Stability Theory of ODE

This chapter we focus on

\[ \begin{equation} \dot{\mathbfit{x}} = \mathbfit{f}(t, \mathbfit{x})\label{eq1} \end{equation} \]

We consider continuous dependence of the solutions on initial condition on a larger interval, i.e. \((\beta,\infty)\), for we have show the dependency holds on finite intervals. However, this is not always as expected. So we have to introduce some basic ideas.

Lyapunov 稳定性 | Lyapunov Stability

We assume that \(\pmb{x}=\pmb{\varphi}(t)\) is a special solution of the ODE \(\ref{eq1}\).

Lyapunov Stability

(i) Lyapunov Stable

If \(\forall \varepsilon>0\), \(\forall t_0>\beta\), \(\exists \delta>0\), \(\forall \pmb{x}_0\), s.t. \(\|\pmb{x}_0-\pmb{\varphi}(t_0)\|<\delta\), ODE \(\ref{eq1}\) with initial value \(\pmb{x}(t_0)=\pmb{x}_0\) has a solution \(\pmb{x}(t;t_0,\pmb{x}_0)\) which exists on \([t_0,+\infty)\) and satisfies

\[ \|\pmb{x}(t;t_0,\pmb{x}_0)-\varphi(t)\|<\varepsilon, \quad \forall t\in [t_0,+\infty) \]

then we call the special solution \(\pmb{x}=\pmb{\varphi}(t)\) is Lyapunov stable, or stable for short.

(ii) Lyapunov Unstable

If \(\exists \varepsilon_0>0\), \(\exists t_0>\beta\), \(\forall \delta>0\), \(\exists \pmb{x}_0\), s.t. \(\|\pmb{x}_0-\pmb{\varphi}(t_0)\|<\delta\), ODE \(\ref{eq1}\) with initial value \(\pmb{x}(t_0)=\pmb{x}_0\) has a solution \(\pmb{x}(t;t_0,\pmb{x}_0)\) which exists on \([t_0, \alpha)\)(\(\alpha<+\infty\)) or there exists \(t_1>t_0\), and it satisfies

\[ \|\pmb{x}(t_1;t_0,\pmb{x}_0)-\varphi(t)\|\geq\varepsilon_0 \]

then the special solution is Lyapunov Unstable.

(iii) Attractive

If \(\exists t_0>\beta\), \(\exists \zeta>0\), \(\forall \pmb{x}_0\), s.t. \(\|\pmb{x}_0-\pmb{\varphi}(t_0)\|<\zeta\), solution of ODE \(\ref{eq1}\) with initial value \(\pmb{x}(t_0)=\pmb{x}_0\) satisfies

\[ \lim_{t\rightarrow +\infty}\|\pmb{x}(t;t_0,\pmb{x}_0)-\pmb{\varphi}(t)\|=0 \]

then the special solution is Attractive.

(iv) Asymptotic stable

If a special solution is Lyapunov stable and attractive, then we call it Asymptotic stable.

To simplify the analysis, we choose to make a translation. That is, if we let

\[ \pmb{y}=\pmb{x}-\pmb{\varphi}(t), \pmb{F}(t,\pmb{y})=\pmb{f}(t,\pmb{y}+\pmb{\varphi}(t))-\pmb{f}(t,\pmb{\varphi}(t)) \]

then we only need to discuss

\[ \begin{align*} \frac{d\pmb{y}}{dt}&=\frac{d\pmb{x}}{dt}-\frac{d\pmb{\varphi}(t)}{dt}\\ &=\pmb{f}(t,\pmb{x})-\pmb{f}(t,\pmb{\varphi}(t))\\ &=\pmb{f}(t,\pmb{y}+\pmb{\varphi}(t))-\pmb{f}(t,\pmb{\varphi}(t))\\ &=\pmb{F}(t,\pmb{y}) \end{align*} \]

with special solution \(\pmb{y}=0\), i.e. \(\pmb{f}(t,\pmb{0})=\pmb{0}\). We call solution \(\pmb{y}\equiv 0\) zero solution.

