Linearising a dynamical system at an arbitrary point (affine dynamical systems)

Usually, one linearises a dynamical system around a fixed point, in order to determine whether the fixed point is stable. That is (assuming the fixed point it at the origin), one approximates

$$\frac{d\mathbf{x}}{dt} = \mathbf{f}(\mathbf{x})$$

with

$$\frac{d\mathbf{x}}{dt} = A\mathbf{x},$$

which is easy to solve in terms of the matrix $A$, and in particular, the eigenvalues of $A$ determine the local asymptotic stability of the fixed point.

However, I wish to linearise a dynamical system around an arbitrary point, not necessarily a steady state, in order to get an idea of the local transient dynamics around this point. Doing this, I obtain the approximation

$$\frac{d\mathbf{x}}{dt} = A\mathbf{x} + \mathbf{b},$$

where the vector $\mathbf{b}$ no longer has all zero entries.

A little research tells me that this is known as an affine dynamical system, which makes sense, but I can't find much information on the properties of affine dynamical systems in multiple dimensions. So my questions are:

My main question is, can one write down an explicit solution for a general affine dynamical system, in terms of the matrix $A$ and vector $\mathbf{b}$?

As a secondary question, where can I read up on (i) affine dynamical systems and their properties, and/or (ii) the general things that can be said about the trajectories of dynamical system in a small region surrounding some arbitrary (non-fixed) point?

• Near a non-fixed point, one can in principle straighten out the vector field, so that case is usually not considered very interesting. (Put differently, the vector field is nearly constant (and nonzero) near that point, so the flow is approximately just constant speed straight line motion there.) – Hans Lundmark Sep 1 '14 at 6:22

Yes, one can write down the general solution to

$\dfrac{d\mathbf x}{dt} = A \mathbf x + \mathbf b, \tag{1}$

just as one may write down the general solution to the homogeneous system

$\dfrac{d\mathbf x}{dt} = A \mathbf x. \tag{2}$

Just as the solution to (2) passing through the point $\mathbf x_0$ at $t = t_0$ is

$\mathbf x(t) = e^{A(t - t_0)} \mathbf x_0, \tag{3}$

so the solution to (1) is

$\mathbf x(t) = e^{A(t - t_0)}(\mathbf x_0 + \int_{t_0}^t e^{-A(s - t_0)} \mathbf b\ ds). \tag{4}$

Both (3) and (4) may be validated by direct differentiation; for example, given (4), we find

$\dfrac{d\mathbf x}{dt} = Ae^{A(t - t_0)}(\mathbf x_0 + \int_{t_0}^t e^{-A(s - t_0)} \mathbf b\ ds) + e^{A(t - t_0)} \dfrac{d}{dt}(\int_{t_0}^t e^{-A(s - t_0)} \mathbf b\ ds)$ $= Ae^{A(t - t_0)}(\mathbf x_0 + \int_{t_0}^t e^{-A(s - t_0)} \mathbf b\ ds) + e^{A(t - t_0)}(e^{-A(t - t_0)})\mathbf b = A\mathbf x(t) + \mathbf b, \tag{5}$

using $e^{A(t - t_0)}(e^{-A(t - t_0)}) = I$, the identity operator. It is also easy to see that this $\mathbf x(t)$ satisfies $\mathbf x(t_0) = \mathbf x_0$, so the initial conditions are also satisfed by (5). Demonstrating that (3) satisfies (2) is, of course, even easier; I leave that to my readership.

There is a vast literature available on (1) and (2) and their solutions. I would check out any good text on ordinary differential equations, and try googling around a little bit, searching for things like inhomogeneous linear ordinary differential equation.

General dynamical systems of the form

$\dfrac{d\mathbf x}{dt} = X(\mathbf x), \tag{6}$

near points $\mathbf x_0$ such that $X(\mathbf x_0) \ne 0$, may always be written in the form

$\dfrac{d\mathbf x}{dt} = (1, 0, 0, \ldots, 0)^T \tag{7}$

upon a suitable change of coordinates; this is possible since we may take the integral curves of $X(\mathbf x)$ themselves to be coordinate lines. I believe a detailed discussion may be found in Abraham and Marsden's Foundations of Mechanics, as well as in many other books addressing differential topology and flows. There is not a lot more which can be said, from a purely differential-topological point of view. Of course, if the domain of definition of (6) is equipped with a Riemannian metric or other distance-measuring apparatus, then the relative motions of neighboring trajectories, whether they draw together or move apart with the passing of time and so forth, may be quantitatively addressed. But we can still write (6) in the form (7) where $X(\mathbf x_0) \ne 0$.

