
# 7.4: Asymptotic Behavior of Continuous-Time Linear Dynamical Systems


A general formula for continuous-time linear dynamical systems is given by

$\frac{dx}{dt} =Ax, \label{(7.40)}$

where $$x$$ is the state vector of the system and $$A$$ is the coefﬁcient matrix. As discussed before, you could add a constant vector a to the right hand side, but it can always be converted into a constant-free form by increasing the dimensions of the system, as follows:

$y =\begin{pmatrix} x\\1\end{pmatrix}\label{(7.41)}$

$\frac{dy}{dt} =\begin{pmatrix} A&a\\0&0\end{pmatrix} \begin{pmatrix} x\\1\end{pmatrix} =By \label{(7.42)}$

Note that the last-row, last-column element of the expanded coefﬁcient matrix is now 0, not 1, because of Eq. 6.3.9. This result guarantees that the constant-free form given in Eq. \ref{(7.40)} is general enough to represent various dynamics of linear dynamical systems. Now, what is the asymptotic behavior of Eq. \ref{(7.40)}? This may not look so intuitive, but it turns out that there is a closed-form solution available for this case as well. Here is the solution, which is generally applicable for any square matrix $$A$$:

$x(t) =c^{At}x(0)\label{(7.43)}$

Here, $$e^{X}$$ is a matrix exponential for a square matrix $$X$$, which is deﬁned as

$e^{X} =\sum ^{\infty}_{k=0}\frac{X^{k}}{k!},\label{(7.44)}$

ith $$X^(0)= I$$. This is a Taylor series-based deﬁnition of a usual exponential, but now it is generalized to accept a square matrix instead of a scalar number (which is a 1 x 1 square matrix, by the way). It is known that this inﬁnite series always converges to a well-deﬁned square matrix for any $$X$$. Note that $$e^(X)$$ is the same size as $$X$$.

Exercise $$\PageIndex{1}$$

Conﬁrm that the solution Eq. \ref{7.43} satisﬁes Eq. \ref{(7.40)}.

The matrix exponential $$e^{X}$$ has some interesting properties. First, its eigenvalues are the exponentials of $$X$$’seigenvalues. Second, its eigenvectors are thesame as $$X$$’s eigenvectors. That is:

$Xv=\lambda{v}\ \Rightarrow \ x^{X}v =e^{\lambda}v\label{(7.45)}$

Exercise $$\PageIndex{2}$$

Conﬁrm Eq. \ref{(7.45)} using Eq. \ref{(7.44)}.

We can use these properties to study the asymptotic behavior of Eq. \ref{(7.43)}. As in Chapter 5, we assume that $$A$$ is diagonalizable and thus has as many linearly independent eigenvectors as the dimensions of the state space. Then the initial state of the system can be represented as

$x(0) =b_{1}v_{1}+b_{2}v_{2}+...b_{n}v_{n},\label{(7.46)}$

where $$n$$ is the dimension of the state space and $$vi$$ are the eigenvectors of $$A$$ (and of $$e^{A}$$). Applying this to Eq. \ref{(7.43)} results in

$x(t) =e^{At}(b_{1}v_{1}+b_{2}v_{2}+...b_{n}v_{n})\label{(7.47)}$

$=(b_{1}e^{At}v_{1}+b_{2}e^{At}v_{2}+...b_{n}e^{At}v_{n})\label({7.48)}$

$=(b_{1}e^{\lambda_{1} {t}}v_{1}+b_{2}e^{(\lambda_{2} {t}}v_{2}+...b_{n}e^{\lambda_{n}{t}}v_{n}).\label{(7.49)}$

This result shows that the asymptotic behavior of $$x(t)$$ is given by a summation of multiple exponential terms of $$e^{λ_{i}}$$ (note the difference—this was $$λ_{i}$$ for discrete-time models). Therefore, which term eventually dominates others is determined by the absolute value of $$e^{λ_{i}}$$ . Because $$|e^{λ_{i}} | = e^{Re(λ_{i})}$$, this means that the eigenvalue that has the largest real part is the dominant eigenvalue for continuous-time models. For example, if $$λ_{1}$$ has the largest real part $$(Re(λ_{1}) > Re(λ_{2}),Re(λ_{3}),...Re(λ_{n}))$$, then

$x(t) =e^{\lambda_{1} {t}}(b_{1}v_{1}+b_{2}e^{(\lambda_{2} -\lambda_{1})t}v_{2}+...b_{n}e^({\lambda_{n} -\lambda_{1})t}v_{n}),\label{(7.50)}$

$\lim_{ t\rightarrow \infty}{x(t)}\approx e^{\lambda_{1} {t}}b_{1}v_{1}.\label{(7.51)}$

Similar to discrete-time models, the dominant eigenvalues and eigenvectors tell us the asymptotic behavior of continuous-time models, but with a little different stability criterion. Namely, if the real part of the dominant eigenvalue is greater than $$0$$, then the system diverges to inﬁnity, i.e., the system is unstable. If it is less than $$0$$, the system eventually shrinks to zero, i.e., the system is stable. If it is precisely $$0$$, then the dominant eigenvector component of the system’s state is conserved with neither divergence nor convergence, and thus the system may converge to a non-zero equilibrium point. The same interpretation can be applied to non-dominant eigenvalues as well.

