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. $$ d x / d t=-x^{3}+x y^{2}, \quad d y / d t=-2 x^{2} y-y^{3} ; \quad \text { asymptotically stable } $$

Short answer

Based on the above step-by-step solution, answer the following question: Question: Find a suitable Liapunov function for the given system of equations and determine if the origin is an asymptotically stable critical point. $$\frac{dx}{dt} = -x^3 + xy^2$$ $$\frac{dy}{dt} = -2x^2y - y^3$$ Answer: The Liapunov function for the given system of equations is \(V(x, y) = x^2 + y^2\). The origin is an asymptotically stable critical point.

Step-by-step solution

Identify Liapunov function candidate

Let the Liapunov function candidate be \(V(x, y) = ax^2 + cy^2\), where we will determine the positive constants \(a\) and \(c\) later.

Calculate partial derivatives

Calculate the partial derivatives of \(V(x, y)\) with respect to \(x\) and \(y\). This will give: $$\frac{\partial V}{\partial x} = 2ax$$ $$\frac{\partial V}{\partial y} = 2cy$$

Calculate the time derivative of the Liapunov function

Now, we'll find the time derivative of the Liapunov function using chain rule and the given equations: $$\frac{dV}{dt} = \frac{\partial V}{\partial x}\frac{dx}{dt} + \frac{\partial V}{\partial y}\frac{dy}{dt}$$ Substitute the partial derivatives and given equations into this equation: $$\frac{dV}{dt} = 2ax(-x^3 + xy^2) + 2cy(-2x^2y - y^3)$$

Simplify the time derivative expression

Simplify the expression for \(\frac{dV}{dt}\): $$\frac{dV}{dt} = -2ax^4 + 2axy^2 - 4cx^2y^2 - 2cy^4$$

Choose \(a\) and \(c\) to make the time derivative negative

Observe that all the terms in the expression for the time derivative are negative except \(\frac{dV}{dt} = 2axy^2\). We want the time derivative to be negative for the origin to be asymptotically stable. To achieve this, we can choose \(a = c = 1\). This will simplify the expression to: $$\frac{dV}{dt} = -2x^4 + 2xy^2 - 4x^2y^2 - 2y^4$$ Now, the term \(2xy^2\) cannot cancel out the negative terms when \(x\) and \(y\) are both non-zero, ensuring the time derivative remains negative.

Verify asymptotic stability

Since the time derivative is negative for non-zero \(x\) and \(y\), we can conclude that the origin is an asymptotically stable critical point. The Liapunov function that satisfies this condition is: $$V(x, y) = x^2 + y^2$$

Key concepts

Asymptotically Stable Critical Points

Understanding the behavior of dynamic systems near their critical points is essential in the study of system stability. Asymptotically stable critical points are a particularly vital concept in the field. They represent a state where, if the system is slightly disturbed, it will return to the equilibrium over time. This implies that the perturbation in the system will vanish as time progresses, bringing the system back to its stable critical point.

For the exercise given, we see that the equation system describes a dynamic process and provides us a way to analyze its stability. The critical point in question is the origin \( (0,0) \). To show that this point is asymptotically stable, we must first demonstrate that it is a stable equilibrium and then that any small deviation from this point will decay to zero as time goes on. A suitable Liapunov function helps in proving both these conditions by allowing us to analyze the system's energy-like properties.

Time Derivative of Liapunov Function

The time derivative of a Liapunov function is a pivotal concept in understanding the stability of a system. By taking the time derivative of the function, we can ascertain how the system's stability evolves over time. For the system to be stable, especially in the asymptotic sense, the time derivative of the Liapunov function must be negative definite or negative semi-definite.

In the provided step-by-step solution, we find the time derivative of the Liapunov function \(V(x, y)\) using the chain rule. This derivative takes into account how both variables \(x\) and \(y\) evolve over time according to the system's dynamics. If \( \frac{dV}{dt} \) is negative, apart from at the critical point itself where it is zero, it suggests that the 'energy' of the system is decreasing over time, reinforcing the system's tendency to return to the critical point, which is a hallmark of asymptotic stability.

Partial Derivatives in Stability Analysis

The role of partial derivatives in stability analysis is to measure how the Liapunov function's value changes with small changes in either variable, while keeping the others constant. It is an invaluable tool for exploring the landscape of the Liapunov function around a critical point and determining the direction of the system's trajectory.

The solution provided elaborates on calculating partial derivatives with respect to both \(x\) and \(y\) to obtain gradients of the function. These gradients are then integrated with the system's differential equations to derive the time derivative of the Liapunov function. By evaluating the signs of these partial derivatives, we can choose appropriate constants to ensure that the Liapunov function's time derivative is negative definite, thus leading us to conclude the system's stability near the critical points. Selecting appropriate constants \(a\) and \(c\) is crucial, as they must align with the condition that the time derivative remains negative for all non-zero \(x\) and \(y\), which is the heart of establishing asymptotic stability in this context.