations.

What are the challenges in solving a System of ODE for Steady State?

There are several challenges in solving a System of ODE for Steady State, such as the complexity of the equations, the sensitivity of the solutions to initial conditions, and the need for specialized software and computational resources. It also requires a good understanding of mathematical concepts and techniques.

How can the solutions of a System of ODE for Steady State be validated?

The solutions of a System of ODE for Steady State can be validated by comparing them with experimental data or with solutions obtained from other methods. It is also important to check for consistency and accuracy, and to consider the assumptions and limitations of the model used to develop the system of equations.

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