An eigenvalue tells us whether a particular component of a system’s state (given by its corresponding eigenvector) grows or shrinks over time. For continuous-time models:

• Re($$λ$$) > 0 means that the component is growing.

• Re($$λ$$) < 0 means that the component is shrinking.

• Re($$λ$$) = 0 means that the component is conserved.

For continuous-time models, the real part of the dominant eigenvalue λd determines the stability of the whole system as follows:

• Re$$(λ_{d}) > 0$$: The system is $$unstable$$, diverging to inﬁnity.

• Re$$(λ_{d}) < 0$$: The system is $$stable$$, converging to the origin.

• Re$$(λ_{d}) = 0$$: The system is $$stable$$, but the dominant eigenvector component is conserved, and therefore the system may converge to a non-zero equilibrium point.

Here is an example of a general two-dimensional linear dynamical system in continuous time (a.k.a. the “love affairs” model proposed by Strogatz [29]):

$\frac{dx}{dt} =\begin {pmatrix} a & b \\c & d\end{pmatrix}x=Ax\label{(7.52)}$

The eigenvalues of the coefﬁcient matrix can be obtained by solving the following equation for $$λ$$:

$\det\begin{pmatrix} a-\lambda &b \\c & d-\lambda\end{pmatrix} =0\label{(7.53)}$

Or:

$(a-\lambda)(d-\lambda)-bc =\lambda^{2} -(a+d)\lambda +ad-bc \label{(7.54)}$

$=\lambda^{2} -Tr(A)\lambda +det(A)=0\label{(7.55)}$

Here, $$Tr(A)$$ is the trace of matrix $$A$$, i.e., the sum of its diagonal components. The solutions of the equation above are
$\lambda = \frac{Tr(A)\pm \sqrt{Tr(A)^{2}-4det(A)}}{2}.\label{(7.56)}$

Between those two eigenvalues ,which one isdominant? Since the radical on the numerator gives either a non-negative real value or an imaginary value, the one with a “plus” sign always has the greater real part. Now we can ﬁnd the conditions for which this system is stable. The real part of this dominant eigenvalue is given as follows:

$Re(\lambda_{d}) = \begin{cases} & \frac{Tr(A)}{2} \text{ if } Tr(A)^{2} <4det(A) \\& \frac{Tr(A) + \sqrt{Tr(A)^{2}-4det(A)}}{2} \text { if } Tr(A)^{2} \geq 4det(A)\end{cases} \label{(7.57)}$

If $$Tr(A)^{2} < 4 \ det(A)$$, the stability condition is simply

$Tr(A) <0.\label{(7.58)}$

If $$Tr(A)^{2} ≥ 4 \ det(A)$$, the stability condition is derived as follows:

$Tr(A) +\sqrt{Tr(A)^{2} -4 \ det(A) }<0\label{(7.59)}$

$\sqrt{Tr(A)^{2}-4 \ det(A)} <-Tr(A)\label{(7.60)}$

Since the radical on the left hand side must be non-negative, $$Tr(A)$$ must be negative, at least. Also, by squaring both sides, we obtain

$Tr(A)^{2}-4det(A)<Tr(A)^{2}\label{(7.61)}$

$-4det(A)<0,\label{(7.62)}$

$det(A) >0.\label{(7.63)}$

By combining all the results above, we can summarize how the two-dimensional linear dynamical system’s stability depends on $$Tr(A)$$ and $$det(A)$$ in a simple diagram as shown in Fig. 7.5. Note that this diagram is applicable only to two-dimensional systems, and it is not generalizable for systems that involve three or more variables.

Exercise $$\PageIndex{1}$$

Show that the unstable points with $$det(A) < 0$$ are saddle points.

Exercise $$\PageIndex{2}$$

Determine the stability of the following linear systems:

$\cdot \frac{dx}{dt}= \begin{pmatrix} -1 & 2\\2 &-2 \end{pmatrix} x$

$\cdot \frac{dx}{dt} =\begin{pmatrix} 0.5 & -1.5\\ 1&-1\end{pmatrix} x$

Exercise $$\PageIndex{3}$$

Conﬁrm the analytical result shown in Fig. 7.4.1 by conducting numerical simulations in Python and by drawing phase spaces of the system for several samples of $$A$$.