Problem 40 - Advanced Engineering Mathematics

Question

(a) Show that \((0,0)\) is an isolated critical point of the plane autonomous system $$ \begin{aligned} &x^{\prime}=x^{4}-2 x y^{3} \\ &y^{\prime}=2 x^{3} y-y^{4} \end{aligned} $$ but that linearization gives no useful information about the nature of this critical point. (b) Use the phase-plane method to show that \(x^{3}+y^{3}=3 c x y\). This classic curve is called a folium of Descartes. Parametric equations for a folium are $$ x=\frac{3 c t}{1+t^{3}}, y=\frac{3 c t^{2}}{1+t^{3}} $$ [Hint: The differential equation in \(x\) and \(y\) is homogeneous.] (c) Use a graphing utility or a numerical solver to obtain solution curves. Based on your phase portrait, would you classify the critical point as stable or unstable? Would you classify the critical point as a node, saddle point, center, or spiral point? Explain.

Step-by-Step Solution

Answer

(0,0) is isolated and linearization is inconclusive. Use phase portrait to hypothesize critical point type.

Step 1: Find Critical Points

To find critical points, solve the system of equations: \[ x^{\prime} = x^{4} - 2xy^{3} = 0 \]\[ y^{\prime} = 2x^{3}y - y^{4} = 0 \]For the first equation, \( x^{4} = 2xy^{3} \). If \( x = 0 \), then \( x^{\prime} = 0 \) is satisfied. Substituting \( x = 0 \) into \( y^{\prime} = 0 \) implies \( y^{4} = 0 \), so \( y = 0 \). Thus, one critical point is \((0,0)\). Similar analysis shows no other critical points exist for \( y eq 0 \), since the equations won't simultaneously hold for another pair that satisfies both.

Step 2: Linearize the System

The linearization involves computing the Jacobian matrix at \( (0,0) \). The Jacobian matrix is:\[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 4x^3 - 2y^3 & -6xy^2 \ 6x^2y & 2x^3 - 4y^3 \end{bmatrix} \]At \((0,0)\), every derivative is zero, hence, \( J(0,0) = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \). The eigenvalues of this matrix do not provide information on stability because both are 0.

Step 3: Explore the Phase Plane

Consider the equation \( x^3 + y^3 = 3cxy \) as given in the problem. This equation defines a family of curves, called the folium of Descartes. The substituting the parametric equations \( x=\frac{3ct}{1+t^3}, y=\frac{3ct^2}{1+t^3} \) into the differential equations confirms they satisfy \( x^{\prime} \) and \( y^{\prime} \) because they match the homogeneous form \( dx^3 + dy^3 - 3cdxdy = 0 \).

Step 4: Graphical Analysis

Using a graphing utility or a numerical simulator, plot the solution curves around \((0,0)\). These typically indicate how solutions behave. For a system like this, the nature of orbits near \((0,0)\) may show closed loops (suggesting a center) or trajectories diverging or converging (suggesting a node or saddle point). Numerical experimentation might be needed for speculation.

Step 5: Conclusions from Phase Portrait

On analyzing the phase portrait (from graphical/numerical methods), determine the nature of the critical point:\- If trajectories loop around \( (0,0) \), it may resemble a center. \- If trajectories are moving away, \( (0,0) \) might be unstable.This step typically requires interpretations based on observed dynamics.

Key Concepts

Critical Points

In a phase-plane analysis, critical points, also known as equilibrium points, are where the system of differential equations become zero. For example, in the given system:\[ \begin{aligned} &x^{\prime}=x^{4}-2 x y^{3} \ &y^{\prime}=2 x^{3} y-y^{4} \end{aligned} \]we set both derivatives to zero. This leads to:

• \( x^{4} - 2xy^{3} = 0 \)
• \( 2x^{3}y - y^{4} = 0 \)

Solving these equations, we find that the point \((0,0)\) is a critical point. This type of point is isolated because it is clear from further analysis that no other points satisfy both equations simultaneously. Isolated critical points can help identify system behavior locally near those points.

Jacobian Matrix

In dynamical systems, the Jacobian matrix is essential for understanding how the system behaves near critical points. The matrix consists of the first-order partial derivatives of the system's equations. For the system provided:\[ x^{\prime}=x^{4}-2 x y^{3}, \quad y^{\prime}=2 x^{3} y-y^{4} \]The Jacobian matrix \( J \) is given by:\[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 4x^3 - 2y^3 & -6xy^2 \ 6x^2y & 2x^3 - 4y^3 \end{bmatrix} \]At the critical point \((0,0)\), this matrix becomes:\[ J(0,0) = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \]This indicates that the linearization gives no useful information about local dynamics. The eigenvalues of this matrix don't provide insights into stability here because they are zero.

Stability Analysis

Stability analysis helps us understand if a system returns to its critical point after a disturbance.Usually, the eigenvalues of a Jacobian matrix at a critical point can help us determine stability.However, in this problem, the Jacobian at \((0,0)\) is a zero matrix, which does not give eigenvalues that aid in concluding the nature of the stability. Thus, we use other methods like a phase-plane analysis to infer stability characteristics. Observation of trajectories can show patterns:

• If they encircle the point, implying a center type, the equilibrium could be neutral stable.
• If trajectories diverge, the point might be unstable.
• If converging, the system could be stable.

Folium of Descartes

The Folium of Descartes is a classic algebraic curve described in the problem by the equation:\[ x^3 + y^3 = 3cxy \]This equation presents a family of curves that can be represented in parametric form as:

• \( x = \frac{3ct}{1+t^3} \)
• \( y = \frac{3ct^2}{1+t^3} \)

These parametric equations form a curve known for its looped shape and symmetry properties. In the context of differential equations, the Folium of Descartes is significant because it demonstrates solutions aligned with trajectories in a phase-plane analysis. This classic curve aids in understanding the behavior of solutions near the critical point. In problems involving homogeneous equations like these, identifying forms like the folium can help illustrate how the system behaves globally beyond local critical points.