Locally Stable but Globally Unstable? In control theory, we talk about the direct and indirect Lyapunov methods, applied to the stability analysis of nonlinear systems.

There are systems that are locally unstable for specific operating points, but that are globally stable.

What i would like to know is if there is any system which is locally stable, for any linearized point, but is globally unstable.

In other words, I would like to know if a system that is guaranteed to be locally stable, thorough linearization, for any point, can be said to be globally stable as well.

In the case it is positive, I would like to know if uniform stability and exponential stability holds as well.
The answer is no. A system might be unstable for some initial conditions even if its linearization is Hurwitz for all $\boldsymbol{x}$.

An example is provided in this lecture, page 4 (Nonlinear Systems and Control — Spring 2015: State-Dependent Riccati Equation Method). Let

$$
\begin{bmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{bmatrix} =
\begin{bmatrix}
\frac{-1}{1 + \epsilon\sqrt{x_1^2 + x_2^2}} &amp; 1 + \sqrt{x_1^2 + x_2^2} \\
0 &amp; \frac{-1}{1 + \epsilon\sqrt{x_1^2 + x_2^2}} \\
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
$$

with $\epsilon &gt; 0.$ The eigenvalues are given by the matrix entries $a_{11}$ and $a_{22}$, which are obviously negative for any $\boldsymbol{x} \in \mathbb{R}^2$. Thus $A(\boldsymbol{x})$ is Hurwitz for all $\boldsymbol{x}  \in \mathbb{R}^2$.

However, take $\epsilon = 1$ and solve the ODE numerically, using for example the Runge-Kutta method, with initial conditions $\boldsymbol{x}_0 = \begin{bmatrix}1 &amp; 1\end{bmatrix}^T$. You will observe that $x_1 \rightarrow \infty$ and $x_2 \rightarrow \text{const}$.
The answer is no. A system might be unstable for some initial conditions even if its linearization is Hurwitz for all $\boldsymbol{x}$.

An example is provided in this lecture, page 4 (Nonlinear Systems and Control — Spring 2015: State-Dependent Riccati Equation Method). Let

$$
\begin{bmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{bmatrix} =
\begin{bmatrix}
\frac{-1}{1 + \epsilon\sqrt{x_1^2 + x_2^2}} &amp; 1 + \sqrt{x_1^2 + x_2^2} \\
0 &amp; \frac{-1}{1 + \epsilon\sqrt{x_1^2 + x_2^2}} \\
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
$$

with $\epsilon &gt; 0.$ The eigenvalues are given by the matrix entries $a_{11}$ and $a_{22}$, which are obviously negative for any $\boldsymbol{x} \in \mathbb{R}^2$. Thus $A(\boldsymbol{x})$ is Hurwitz for all $\boldsymbol{x}  \in \mathbb{R}^2$.

However, take $\epsilon = 1$ and solve the ODE numerically, using for example the Runge-Kutta method, with initial conditions $\boldsymbol{x}_0 = \begin{bmatrix}1 &amp; 1\end{bmatrix}^T$. You will observe that $x_1 \rightarrow \infty$ and $x_2 \rightarrow \text{const}$.