Stability for a system of nonlinear ODEs

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  • #1
hunt_mat
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Hi,

I am looking at the following system of ODEs:
[tex]
\begin{eqnarray*}
\dot{\omega}_{3}+\alpha\omega_{3} & = & \frac{\beta_{1}+\beta_{3}}{\rho_{0}}J_{3} \\
\dot{J_{3}}+2(\alpha_{2}-\alpha_{1})\beta_{2} & = & 0 \\
\dot{\beta}_{1}+\omega_{3}\beta_{2} & = & 0 \\
\dot{\beta}_{2}-\frac{\bar{\alpha}}{2}J_{3}+\frac{1}{2}\omega_{3}(\beta_{3}-\beta_{1}) & = & 0 \\
\dot{\beta}_{1}+\dot{\beta}_{3} & = & 0
\end{eqnarray*}
[/tex]

With suitable scaling I can reduce the system down to the following:
[tex]
\begin{eqnarray*}
\dot{\omega} & = & -\omega+CJ \\
\dot{J} & = & -\beta_{2} \\
\dot{\beta}_{1} & = & -\omega\beta_{2} \\
\dot{\beta}_{2} & = & \hat{\alpha}J-\frac{1}{2}\omega(C-2\beta_{1})
\end{eqnarray*}
[/tex]
Where [itex]\beta_{1}(t)+\beta_{3}(t)=C[/itex]. As I said in the title, I am interested in the stability of this system, so the first thing I so is look for the equilibrium points. I find there are two such points, one rather trivial one which is easy to analyse is [itex](\omega,J,\beta_{1},\beta_{2})=(0,0,C/2,0)[/itex], and another one which is: [itex](\omega,J,\beta_{1},\beta_{2})=(CJ_{0},J_{0},(C-2\hat{\alpha})/2),0)[/itex].

So my question is this: How do I determine [itex]J_{0}[/itex]? Do I consider it a parameter and look at different cases?
 

Answers and Replies

  • #2
epenguin
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Your last equation seems to be saying β2 = 0 is a steady state condition (which it is) but doesn't that just lead to all variables = 0 as steady state condition and the only one?

You might look to see whether you have in there a possible reduced variable case, I.e. some but not all of the variables be constant at particular values, and a reduced set of d.e.s , maybe that is what you are getting at.

If you are going to present much more it might be better to rename your reduced variables as something less confusing like x, y, z, w , or x1, x2, x3, x4.
 
  • #3
epenguin
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Your expression for a second steady state does not look right to me so please set out more working.
 
  • #4
hunt_mat
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Perhaps if I non-dimensionalise them to get rid of the constant C, that will make things easier for you:
[tex]
\begin{eqnarray}
\dot{x}_{1} & = & -x_{1}+x_{2} \\
\dot{x}_{2} & = & -x_{4} \\
\dot{x}_{3} & = & -x_{1}x_{4} \\
\dot{x}_{4} & = & x_{2}-\gamma x_{1}(1-x_{3})
\end{eqnarray}
[/tex]
I realised that I have not one but two families of equilibria:
[tex]
(0,0,x_{3},0)\quad (x_{1},x_{2},(1-\gamma^{-1},0))
[/tex]
This can be seen because if [itex]\dot{x}_{1}=0[/itex] then this implies that [itex]x_{1}=x_{2}[/itex] and if [itex]\dot{x}_{3}=0[/itex] then [itex]x_{1}[/itex] can be anything. If it's zero then automatically [itex]x_{0}=0[/itex], but I can have arbitrary [itex]x_{3}[/itex]. However [itex]x_{1}=x_{2}\neq 0[/itex] then I have to have that [itex]x_{3}=1-\gamma^{-1}[/itex].

You can go on to compute the characteristic equations for the two families, they are:
[tex]
\begin{eqnarray}
\lambda(\lambda^{3}+\lambda^{2}+\lambda-\hat{\alpha}(1-\beta_{1})) & = & 0 \quad\textrm{first family} \\
\lambda(\lambda^{3}+\lambda^{2}+(1-\gamma (x_{2}^{0})^{2})\lambda-\gamma (x_{2}^{0})^{2}) & = & 0\quad\textrm{second family}
\end{eqnarray}
[/tex]
I hope that explains a little more.
 
Last edited:
  • #5
epenguin
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I realised what you had done shortly after closing computer last night but it was too late to restart.

However it seems to me now that steady-state points are
x1 = x2 = x3 = 0 and then x3 can be anything, as you say.

Plus all points (x1, Cx1, (α/C - C/2), 0) , i.e. all points on a line in 4-D space. We can visualise at least the 3-D subspace x4 = 0 , it is the intersection of the plane x1 = Cx2 with the plane x = (α/C - C/2) if I have made no mistake.

 
  • #6
hunt_mat
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When I solve the equations numerically I find that I have a centre. I don't see that from the analysis I've performed. Is there any way of seeing linear stability from the system?
 
  • #8
hunt_mat
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I computed the Jacobian at the equilibrium points and found the corresponding eigenvalues. There are two families of equilibrium points, these are where the two polynomials in [itex]\lambda[/itex] come from.

I need some assistance in analysing the polynomials.
 
  • #9
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I'd use a graphic approach. For example, the first eigenvalue equation
\begin{equation*}
\lambda(\lambda^{3}+\lambda^{2}+\lambda-\hat{\alpha}(1-\beta_{1})) = 0
\end{equation*}
has the obvious 0 eigenvalue. The other three eigenvalues are the solution to
\begin{equation*}
\lambda^2(\lambda+1) = -\lambda+\delta
\end{equation*}
which are the crossing points of a cubic and a horizontal line. Plot the cubic and the line in order to understand how the roots change as a function of ##\delta##. Hope it helps...
 
  • #10
epenguin
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Did you compute the Jacobian? It's a standard technique in the study of nonlinear ODEs. You can find a description here: https://sites.ualberta.ca/~pass/math372/linearstability.pdf.
Yes It is standard and you can find many treatments. They are mostly silent about zero eigenvalues.:oldbiggrin:

When I solve the equations numerically I find that I have a centre. I don't see that from the analysis I've performed. Is there any way of seeing linear stability from the system?

Interesting. Could you show some? Did you do as a function of time, or also the more interesting one variable as a function of the other(s)? There are four variables here, I'd guess that one or two of them are not varying very much. If you can't show 3-D plots please show families in 2-D plot, but please show plots with different pairs of variables. What is the centre? Does it correspond to the stationary point already calculated? What happens around the other one? ( well one of them is not a point, but a line, I think we agree on that.) Also helpful to draw in isocline lines or planes where variables change direction re coordinate axes.

Is this research on a Biological or other model? Ordinary non-linear differential equation in 4 variables seems more than a student exercise. Have you incorporated all known or plausible constraints? Can you tell us if nature of problem restricts range of variables, e.g. any have to be positive?

What I think is oten done with such problems is to try and reduce the complication by reducing the number of variables using plausible approximations. Which may be valid generally, or within an interesting range. For example dx1/dt = 0 , i.e. x1 = Cx2 is a typical approximation that can come up in this sort of problem, corresponding to one process reaching an equilibrium faster than other processes. Another approximation that might be worth trying looking at would be since x3 can be anything in one steady state, make it large enough to swamp the linear terms in the fourth equation. With approximations like these you might be able to solve the equations analytically. In these ways you build up an understanding of the whole system.
 
Last edited:

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