Stability analysis by local linearisation

Ordinary differential equations: Asymptotics and stability

Stability analysis by local linearisation

First we discuss the local linearisation technique on the basis of known examples and then we explain the general case.

Consider the ODE of exponential growth $\frac{\dd y}{\dd t}=r\cdot y$ with relative growth rate constant $r$ . If $r\lt 0$ , then the equilibrium $y=0$ is attracting; if $r\gt 0$ , then the equilibrium $y=0$ is repelling. The parameter $r$ also comes to the fore when you compute the derivative of the right-hand side with respect to $y$ in the equilibrium $y=0$ : $\frac{\dd (r\cdot y)}{\dd y}(0)=r\text.$ The behaviour of solutions around the equilibrium $y=0$ depends on the value of $r$ and the value $r=0$ marks a transition from an attracting to a repelling equilibrium, and vice versa. This change in behaviour around an equilibrium depending on a parameter value is called a bifurcation, and the parameter value at which this change occurs, in this case $r=0$ , is called a bifurcation value. We will discuss this in more depth in another section.

Consider the ODE of logistic growth $\frac{\dd y}{\dd t}=r\cdot y\cdot \left(1-\frac{y}{a}\right).$ equilibria are $y=0$ and $y=a$ . The equilibrium $y=0$ is attracting if $r<0$ and repelling if $r>0$ . We can understand this as follows. Consider first an initial value $y_0$ which is very close to 0 and in absolute value much closer to 0 than to 1. For such a number it is true that the square $y_0^2$ is much smaller than $|y_0|$ . To illustrate this: if $a=0.001$ then $a^2=0.00001.$ For points $(t,y(t))$ near $(0,y_0)$ therefore holds that $y(t)^2\lt |y(t)|$ . In short, the initial value problem written down as $\frac{\dd y}{\dd t}=r\cdot y-\frac{r}{a}\cdot y^2,\quad y(0)=y_0,$ has a second term in the right-hand side that can be neglected for $t$ in the neighbourhood of 0 with respect to the first term in case $y_0$ is close to 0. In other words, near the equilibrium the solution of the initial value problem will exhibit the same behaviour as the solution of the initial value problem $\frac{\dd y}{\dd t}=r\cdot y,\quad y(0)=y_0$ and we already know how this depends on the sign of $r$ . This approach to characterise the type of an equilibrium is called the method of local linearisation of the original problem.

The way we approach this from now on is as follows. The derivative of $\varphi(y)=r\cdot y\cdot \left(1-\frac{y}{a}\right)$ with respect to $y$ can be determined by means of the rules for differentiation: $\frac{\dd \varphi(y)}{\dd y}=\frac{\dd\!\left(r\cdot y \cdot \left(1-\frac{y}{a}\right)\right)}{\dd t}= r\cdot \left(1-\frac{y}{a}\right) - r\cdot \frac{y}{a}=r\cdot \left(1-\frac{2y}{a}\right)\text.$ Substitution of the equilibrium value $y=0$ gives the parameter $r$ .

In the same we you can study the equilibrium $y=a$ . Substitution of the equilibrium value $y=a$ in the above derivative gives $-r$ . The stability of $y=a$ is based on the same reasoning: the equilibrium is attracting if $-r\lt 0$ and repelling if $-r\gt 0$ . In other words, the equilibrium $y=a$ is attracting if $r>0$ and repelling as $r<0$ .

Again $r=0$ is a bifurcation value. In technical jargon, this is a transcritical bifurcation: for all other values of $r$ there are an attracting and repelling equilibrium, but when passing the bifurcation value the nature of the equilibria switches. Some people say that an exchange of stabilities takes place between the equilibria when passing the bifurcation value.

The general recipe for determining the stability of an equilibrium of an autonomous differential equation of the form $\frac{\dd y}{\dd t}=\varphi(y)$ is as following.

Stability analysis via local linearisation

Determine all equilibria by solving the equation $\varphi(y)=0$ . Suppose that $\varphi(\eta)=0$ , i.e., let $y=\eta$ be an equilibrium of the differential equation.
Differentiate $\varphi(y)$ with respect to $y$ . Call $\frac{\dd(\varphi(y))}{\dd y}$ in future $\varphi'(y)$ .
Determine the value of $\varphi'(\eta)$ . Henceforth we symbolise $\varphi'(\eta)$ with $c$ .
If $c \lt 0$ , then $y=\eta$ is an attracting equilibrium and a solution moves close to the neighbourhood of the steady state exponentially towards it with growth factor $e^c$ .
If $c \gt 0$ , then $y=\eta$ is a repelling equilibrium and a solution moves close to the neighbourhood of the steady state exponentially away from it with growth factor $e^c$ .

The above recipe is not really mysterious. Suppose that the function $\varphi(y)$ has a zero in $y=\eta$ . For a sufficiently smooth function $\varphi$ we can approximate the graph of the function $y$ in the neighbourhood of $\eta$ by the tangent line through $(\eta,\varphi(\eta))$ and we can write down $\varphi(y)\approx \varphi(\eta)+ \varphi'(\eta)\cdot (y-\eta)= \varphi'(\eta)\cdot (y-\eta).$ In other words, the original ODE is near $y=\eta$ similar to the ODE $\frac{\dd y}{\dd t}=c\cdot (y-\eta)$ with $c=\varphi'(\eta).$

The recipe is now nothing more nor less than a reformulation of the Hartman-Grobman theorem:

Solutions of a non-linear one-dimensional dynamical system $\frac{\dd y}{\dd t}=\varphi(y)$ behave sufficiently near an equilibrium $y=\eta$ in the same way as solution of the linear system $\frac{\dd y}{\dd t} =c\,(y-\eta)$ provided that $c=\varphi'(\eta)\neq 0$ , that is, the slope of $\varphi(y)$ in the equilibrium is nonzero.

Introduce the deviation $u(t)=y-\eta$ . This function satisfies the ODE for exponential growth: $\frac{\dd u}{\dd t}=\frac{\dd(y-\eta)}{\dd t}=\frac{\dd y}{\dd t}=c\cdot u,$ whose the following solution $u(t)=e^{c\cdot t}=g^t, \text{ with }g=e^c.$ If $c\lt 0$ , then there is exponential decay with growth factor $e^c$ ; the deviation goes to zero and the equilibrium is attracting. If $c\gt 0$ , then there is exponential growth factor $e^c>1$ and the deviation blow up so that the equilibrium is repelling.

In case $c=0$ , local linearisation does not help describe the behaviour of the non-linear dynamical system. Typically, this happens when the system undergoes a bifurcation.