Finding the characteristic timescale of a first-order nonlinear ODE

Question

I know that to find the timescale of a first order linear equation

$$\frac{dX(t)}{dt} + aX(t) = b$$

you just take the inverse of the integrating factor, so

$$t_x = \frac{1}{a}$$

Henning and joriki provide mathematical definitions of the characteristic timescale.

I have a system of coupled ODEs. I want to find the timescale for the generation of one of the species, $X$, whose derivative is of the form

$$\frac{dX(t)}{dt} + aX(t) = bY(t)Z(t)$$

Firstly, am I right in saying this is a first order non-linear ODE?

Secondly, does anyone have any pointers for how to find its timescale?

Many thanks.

Answer 1

Before addressing the question actually asked, we need to first clarify what it means for a system to have a characteristic timescale, and explore how we can compute it in even in the linear scenario given in the original question.

Characteristic timescale

As joriki explained, to speak of a characteristic timescale we assume

1. we are dealing with a dynamical system (system of ordinary differential equations; we can allow $X$ to be vector valued) $$ \frac{\mathrm{d}}{\mathrm{d}t} X(t) = F(X) \tag{1}$$
2. this system has an equilibrium $X_0$ $$ F(X_0) = 0 \tag{2}$$
3. the system tends to restore to equilibrium exponentially, that is to say that solutions with initial data in some open neighbourhood of $X_0$ have the asymptotic behaviour, as $t\to +\infty$ $$ \|X(t) - X_0 \| = o(e^{- t / k}) \tag{3}$$ for some $k > 0$.

The characteristic timescale is then the supremum over all $k$ for which equation (3) holds.

Important: if the system has no equilibria (equation (2) never holds), or if the system does not restore to equilibria exponentially (equation (3) fails), we cannot talk about a characteristic timescale.

Example: linear system

Your linear system can be cast into the form of equation (1) as $$ \dot{X} = F(X) = - a X + b $$ It has a unique equilibrium point $$ F(X_0) = 0 \implies -a X_0 + b = 0 \implies X_0 = b/a $$

So we want to consider initial data that is in some open neighbourhood of this $X_0$. Thanks to the linearity, however, statements that hold for an open neighbourhood will also hold for any initial data. (Note that this fact is not true in general for nonlinear systems.) So we need to derive some sort of control a la equation (3). To do so, it becomes convenient to rewrite your system in terms of its deviation from equilibrium, in this case

$$ Y = X - X_0 $$

The equation for $Y$ can be derived from the equation for $X$ and it is

$$ \dot{Y} = - a (Y + X_0) + b = - a Y $$

and we see that $Y$ can be solved explicitly as

$$ Y(t) = e^{- a t} Y(0) $$

so we see that for any $k < 1/a$,

$$ |Y(t)| = o(e^{-t/k}) $$ and hence $k = 1/a$ is the characteristic timescale.

Example: linear systems without characteristic timescales

To think about equations for which (2) fails, consider the ODE $$ \dot{X} = b $$ where $b \neq 0$. The function $F(X) = b$ has no roots, and hence there are no equilibrium points.

To think about equations for which (3) fails but (2) holds, consider the ODE $$ \dot{X} = a X $$ where $a \geq 0$. The point $X_0 = 0$ is a fixed point of this ODE. However, for any $k > 0$ it cannot hold that $$ |X(t)| = o(e^{-t/k})$$ When $a = 0$ every solution is constant, and hence for initial data not equal to 0, the solution does not converge back to 0. For $a > 0$ the solutions diverge exponentially.

Stability of dynamical systems

Now, given a system of form (1), and an equilibrium point of form (2), there is a simple criterion to tell whether an exponential restoration to equilibrium holds. This is by examining the linearisation of the equation.

Suppose that $F(X)$ is differentiable at $X_0$, the equilibrium point. By definition this means that there exists a linear operator $A$ such that $$ F(X) = A(X - X_0) + o(\|X - X_0\|) $$ as $X$ approaches $X_0$. This means that for initial data that is not $X_0$, but close to $X_0$, we can expect the solution to behave similar to the linear system $$ \dot{X} = A(X - X_0) $$

Since for the constant solution $X_0$ we have that $\dot{X_0} = 0$, we can again rewrite the equation in terms of the deviation $Y = X - X_0$ as $$ \dot{Y} = A Y + o(|Y|).$$ So as it turns out that a necessary and sufficient condition for $Y$ to always decay exponentially to 0, for any initial data in a small neighbourhood of the origin, is that

The real parts of the eigenvalues of the matrix $A$ are all negative. (Since $A$ is not necessarily symmetric, the eigenvalues can be complex.)

Let $\lambda_k = -a_k + b_k i$ be the eigenvalues. And let $Y_k$ be the corresponding eigenvectors of $A$. We can then say that asymptotically $Y_k$ is a perturbation of characteristic timescale $1 / a_k$, and that the characteristic timescale of the entire system is $1/a_1$ where $a_1$ is the smallest among the $a_k$.

So in short: in order to determine the characteristic timescale of a general nonlinear system, you need to

1. First find the relevant equilibrium point(s) of the system.
2. Linearize the system around said equilibria, and compute the eigenvalues of the linearisation. From the eigenvalues you can read off the characteristic timescales of different components of the system (corresponding to the eigenvectors).

Note in particular without further information, just knowing one ODE of the entire system is insufficient to determine the equilibrium points, nor the eigenvalues and eigenvectors. So based on the information you have given the question is not answerable.