线性近似判别法 | Linear Approximate Method

If we represent ODE \(\ref{eq1}\) as

\[ \begin{equation} \frac{d\pmb{x}}{dt}=\pmb{A}(t)\pmb{x}+\pmb{R}(t,\pmb{x})\label{eq3} \end{equation} \]

where

\[ \pmb{A}(t)=\frac{\partial \pmb{f}}{\partial \pmb{x}}\Bigg|_{\pmb{x}=\pmb{0}} \]

and \(\pmb{R}(t,\pmb{x})\) is the summation of higher order items, i.e.

\[ \lim_{\|\pmb{x}\|\rightarrow 0}\frac{\|\pmb{R}(t,\pmb{x})\|}{\|\pmb{x}\|}=0. \]

Lyapunov Stability of LODEs

Firstly, we discuss the Lyapunov stability of LODEs

\[ \begin{equation} \frac{d\pmb{x}}{dt}=\pmb{A}(t)\pmb{x}. \label{eq2} \end{equation} \]

Theorem of Lyapunov Stability of LODEs

Assume \(\pmb{\varPhi}(t)\) is a basic solution matrix of LODEs \(\ref{eq2}\), then its zero solution is

(i) stable iff \(\forall t_0>\beta\), \(\exists K(t_0)>0\), s.t.

\[ \|\pmb{\varPhi}(t)\| \leq K,\quad, t\geq t_0 \]

(ii) asymptotic stable iff

\[ \lim_{t\rightarrow +\infty}\|\pmb{\varPhi}(t)\|=0 \]

For time-varying system, it is hard to solve a basic solution matrix. But for time-invariant system, we can not only solve the basic solution matrix, but also deduce a better result.

Theorem for Lyapunov Stability of time-invariant System

Consider const-coeffient system

\[ \frac{d\pmb{x}}{dt}=\pmb{A}\pmb{x} \]

its zero solution is

(i) asymptotic stable iff all the eigenbalues of \(\pmb{A}\) are negative.

(ii) stable iff all the eigenvalues of \(\pmb{A}\) are not positive, and for eigenvalue whose real part is zero, its corresponding Jordan block is one roder (or can be diagonalized).

(iii) unstable iff there exists one positive eigenvalue or for eigenvalue whose real part is zero, its corresponding Jordan block is one roder(or cannot be diagonalized).

Lyapunov Stability of Non-linear ODEs

For non-linear system, we can have some results that is similar to the above. To simplify the problem, we let the linear part \(\pmb{A}(t)\equiv \pmb{A}\) in ODE \(\ref{eq3}\) be a constant matrix. Then, we can have the following Theorem

Theorem for non-linear system using Linear Approximatie Method

(i) If all the eigenvalues of \(\pmb{A}\) are negative, then the zero solution of ODE \(\ref{eq3}\) is asymptotic stable.

(ii) If there exists an eigenvalue of \(\pmb{A}\) which is positive, then the zero solution of ODE \(\ref{eq3}\) is unstable.

HintsProof

Use Picard theorem and extension theorem to get a solution on \([0,t*)\), then show that solution can tend to \(0\) by using Gronwall inequation.

第二法 | Second Method of Lyapunov

The first method does not carry on because it makes use of power series. But his second method did gain ground, which employs an energy function which characterizes the solution. So this method is also called Direct Method.

Here, we only consider autonomous ODEs, that is,

\[ \begin{equation} \frac{d\pmb{x}}{dt}=\pmb{f}(\pmb{x})\label{eq4} \end{equation} \]

with the right side of ODE not containing variable \(t\).

The basic idea can be shown in the following diagram.

\[ \begin{align*} \|\pmb{x}_0\|&\ll 1\\ &\Downarrow \quad \text{(by continuity of $V$)} \\ V(\pmb{x}_0)&\ll 1\\ &\Downarrow \quad \text{(by $\frac{d V}{dt}< 0$)}\\ V[\pmb{x}(t)]&\leq V[\pmb{x}(0)]\ll 1\\ &\Downarrow \quad \text{(by monotony)}\\ \|\pmb{x}(t)\|&\leq\|\pmb{x}(0)\|\ll 1 \end{align*} \]

Definition of Definite Sign Function

Assume \(h>0\), function \(V(\pmb{x})\in C^1(\|\pmb{x}\|\leq h)\), if \(V(\pmb{0})=0\) and

(i) \(V(\pmb{x})>0\)(or \(<0\)) on \(0<\|\pmb{x}\|\leq h\), then we call \(V(\pmb{x})\) is a definite positive(or negative) function.