Note Added in Response to Nathaniel's comment of 11 September 2014; Thursday 11 September 2014 10:03 PM PST: A few words about equation (4); these in response to Nathaniel's comment; being a bit long for an in-kind response, I decided to edit them into my answer. Equation (4) indeed yields the correct result, $\mathbf x_0$, for the value of $\mathbf x$ when $t = t_0$, that is, of $\mathbf x(t_0)$, for taking $t = t_0$ in (4) yields

$\mathbf x(t_0) = e^{A(t_0 - t_0)}(\mathbf x_0 + \int_{t_0}^{t_0} e^{-A(s - t_0)} \mathbf b\ ds); \tag{8}$

but $e^{A(t_0 - t_0)} = e^{A(0)} = e^0 = I$, and the integral vanishes since its upper and lower limits are both $t_0$; thus we are left with

$\mathbf x(t_0) = I\mathbf x_0 = \mathbf x_0; \tag{9}$

the initial condition is in fact sensible, consistent, and as advertised. Next, suppose the real part of each eigenvalue of $A$ is negative, that is, $\Re(\lambda) < 0$ for $\lambda$ an eigenvalue of $A$. Then $e^{A(t - t_0)} \to 0$ as $t \to \infty$, as is well-known. Thus, writing (4) as

$\mathbf x(t) = e^{A(t - t_0)} \mathbf x_0 + e^{A(t - t_0)}\int_{t_0}^t e^{-A(s - t_0)}\mathbf b ds, \tag{10}$

we see that the first term on the right $e^{A(t - t_0)} \mathbf x_0 \to 0$ as $t$ grows without bound, leaving us only to evaluate the second. As for the integral, we have

$\int_{t_0}^t e^{-A(s - t_0)} \mathbf b ds = (-A^{-1}(e^{-A(s - t_0)} \mathbf b \mid_{t_0}^t, \tag{11}$

as may be verified by direct differentiation of $-A^{-1}e^{-A(s - t_0)} \mathbf b$; furthermore

$(-A^{-1}(e^{-A(s - t_0)} \mathbf b \mid_{t_0}^t = -A^{-1}(e^{-A(t - t_0)} - I) \mathbf b, \tag{12}$

and thus, combining (11) and (12), we have

$e^{A(t - t_0)}\int_{t_0}^t e^{-A(s - t_0)} \mathbf b ds = -A^{-1}(I - e^{A(t - t_0)}) \mathbf b \to -A^{-1} \mathbf b \; \text{as} \; t \to \infty; \tag{13}$

this shows all solutions of (1), given by (4), tend to $-A^{-1}\mathbf b$ as $t$ increases without bound. We note that, at the point $\mathbf x = -A^{-1} \mathbf b$, $\dot {\mathbf x} = 0$ since

$\dot {\mathbf x}(-A^{-1}\mathbf b) = A(-A^{-1}\mathbf b) + \mathbf b = 0. \tag{14}$

$-A^{-1}\mathbf b$, the unique zero of the vector field $\dot {\mathbf x} = A\mathbf x + \mathbf b$, is the limit of every trajectory as $t \to \infty$. It is easy to see using the above equations that taking $\mathbf x_0 = -A^{-1}\mathbf b$ yields the solution $\mathbf x(t) = \mathbf x_0$ for all $t$; $\mathbf x(t)$ is then an equilibrium trajectory of the system. End of Note.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

• Thanks, I guess that pretty much answers it. (I like to give some time before accepting.) I suppose I should have been a bit more specific about my "what can be said..." question. I'm interested in applications along the lines of ecology, where there isn't an obvious metric defined on the space, but the variables themselves have meaning as the populations of biological species etc. So the questions of interest are things like "how will the trajectory of $x_j$ change if a small perturbation is made to $x_i$ at time 0." – Nathaniel Sep 1 '14 at 13:31
• @Nathaniel: if you replace $\mathbf x_0$ in (4) by $\mathbf x_0 + \mathbf z_0$ and subtract (4) from the the result, you will see that $\mathbf z(t)$ satisfies (3). This will in principle allow you to calculate the change in $\mathbf x_j(t)$ induced by a perturbation of $\mathbf x_i(t_0)$. – Robert Lewis Sep 1 '14 at 17:33
• @Nathaniel: the general topic is known as variational equations; chapter III of J.K. Hale's Ordinary Differential Equations explains it in great detail though from a rather advanced point of view. – Robert Lewis Sep 1 '14 at 17:36
• @Nathaniel: in any event, glad to help out and thanks for the good word. Regards and a Good Holiday. – Robert Lewis Sep 1 '14 at 17:40
• Great, thanks, that's exactly what I wanted. Many thanks, and a good holiday to you too! – Nathaniel Sep 2 '14 at 0:47