To understand how to derive the condition \(V'(x)f(x) \leq 0\), let’s break it down step by step within the context of Lyapunov stability theory, which is commonly used to analyze the stability of dynamic systems.

1. Lyapunov Function \(V(x)\):

  • The function\(V(x)\) is chosen or constructed to be a measure of the “energy” or “potential” of the system. For stability analysis, we typically require that: -\(V(x) > 0\) for all\(x \neq 0\) (positive definiteness). -\(V(x) = 0\) at the equilibrium point\(x = 0\).

2. System Dynamics \(\dot{x} = f(x)\):

  • Consider a dynamic system described by the differential equation\(\dot{x} = f(x)\), where\(\dot{x}\) is the time derivative of the state\(x\), and\(f(x)\) describes the system’s dynamics.

3. Time Derivative of the Lyapunov Function:

  • To analyze the stability of the system, we examine how the Lyapunov function\(V(x)\) changes over time as the system evolves.

  • The rate of change of\(V(x)\) with respect to time is given by the total derivative: \(\frac{dV(x)}{dt} = \frac{\partial V(x)}{\partial x} \cdot \frac{dx}{dt} = V'(x) \cdot \dot{x},\) where\(V'(x) = \frac{\partial V(x)}{\partial x}\) is the gradient of\(V(x)\) with respect to\(x\), and\(\dot{x} = f(x)\).

  • Substituting\(\dot{x} = f(x)\) into this equation, we get: \(\frac{dV(x)}{dt} = V'(x) \cdot f(x).\)

4. Stability Condition:

  • For the system to be stable, we require that the Lyapunov function\(V(x)\) does not increase over time, meaning that\(\frac{dV(x)}{dt} \leq 0\).
  • This condition implies: \(V'(x) \cdot f(x) \leq 0.\)
  • If\(V'(x)f(x) \leq 0\), then\(V(x)\) is non-increasing along the trajectories of the system. This non-increasing behavior suggests that the system’s “energy” is either decreasing or staying the same, which corresponds to the system being stable or moving toward a stable state.

5. Derivation Summary:

  • The condition\(V'(x)f(x) \leq 0\) is derived by considering the time derivative of the Lyapunov function\(V(x)\).
  • If\(V'(x)f(x) \leq 0\), it guarantees that\(V(x)\) does not increase over time, meaning the system (or algorithm) is stable, as it tends to reduce or maintain its “energy” over time.

Example (Illustration):

Let’s consider a simple system:

\(\dot{x} = -x,\) and define the Lyapunov function\(V(x) = \frac{1}{2}x^2\).

  • The gradient\(V'(x) = x\).
  • The time derivative of\(V(x)\) along the trajectory of the system is: \(\frac{dV(x)}{dt} = V'(x) \cdot \dot{x} = x \cdot (-x) = -x^2 \leq 0.\)

Here,\(V'(x)f(x) \leq 0\) because\(-x^2\) is always non-positive. This means\(V(x)\) decreases over time, which indicates that the system is stable.

Conclusion:

The condition\(V'(x)f(x) \leq 0\) is derived as a requirement for the Lyapunov function to be non-increasing, which in turn ensures the stability of the system. It indicates that the system is not gaining energy and is either stable or moving towards stability.