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I need help on how to linearize the following ODE equation so that I am able to do Stability analysis for the equation. Thanks for the help.

$\frac{dQ}{dz} = 2aM^{1/2}$

$\frac{dM}{dz} = \frac{QF}{M}$

$\frac{dF}{dz} = bQ$

Where a and b are constants. Initial points are $Q = M = 0, F = 1$ at z = 0.

Non-dimensional form of the above equation is:

$\frac{d \hat{Q}}{d\hat{z}} = \hat{M}^{1/2}$

$\hat{M}\frac{d \hat{M}}{d\hat{z}} = \hat{F}\hat{Q}$

$\frac{d \hat{F}}{d\hat{z}} = -\hat{Q}$

Case 1: \begin{equation} \begin{pmatrix} \frac{dQ}{dz} \\ \frac{dM}{dz} \\ \frac{dF}{dz} \end{pmatrix}=\begin{pmatrix} 0 & \frac{1}{M^{1/2}} & 0 \\ 0 & 0 & \frac{Q}{M} \\ -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} Q \\ M \\ F \end{pmatrix} \end{equation}

Case 2:

\begin{equation} \begin{pmatrix} \frac{dQ}{dz} \\ \frac{dM}{dz} \\ \frac{dF}{dz} \end{pmatrix}=\begin{pmatrix} 0 & \frac{1}{M^{1/2}} & 0 \\ \frac{F}{M} & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} Q \\ M \\ F \end{pmatrix} \end{equation}

Dereje
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  • Welcome to MSE. Are you sure that you put everything right? For your initial values, in the second equation you have indeterminate $0/0$. – user539887 Feb 08 '19 at 10:30
  • Yes actually M can not be zero. But it could be assumed that very small value. – Dereje Feb 09 '19 at 11:54
  • I still do not understand. Do you mean linearization around an equilibrium? But there are no equilibria, since by the first equation $M$ must be zero, which is incompatible with the second equation. – user539887 Feb 09 '19 at 13:02
  • Yes you are right, may be if try to convert the equations into non-dimensional equations, the indeterminate might disappear. – Dereje Feb 09 '19 at 20:39
  • I've done the non-dimensional form of the above equation(I wrote it above in my original question) but it doesn't make as such a difference. – Dereje Feb 09 '19 at 22:15
  • Perhaps Singular perturbation, or Singular perturbation theory? – user539887 Feb 10 '19 at 09:33
  • Singular perturbation is a bit complex for me but instead for this case, I just changed the equilibrium point into (1,1,1) and have done the linear transformation matrix in two cases. Both have the same output( Have a look above). When I check for stability by calculating the Eigen values the first one is stable but the second case is unstable. I don't understand how this difference happened. Need your help. – Dereje Feb 11 '19 at 00:46
  • $(1,1,1)$ is not an equilibrium point. Analysis of stability via the signs of eigenvalues of the linearization is valid for equilibrium points only. – user539887 Feb 11 '19 at 10:23
  • (1,1,1) is the value of the initial values of Q, M and F at z= 0. Still not ? – Dereje Feb 11 '19 at 13:17
  • The right-hand sides of the equations at $(1,1,1)$ are not equal to zero. – user539887 Feb 13 '19 at 06:48

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