Limiting behavior of a system of differential equations

Question

Consider this system of differential equations

\begin{equation} \label{eq:9} x^{'}_i(t) = \sum_{j=1}^{n} a_{i,j}\,(x_{j}(t)- x_{j}(t))= \sum_{j=1}^{n} a_{i,j}\,x_{j}(t) - x_i(t)\,\sum_{j=1}^{n} a_{i,j} \end{equation}

The solution can be written in the matrix exponential form:

Let $x(t) = \begin{bmatrix} x_1(t) \\ \vdots \\ x_n(t) \end{bmatrix}$ and $A=\begin{pmatrix} -\sum_{j} a_{1,j} & a_{1,2} & \dots & a_{1,n}\\ a_{2,1} & - \sum_{j} a_{2,j} & \dots & a_{2,n}\\ \vdots & \vdots & \vdots & \vdots \\ a_{n,1} & a_{n,2}& \dots & - \sum_{j} a_{n,j} \end{pmatrix} $

Then, the above system can be written in a matrix form

\begin{equation} \label{eq:10} x^{'}(t) = A\,x(t). \end{equation}

The solution to the above matrix differential equation is

\begin{equation} \label{eq:11} x(t) = e^{A\,t}\,x(0) = \sum\limits_{k=1}^{n} \frac{A^k\,t^k}{k!}\,x(0) \end{equation}

The exponential solution is not very practical. But it's fine! I am not looking for an explicit solution for $x_i$ (It would be perfect to have it though). What I am looking for is the limiting behavior of $x_i(t)$, $$ \lim_{t \to \infty} x_i(t)=?.$$

It's not difficult to show that matrix A, has eigenvalues that are either $0$ or complex numbers with negative real part and therefore, there is a limit for each $x_i(t)$. Moreover, the limit is the same for all $x_i(t)$, i.e., there exists $c$ such that for all $i$,

$$ \lim_{t \to \infty} x_i(t)=c.$$

However, I haven't found an explicit formula for $c$ in terms of constants $a_{i,j}$ and initial values $x_i(0)$. While I am trying to find it, any help would be appreciated.

In case $n=2$, we have,

$$c = \frac{a_{21}\,x_1(0)+a_{12}\,x_2(0)}{a_{12}+a_{21}} $$

Answer 1

For $n>2$, a nice formula for the quantities $\lim_{t\to\infty} x_i(t)$ doesn't seem to be possible for general matrices $A$. In fact, these limits may not even exist, depending on the values of its entries and the initial conditions. However, it is relatively easy to find the steady state values, by using the Jordan decomposition of $A$, when they do exist, and extract them at least numerically.

It is easy to see that $A$ has at least one null eigenvector for any $n\geq 2$, and hence is non-invertible:

$$A_{n}\begin{pmatrix}1\\1\\\vdots\\1\end{pmatrix}=0$$

Suppose that the matrix is potentially non-diagonalizable, but is diagonalizable in the subsector corresponding to the eigenvalue $\lambda=0$ (in other words, there are as many eigenvectors corresponding to that eigenvalue as it's multiplicity) and has no purely imaginary eigenvalues. Then, given that all non-zero eigenvalues have a negative real part and using the formula for exponentiation of Jordan blocks, we see that in the limit $t\to\infty$, the only modes that survive the long time exponentiation is exactly the ones belonging to the zero subsector (one can see this relatively easily from the property $\exp(tA)=S\exp(tJ)S^{-1}$ and the fact that the Jordan normal form is block-diagonal, exponentiating the zero rows of that matrix results in 1's in the diagonal). Hence, one can write down the formula

$$\lim_{t\to\infty} x_i(t)=\sum_{k=1}^n\sum_{j=1}^rS_{ij}S^{-1}_{jk}x_k(0)~,~ r<n, ~~i=1,...,n$$

where $r$ is the number of eigenvectors in the zero subsector. This is the most general case which allows steady states for all components.

To see what happens when one of the conditions above is not met, suppose now that the zero subsector also contains at least one non-trivial Jordan subblock of size $m$. Exponentiating this subblock gives

$$\exp \begin{pmatrix}0&t&0&0\\0&0&t&0\\0&0&0&t\\&\vdots&&\ddots\end{pmatrix}=\begin{pmatrix}1&t&t^2/2!&...\\0&1&t&t^2/2!\\0&0&1&t\\&\vdots&&\ddots\end{pmatrix}$$

which means that there are at least some $x_i(t)$ that never reach a steady state and instead diverge.

Another instance where the limits may diverge is if some of the eigenvalues are purely imaginary, in which case the orbits of the $x_i(t)$ will be oscillating. An example is provided by the following matrix

$$A_3=\begin{pmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{pmatrix}$$

which admits eigenvalues $\{0,\pm i\sqrt{3}\}$ and hence produces oscillating solutions with period $2\pi/\sqrt{3}$.