(ii) \(V(\pmb{x})\geq0\)(or \(\leq0\)) on \(0<\|\pmb{x}\|\leq h\), then we call \(V(\pmb{x})\) is a constant positive(or negative) function.

Denifite positive function has a property.

Property of Definite Positive Function

\(\forall \varepsilon>0\), \(\exists \delta>0\), \(\forall \pmb{x}\), s.t. \(V(\pmb{x})<\delta\), then \(\|\pmb{x}\|<\varepsilon\).

HintsProof

Pay attention to the positive definite function.

\(\forall \varepsilon>0\), let

\[ \delta=\min_{\varepsilon\leq\|\pmb{x}\|\leq h}V(\pmb{x}) \]

then

\[ \|\pmb{x}\|\geq \delta \Rightarrow V(\pmb{x})\geq \min_{\varepsilon\leq\|\pmb{x}\|\leq h}V(\pmb{x}) = \delta. \]

So

\[ V(\pmb{x})<\delta\Rightarrow \|\pmb{x}\|<\varepsilon \]

Theorem of Stability test

In ODE \(\ref{eq4}\), assume \(h>0\). If there exists a definite positive function \(V(\pmb{x})\) on \(\|\pmb{x}\|\leq h\), whose total derivative

\[ \frac{dV}{dt}=\nabla^{\pmb{x}}V(\pmb{x})\cdot \frac{d\pmb{x}}{dt}=\sum_{i=1}^n\frac{\partial V}{\partial x_i}\frac{d x_i}{dt}=\sum_{i=1}^n\frac{\partial V}{\partial x_i}f_i \]

(i) is a constant negative function, then the zero solution to ODE \(\ref{eq4}\) is Lyapunov stable.

(ii) is a definite negative function, then the zero solution to ODE \(\ref{eq4}\) is Lyapunov asymptotic stable.

(iii) is a definite positive function, then the zero solution to ODE \(\ref{eq4}\) is Lyapunov unstable.

For (i) in Theorem of Stability Test, if we find another condition, then the system can also be asymptotic stable, and this is exactly the following theorem.

LaSalle's invariance principle

If ODE \(\ref{eq4}\) satisfies

For (iii) in Theorem of Stability Test, we can have a weaker theorem.

Theorem for Lyapunov Unstable

For ODE \(\ref{eq4}\), if there exists a open region \(\mathcal{N}\subset B_h(\pmb{0})\), satisfies

(i) \(V(\pmb{x})\in C^1(\mathcal{N})\).

(ii) \(\pmb{0}\in \partial \mathcal{N}\), \(\exists \delta>0\), \(\forall \pmb{x}\in \partial \mathcal{N}\cap B_\delta(\pmb{0})\), such that

\[ V(\pmb{x})=0 \]

(iii) \(\forall \pmb{x}\in \mathcal{N}\cap B_\delta(\pmb{0})\), \(V(\pmb{x})\) and \(\frac{dV}{dt}\) are both definite positive function.

For a specific problem, we can choose first quadrant of the plane

\[ \mathcal{N}=\{(x,y): x>0,y>0\} \]

with a typical \(V(\pmb{x})=xy\), then test if \(\frac{dV}{dt}\) is definite positive on \(\mathcal{N}\).

Example1. Determine the Lyapunov stability of the following system.

\[ \begin{cases} x'=x^5+\sin{x^5}+2y+6y^5\\ y'=-e^x+8x^3+e^{y^3} \end{cases} \]

Answer

use Hamilton system approximation.

Example2. Determine the Lyapunov stability of the following system.

\[ \begin{cases} x'=-3x^3y^4\\ y'=x^4y^3 \end{cases} \]

Answer

The above one cannot use Hamilton system approximation for it only has one item. But we can divide one by two, and get

\[ \frac{dy}{dx}=-\frac{x}{3y} \]

which is a homogeneous equation. So the solution is \(x^2+3y^2=C\). We can choose Lyapunov function \(V=0.5x^2+1.5y^3\), then \(dV/dt\equiv 0\), which is Lyapunov stable but not asymptotic stable.