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Taking the example of a two dimensional system, desribred by the following ODE's:

\begin{align} \frac{dx_1}{dt}&=f_1(x_1,x_2)\\ \frac{dx_2}{dt}&=f_2(x_1,x_2) \end{align}

The Jacobian Matrix JM is then given by:

$$JM=\left( \begin{array}{cc} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{array}\right)$$

Now quoting from scholarpedia:

The stability of typical equilibria of smooth ODEs is determined by the sign of real part of eigenvalues of the Jacobian matrix. These eigenvalues are often referred to as the 'eigenvalues of the equilibrium'. The Jacobian matrix of a system of smooth ODEs is the matrix of the partial derivatives of the right-hand side with respect to state variables where all derivatives are evaluated at the equilibrium point x=xe . Its eigenvalues determine linear stability properties of the equilibrium.

An equilibrium is asymptotically stable if all eigenvalues have negative real parts; it is unstable if at least one eigenvalue has positive real part.

  • Is there an intuitive way of understanding why the sign of eigenvalues of JM imply the state of stability of the system?
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    $\begingroup$ Hint: Think about the one-dimensional case, and how the graph of $f$ looks at the equilibrium point when the derivative (or 1D Jacobian) $f'$ is positive, and how it looks when it is negative. The multi-dimensional statement is much the same. $\endgroup$
    – ACuriousMind
    Commented Oct 28, 2014 at 16:08
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    $\begingroup$ The intuition is that the real part of the eigenvalues corresponds to exponential growth (decay) for positive (negative) sign, which leads to unstable (stable) behaviour, respectively. $\endgroup$
    – Qmechanic
    Commented Oct 28, 2014 at 16:28

2 Answers 2

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Let us look at a one-dimensional example:

Equilibrium plot

Recall that $f(x) = \dot{x}$, so $f$ encodes the time evolution of $x$. If $f < 0$, then $x$ will move to the left. If $f > 0$, then $x$ will move to the right. If $f = 0$, $x$ will not move at all, this is why $f(x_0) = 0$ is the equilibrium condition.

Now, look what happens if you perturb the equilibria $x_i$ slightly to the right: If $f'(x_i) < 0$, then, in a small surrounding of $x_i$, $f(x) < 0$ for $x > x_0$ in that surrounding, i.e. to the right of a equilibrium with $f'(x_i) < 0$, $x$ will move to the left - returning to equilibrium! Conversely, if $f'(x_i) > 0$, then $f(x) > 0$ for all $x$ to the right in the small surrounding, meaning that, after a small nudge to the right, $x$ will move even further right, leaving equilibrium!

The same line of reasoning can be applied to perturbations to the left, altogether showing that, if $f'(x_i) < 0$, then a small perturbation around $x_i$ will always move back into $x_i$, and if $f'(x_i) > 0$, then a small perturbation will become bigger and bigger, moving away from $x_i$.

The multi-dimensional case is less graphic, but the intuition is the same - negative eigenvalues of the Jacobian mean that the time evolution points back into equilibrium, positive eigenvalues mean that it points away from equilibrium.

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  • $\begingroup$ Thanks very clear now. So to sum up, if we look for stable equilibrium points, we need to find $x_0$'s ($f(x_0)=0$), for which the 2nd derivative of $x$ to the right and left is both negative, right? (so it means $x_0$'s that are maximums of our trajectory? how can a maximum be the point of stability) $\endgroup$
    – user929304
    Commented Oct 30, 2014 at 15:18
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    $\begingroup$ @user929304: $x$ has no derivatives "to the right and left" ("left" and "right", as I used them, refer to "smaller $x$" and "larger $x$"). $x$ has only time derivatives. A stable equilibrium is one where $f'(x_0) < 0$, so that $f$ is positive to the left of the equilibrium and negative to the right - because then it moves left when nudged to the right, and right when nudged to the left. $\endgroup$
    – ACuriousMind
    Commented Oct 30, 2014 at 15:24
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  1. $f=-kx$ is stable whereas $f=kx$ is unstable.
  2. You can usually rewrite a matrix $A$ as $A=PDP^{-1}$ where $P$ is a matrix of eigenvectors and $D$ is a diagonal matrix of eigenvalues.
  3. If $F=Ax$, then by the above, $(P^{-1}F)=D(P^{-1}x)$. Now you have $n$ independent equations exactly of the form $f=kx$ or $f=-kx$. If any one of them is like $f=kx$, the solution will blow up to infinity under a tiny perturbation.

