Understand Kinetics Modeling Equations/Derivation

  • Context: Graduate 
  • Thread starter Thread starter Incognition
  • Start date Start date
  • Tags Tags
    Kinetics Modeling
Click For Summary
SUMMARY

The discussion focuses on the derivation and stability analysis of a system modeled by six coupled first-order differential equations, specifically the equations for \( m_i \) and \( p_i \). The unique steady state becomes unstable under the condition \(\frac{(\beta+1)^2}{\beta}<\frac{3X^2}{4+2X}\), where \( X \) is defined as \( X=-\frac{\alpha n p^{(n-1)}}{(1+p^n)^2} \). The steady state solution is derived by setting the derivatives to zero, leading to the conclusion that \( p_i = m_i \) at equilibrium. The challenge lies in determining the stability of this steady state solution, which requires careful analysis of the behavior of the system around the equilibrium point.

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with stability analysis in dynamical systems
  • Knowledge of equilibrium solutions and their significance
  • Basic grasp of mathematical modeling concepts
NEXT STEPS
  • Study the derivation of steady state solutions in coupled differential equations
  • Learn about stability criteria for equilibrium points in nonlinear systems
  • Explore numerical methods for solving differential equations
  • Investigate applications of Kinetics Modeling in biological systems
USEFUL FOR

Researchers, mathematicians, and engineers interested in mathematical modeling, particularly those working with dynamical systems and stability analysis in fields such as biology, chemistry, and physics.

Incognition
Messages
8
Reaction score
0
In a paper, I encountered a system modeled by six coupled first-order differential equations like so:

[tex]\frac{dm_i}{dt}=-m_i+\frac{\alpha}{1+p^n_j}+\alpha_0[/tex]

[tex]\frac{dp_i}{dt}=-\beta(p_i-m_i)[/tex] , where i=1,2,3 and j=3,1,2.

According to the paper, the system has a unique steady state which becomes unstable when [tex]\frac{(\beta+1)^2}{\beta}<\frac{3X^2}{4+2X}[/tex], where X is defined [tex]X=-\frac{\alpha n p^(n-1)}{(1+p^n)^2}[/tex]and p is the solution to [tex]p=\frac{\alpha}{1+p^n}+\alpha_0[/tex].

Lacking a textbook, I have had very little success in seeing how the steady state was derived. I intend to model a similar system. Can someone point me in the right way to understand these equations or show the derivation outright?

Thank you in advance.
 
Physics news on Phys.org
Finding the steady state solution (also called an "equilibrium solution") is very simple. The "steady state" solution is one that doesn't change which means that the derivatives are 0. You must have
[tex]\frac{dm_i}{dt}=-m_i+\frac{\alpha}{1+p^n_j}+\alpha_0= 0[/tex]
and
[tex]\frac{dp_i}{dt}=-\beta(p_i-m_i)= 0[/tex]
From the second equation, we obviously have [itex]p_i= m_i[/itex]
Putting that into the second equation, we have
[tex]-p_i}+\frac{\alpha}{1+p^n_j}+\alpha_0= 0[/tex]
If we let p be the function that satisfies
[tex]-p}+\frac{\alpha}{1+p^n}+\alpha_0= 0[/tex]
(just the equation above with pn replaced by p)
then the steady state solution is pi(t)= mi(t)= p for all i and t.

The "hard" part is determining when the steady state solution is stable or unstable. It is stable if, when the solution is slightly off the steady state solution, it tends to go toward it, and unstable if it tends to move away. In particular, taking p to be the solution to the equation above,
if pi>p, [itex]\frac{dp_i}{dt}[/itex] must be negative so the solution will tend downward toward p. Conversely, if pi< p, then the derivative must be positive. Of course, as long as the solution is not the steady state solution, we cannot assume that all the pi(t) are the same not that pi= mi. You can perhaps see from the form of the condition- a complicated equation involving X and then a complicated function defining X itself- that determining when the steady state solution is stable or unstable is not a simple problem!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K