Relation between stability of fixed points of discrete and continuous dynamical systems

Question

Let $f:\mathbb R^d\to\mathbb R^d$ be a diffeomorphism, $x_0$ be a fixed point of $$\frac{\rm d}{{\rm d}t}\varphi_t(x)=f\left(\varphi_t(x)\right)\tag1$$ (i.e. $f(x_0)=0$) and $A:={\rm D}f(x_0)$.

It is a well-known result that if the largest eigenvalue $\lambda_1$ of $A$ is negative, then $x_0$ is asymptotically stable.

Now, on the other hand, if $x_0$ is a fixed point of the discrete system $$x_n=f(x_{n-1})\tag2,$$ we may consider the linearization $$h_n=Ah_{n-1}\tag3$$ and it is well-known that $x_0$ is asymptotically stable if $|\lambda_1|<1$.

How are these stability results related?

I've read that we somehow need to "integrate" $(2)$ over time and then use that if $\lambda$ is an eigenvalue of $A$, then $e^\lambda$ is an eigenvalue of $e^A$, but I neither know how precisely we need to transform $(2)$, nor why $e^A$ is occurring then.

Answer 1

The link between the two results comes from linear system analysis.

1. From continuous-time dynamic system(CTDS) to discrete-time dynamic system(DTDS), linear case

Consider the state-space representation of a CTDS, $\dot{x}=Ax,x(0)=x_0$ where $x,x_0\in\mathbb{R}^d,A\in\mathbb{R}^{d\times d}.$ We can explicitly solve this: $$x(t)=e^{At}x(0).$$

Let $x_n=x(nT)$ where $T$ is a positive constant, we have $$x_{n+1}=e^{AT}x_n=e^Te^Ax_n.$$ That is to say, we obtain a DTDS by sampling a CTDS. This is how $e^A$ occurs.

2. Stability and stability criteria of the two linear systems

Let $x_0$ be a fixed point of the CTDS, then $Ax_0=0$. Since $$e^A=\sum_{n=0}^\infty\frac{A^n}{n!},$$ we have that $$e^Ax_0=0.$$

Hence $x_0$ is also a fixed point of the DTDS.

Using the definition for stability, we can easily check that the (in)stability of the CTDS at $x_0$ is equivalent to the (in)stability of the DTDS at $x_0$ and the factor $e^{T}$ doesn't matter.

As you said, if $λ$ is an eigenvalue of $A$, then $e^λ$ is an eigenvalue of $e^A$, therefore all eigenvalue of $A$ is negative is equivalent to all eigenvalue of $e^A$ is inside the unit circle.

3. Linearization keeps the stability invariant

See Hartman–Grobman theorem.

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Edit 1:

The full logic goes like this: the stability of a nonlinear system (continuous-case or discrete-case) at some equilibrium point can be partly inferred by analyzing the linear system obtained by linearization at that point (Hartman–Grobman theorem for the continuous-case and Banach fixed-point theorem for the discrete-case). Therefore, we can focus on the linearized version of the two systems, and the stability analysis still holds.

To put it simply, We are actually dealing with four systems here: $$\dot{x}=f(x),\tag{1}$$ $$\dot{x}=Ax,A:={\textrm D}f(x_0),\tag{1'}$$ $$x_n=f(x_{n-1}),\tag{2}$$ $$x_n=Ax_{n-1},A:={\textrm D}f(x_0),\tag{2'}$$ System (1') and (2') are the linearized version of systems (1) and (2), respectively. As you mentioned, if the largest eigenvalue $\lambda_1$ of $A$ is negative, then (1') and (1) are both asymptotically stable at $x_0$. Similarly, if $|\lambda_1|<1$, then (2') and (2) are both asymptotically stable at $x_0$.

Since (1') is a linear time-invariant system, we can easily explicitly solve it: $x(t)=e^{At}x(0)$. By sampling, we get another DTDS: $$x_{n+1}=e^Te^Ax_n\tag{3}$$

It is by connecting (1') with (2') using (3) that we get the relationship of the two criteria. The results are shown above.

Answer 2

It seems to me that you confuse yourself a little by choosing the same $f$ for both continuous and discrete versions.

Let me try to address your question from a slightly different perspective.

Consider continuous dynamical system $$ \dot x=f(x), $$ such that $0$ is an equilibrium. Assume that we know that for this equilibrium to be stable all the eigenvalues of linearization $Df(0)$ must have negative real parts.

Now together with continuous system consider its discretization $$ x'=x+hf(x), $$ which clearly has the same equilibrium $0$. You can consider this discretization as the simplest numerical method to solve ODE system.

Now, linearization of the discrete system is $$ I+h Df(0), $$ whose eigenvalues are $1+h\lambda$ if $\lambda$ is an eigenvalue of $Df(0)$.

For 0 to be stable in discrete version one must have $$ |1+h\lambda|<1 $$ i.e., lie in the disk with the center at $-1/h$ and radius $1/h$.

Hence, to answer

How are these stability results related?

we can say that the stability condition for the discrete version is more restrictive compare to the continuous one. If 0 is stable for the continuous system, it will be stable for the discrete version only if $h$ sufficiently small. If 0 is a center for the linearization of continuous system, it will be unstable for the discrete version no matter the size of $h$, although it still could be asymptotically stable for the continuous system.