For example: $ \left( \begin{array}{cc} 2 & 2 \\ -1 & 1 \\ \end{array} \right)\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)\left( \begin{array}{cc} 2 & 2 \\ -1 & 1 \\ \end{array} \right)^{-1}=\left( \begin{array}{cc} 0 & -2 \\ -\frac{1}{2} & 0 \\ \end{array} \right)=A$

Just from looking at the $PDP^{-1}$ diagonalization, I can tell - if $F=Ax$,then any $x$ going in the direction $(2,-1)$ will tend to be pushed away from the origin, and any $x$ going in the diretion $(2,1)$ will tend to be pushed back towards the origin. You should understand $P^{-1}x$ as a change of basis into a coordinate system where the axes are some $n$ arbitrary directions which act as lines of force.

Here is a plot of that system $F=\left( \begin{array}{cc} 0 & -2 \\ -\frac{1}{2} & 0 \\ \end{array} \right)x$.

pushing away from the origin

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    $\begingroup$ It seems to me that, in your answer, you are assuming that $JM$ can always be diagonalized. It is false also in $\mathbb C$...Actually you wrote "usually". What happens if it is not the case? $\endgroup$ Commented Oct 28, 2014 at 16:33
  • $\begingroup$ @ValterMoretti fair point, could you clarify upon this in a general manner?(i.e. whether it can or cannot be diagonalized). It would be definitely nice to have your take on this post. $\endgroup$
    – user929304
    Commented Oct 29, 2014 at 11:53
  • $\begingroup$ @user929304 Simply you should (a) classify the non-$\mathbb C$-diagonalizable real $2\times 2$ matrices $A$, (b) study the behaviour of the solutions of $d\vec{x}/dt=A\vec{x}$ around the critical point $\vec{x}=0$ and finally (c) exploiting a (non-trivial) argument relating the behaviour of the solution of your dynamical system with the those of the associated linearised system. The latter is defined by chosing $A=JM_{\vec{x}_0}$ computed at the critical point $\vec{x}_0$ such that $f_i(\vec{x}_0)=0$, $i=1,2$. Obviously the coordinates have to be centered on $\vec{x}_0$. $\endgroup$ Commented Oct 29, 2014 at 12:20
  • $\begingroup$ @ValterMoretti I couldn't figure out any intuition behind it! I guess you still have the eigenvalues of $A$ have to be negative (except in the worst posed problems you'll still have a characteristic equation with $n$ solutions counting multiplicity) but it looks like in the $x''=Ax$ case you can have your eigenvalue negative but your solution blowing up to infinity like $x \sin(x)$! But all of that is really far from an intuitive understanding, so I guess it doesn't belong in the answer. I'd rather just sweep it under the rug. $\endgroup$
    – user12029
    Commented Oct 31, 2014 at 22:34
  • $\begingroup$ Indeed I did not wand to give an intuition. What I wanted to say is that the linearized case has the same qualitative behavior as the non-linearized case. This is a (difficult) theorem. However once you know it you can concentrate on the linear case only and you can classify the various cases since explicit solutions can be written down in that case, for 2x2 real matrices. There are only a few of possible types. See for instance en.wikipedia.org/wiki/Phase_plane $\endgroup$ Commented Oct 31, 2014 at 23:06

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