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This is probably a pretty easy question to someone working in the field of ODEs (which I generally do not), but nonetheless, I've haven't been able to find clear a answer in the textbooks I've looked in. Consider the ODE

$X'(t) = A + BX(t)$

with some initial condition $X(0)$. Here, A is a vector of dimension $n$, and $B$ is an $n\times n$ matrix. I know that this equation has a unique solution given by

$X(t) = \exp(tB)X(0) + \int_0^t \exp((t-s)B)A ds$

with $\exp(tB)$ denoting the matrix exponential. I also know that if all eigenvalues of $B$ have strictly negative real part, then $B$ is in particular invertible, and it holds that $X(t)$ converges to $-B^{-1}A$ as $t$ tends to infinity.

My question is this: What can be said about the limit of $X(t)$ when the eigenvalues of $B$ do not all have strictly negative real parts? Ideally, I would like to know about a criterion for which coordinates of $X(t)$ tend to infinity, minus infinity or a finite limit as a function of, say, the eigenvalues of $B$.

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If matrix $B$ is invertible, then you can always make a substitution $Y=X+B^{-1}A$ for which you have $$ \dot Y=BY. $$ If there exist an eigenvalue $\lambda>0$ of $B$ then both $Y$ and $X$ tend to infinity. If all eigenvalues have negative real parts then $Y\to 0$ and $X\to -B^{-1}A$. If the origin is Lyapunov stable for $Y$ then the solutions for $X$ will stay bounded.

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Thanks for the answer! Could you elaborate on a few things? First, in the case of the existence of a positive eigenvalue $\lambda$ (I assume that you mean an eigenvalue with positive real part), can you say something about which coordinates (or linear combinations of coordinates) will tend to infinity? And is it possible to say something about the non-invertible case, or the case where there exists an eigenvalue with zero real part? –  Alexander Sokol Apr 12 '13 at 18:13
    
Yes, if the real part is positive. Which coordinates will tend to infinity depend on the structure of $B$. If there are eigenvalues of the form $\pm \omega i$, then $\dot Y=BY$ is Lyaponov stable, hence the solutions for X are bounded. If $B$ is non-invertible, we have two cases: $BX=A$ has no solutions, in this case $X$ will tend to infinity, or has a linear manifold of solutions, which can be either attracting or repelling. –  Artem Apr 12 '13 at 19:26
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