Question

Construct a suitable Liapunov function of the form \(a x^{2}+c y^{2}\) where \(a\) and \(c\) are to be determined. Then show that the critical point at the origin is of the indicated type. $$ d x / d t=-\frac{1}{3} x^{3}+2 x y^{2}, \quad d y / d t=-y^{3} ; \quad \text { asymptotically stable } $$

Short answer

Answer: An appropriate Liapunov function for the given system is \(V(x, y) = 3x^2 + y^2\). The critical point at the origin is asymptotically stable.

Step-by-step solution

Construct a Liapunov function

A Liapunov function has the form \(V(x, y) = a x^2 + c y^2\). Here, a and c are constants we need to determine. Let's write it down: $$ V(x, y) = a x^2 + c y^2 $$

Compute the time-derivative of the Liapunov function

We need to compute \(\frac{dV}{dt}\) to check the stability of the critical point at the origin. The time-derivative of V is given by: $$ \frac{dV}{dt} = \frac{\partial V}{\partial x} \frac{dx}{dt} + \frac{\partial V}{\partial y} \frac{dy}{dt} $$ Find the values for the partial derivatives of V: $$ \frac{\partial V}{\partial x} = 2ax, \quad \frac{\partial V}{\partial y} = 2cy $$ Now substitute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) from the given exercise, and the partial derivatives: $$ \frac{dV}{dt} = (2ax)(-\frac{1}{3} x^{3} + 2xy^{2}) + (2cy)(-y^{3}) $$

Determine the constants a and c

In order to make the critical point at the origin asymptotically stable, we need \(\frac{dV}{dt}\) to be negative for all x and y except at the origin (x = 0, y = 0). Let's simplify the expression for \(\frac{dV}{dt}\): $$ \frac{dV}{dt} = -\frac{2}{3} ax^4 + 4axy^2 - 2cy^4 $$ We notice that a possible choice for a and c is a = 3 and c = 1, which gives us: $$ \frac{dV}{dt} = -6x^4 + 12x^2 y^2 - 2y^4 $$ With these values of a and c, we find that \(\frac{dV}{dt}\) is always negative for x ≠ 0 and y ≠ 0, and thus, the critical point at the origin is asymptotically stable.

Conclusion

We have successfully constructed a Liapunov function \(V(x, y) = 3x^2 + y^2\) for the given system. By calculating the time-derivative \(\frac{dV}{dt}\) and selecting a = 3 and c = 1, we have shown that the critical point at the origin is asymptotically stable.

Key concepts

Critical Point

A critical point in dynamical systems refers to a state where the system experiences no change at that specific point. In mathematical terms, this is a point where all derivatives, generally the first derivatives, equal zero. For the system described in the exercise, the critical point is at the origin,

• Where: \(x = 0\) \(y = 0\)
• The derivatives \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\) at this point.

The primary goal is to analyze this point to comprehend the system's behavior. By examining the stability of the critical point using the Liapunov function, we can determine whether small perturbations will die out or escalate, impacting the system in the long term.

Asymptotic Stability

Asymptotic stability is a concept telling us how a system behaves as time progresses. When a critical point is asymptotically stable, it means if the system starts near the critical point, it will return to it over time. This indicates that the system's deviations diminish, leading back towards equilibrium. The exercise shows that the critical point at the origin is such a point because, by choosing the correct constants in the Liapunov function,

• \(\frac{dV}{dt} \) becomes negative for values close to, but not equal to, zero.
• With \(a = 3\) and \(c = 1\), it ensures the point returns to equilibrium, confirming asymptotic stability.

Thus, asymptotic stability assures that the system is not only stable but also that any deviations from equilibrium will eventually dissipate entirely.

Time-Derivative

Understanding the time-derivative of a function, particularly the Liapunov function in this exercise, is key to analyzing system stability. The time-derivative \(\frac{dV}{dt} \) indicates how the Liapunov function evolves over time relative to the system's state changes.

The Role of Partial Derivatives

Partial derivatives \(\frac{\partial V}{\partial x}\) and \(\frac{\partial V}{\partial y}\) represent the changes in the Liapunov function with respect to each variable independently. Through these, we form the expression:

• \(\frac{\partial V}{\partial x} = 2ax\)
• \(\frac{\partial V}{\partial y} = 2cy\)

Substituting Into the System's Derivative Equations

By substituting these partial derivatives back into the system, combined with \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), we compute \(\frac{dV}{dt}\). A negative value indicates energy wastage, suggesting the system's position diminishes over time, leaning toward stability. The choice of \(a = 3\) and \(c = 1\) allows \(\frac{dV}{dt}\) to mostly negative statusments, proving the system’s stability.