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Given the ordinary differential equation

$$ \begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \end{bmatrix} = \begin{bmatrix} -x_1+x_2+1 \\ 2x_1-3x_2-2 \\ \end{bmatrix} $$ i have to check the stability of the equilibrium point. Solving the RHS for $0$ gives us the equilibrium $$\hat x=\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$$ and now i want to use the usual eigenvalue stability criterion for linear ODEs, given that our original equation rewrites to

$$ \begin{bmatrix} \dot x_1 \\ \dot x_2 \\ \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 2 & -3 \\ \end{bmatrix}x + \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}, $$ so we basically have a linear ODE which got "shifted" by a constant, so is not of the usual form $\dot x=Ax$, for which we had formulated the eigenvalue stability criterion, but of the form $\dot x = Ax +b$. Checking the eigenvalues of our matrix, we see that all eigenvalues are either negative or semisimple, hence making $\hat x$ a stable equilibrium point.

The question is: How can i justify the use of this stability test?

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  • $\begingroup$ I don't know what is semisimple eigen value but your matrix has 2 eignevalues $-2\pm\sqrt{6}$. One of the value is positive so you cannot claim anything about stability of your stationary point. $\endgroup$ – Alexander Vigodner Mar 6 '15 at 21:10
  • $\begingroup$ @AlexanderVigodner the eigenvalues are negative, see here for example. $\endgroup$ – ChocolateBar Mar 6 '15 at 21:17
  • $\begingroup$ Sorry, of course, you are correct, my bad. So 2 eigenvalues $-2\pm \sqrt{3}$ are negative. I still don't know what is semisimple is and why you use this classification for stabiliy analysis. IMHO the negativity of real part of eigenvalue is quite enough. $\endgroup$ – Alexander Vigodner Mar 6 '15 at 21:31
  • $\begingroup$ The theorem i was referring to in the question states, that for linear equations $\dot x = Ax$, either negative real part of the eigenvalues of $A$ or in case of a real part equal to $0$ the one-dimensionality of the corresponding Jordan Block (that is meant by semisimple eigenvalue) suffices for stability. $\endgroup$ – ChocolateBar Mar 6 '15 at 21:39
  • $\begingroup$ Yes I understood now what you mean, but in your case both eigenvalues are negative, so no need to mention semisimple property at all. Formally, you are correct, however, if you don't mean asymptotic stability. And you have the full right to use this therorem becuase after simple change of variables $y=x-[1,0]^T$ your system is reduced to the standard one $\dot y= Ay$. $\endgroup$ – Alexander Vigodner Mar 6 '15 at 21:46
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Reformulate the system in a new coordinate system $\boldsymbol{x} = \boldsymbol{x}_\text{solution}+\boldsymbol{z}$. This will transfrom

$$\dot{\boldsymbol{x}} = \boldsymbol{Ax}+\boldsymbol{b}.$$

to $$\dfrac{d}{dt}\boldsymbol{x}_\text{solution}+\dfrac{d}{dt}{\boldsymbol{z}}=\boldsymbol{A}\left[ \boldsymbol{x}_\text{solution}+\boldsymbol{z}\right]+\boldsymbol{b}$$ $$\implies \dfrac{d}{dt}\boldsymbol{x}_\text{solution}+\dfrac{d}{dt}{\boldsymbol{z}}=\boldsymbol{Az}+\left[\boldsymbol{A} \boldsymbol{x}_\text{solution} +\boldsymbol{b}\right].$$

As $\boldsymbol{x}_\text{solution}$ is a solution (note that is is not necessary that it is an equilibrium point) to $\dot{\boldsymbol{x}} = \boldsymbol{Ax}+\boldsymbol{b}$ we can simplify the previous equation to

$$\dot{\boldsymbol{z}}=\boldsymbol{Az}.$$

This equation has the trivial equilibrium point in the origin. The system matrix $\boldsymbol{A}$ did stay invariant under the substitution. This is why we can simply calculate the eigenvalues of the linear system in order to perform stability analysis.

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