Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be an $n\times n$ real matrix, and let's consider the linear system of ODEs $x'=Ax$.

I'm trying to characterize the Lyapunov stability of the origin according to the real part of the eigenvalues of $A$. Gerard Teschl's Ordinary Differential Equations and Dynamical Systems, p.68, reads:

0 is stable iff every eigenvalue $\lambda$ of $A$ satisfies $Re(\lambda)\leq 0$, and for all eigenvalues such that $Re(\lambda)=0$ the corresponding algebraic and geometric multiplicities are equal.

0 is asymptotically stable iff every eigenvalue $\lambda$ of $A$ satisfies $Re(\lambda)<0$.

I'm not convinced at all of his proof; I find it somewhat sketchy and the author excessively chatty for my taste. So what I ask in this question is

1) a full proof of this statement.

I have tried the following: I've found the following

Lemma: if $\lambda=a+ib$ is an eigenvalue of $A$, then there exist $u,v\in \mathbb{R}^n$ non-zero such that (1): $e^{At}u=e^{at} (\cos(bt)u+\sin(bt)v)$, $e^{At}v=e^{at}(\cos(bt)v-\sin(bt)u)$.

which I was not able to prove, so I also ask of you

2) a full proof of this lemma.

Using this lemma, it's not hard to prove that if 0 is stable, then every eigenvalue $\lambda$ of $A$ must verify $Re(\lambda)\leq 0$.

Indeed, suppose there exists $\lambda=a+ib$ eigenvalue of $A$ with $a>0$. Then, by the lemma there are vectors $u,v$ satisfying (1). By scaling the equation (1) we can get vectors $u,v$ of arbitrarily small norm to satisfy (1). So we get solutions arbitrarily close to the origin which are of the form $e^{at} (\cos(bt)u+\sin(bt)v)$. Since $a>0$, by taking limits we see that the norm of this solution tends to infinity as $t$ grows larger, thus $0$ would be unstable.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

I give you some hints, I hope it helps. For the lemma: If $\lambda$ is an eigenvalue, then there is a complex eigenvector. Your $u$ and $v$ are exactly the real and imaginary parts of this eigenvectors.

For the Theorem. It is easy if you allow a bit more linear algebra. So assume that everything is complex. Then we can assume without loss of generality that the matrix $A$ is in Jordan normal form.

The solution of the ODE can be represented in general using the matrix exponential function: if your initial value was $x_0\in\mathbb{R}^n$, then the solution wil be $$x(t)=e^{At}x_0.$$ The exponential is usually defined by the power series. Now, if your matrix is in Jordan normal form, then you can calculate this power series. More precisely, if $$A=\begin{pmatrix} \lambda & 1 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & \cdots & 0 \\ \vdots & & & & \vdots \\ 0 &\cdots & &\cdots & \lambda\end{pmatrix} = \text{diag}(\lambda) + N,$$ where $N$ is nilpotent, then $$e^{tN} = \begin{pmatrix} 1 & t & \frac{t^2}{2} & \cdots & \frac{t^{n-1}}{(n-1)!} \\ 0 & 1 & t & \cdots & \frac{t^{n-2}}{(n-2)!} \\ \vdots & & & & \vdots \\ 0 & \cdots & & \cdots & 1 \end{pmatrix}$$, and $$e^{tA} = e^{t\lambda}e^{tN}.$$

You can see now that

  • If you have $\Re\lambda<0$, then you have asymptotic stability.
  • If you have $\Re\lambda>0$, then you have no hope to get stability.
  • Hence, it is necessary to have $\Re\lambda\leq 0$ for stability.
  • You can immediately see from the formulae that if $\Re\lambda=0$, then you cannot allow a Jordan block which is greater then 1, because polynomial terms appear in the exponential. This shows the point on the geometric and algebraic multipicities.

I hope this helped.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.