Chapter 10: Problem 15
The graph of a nullcline is not always a line, and we do not have to restrict ourselves to the first quadrant when sketching phase portraits. Perform each of the tasks (i)-(iv) required, only this time, sketch the phase portrait in all four quadrants of the \(x y\) phase plane. $$ \begin{aligned} &x^{\prime}=y-x^{3} \\ &y^{\prime}=x-y \end{aligned} $$
Short Answer
Step by step solution
Find the Nullclines
Determine Intersection Points
Analyze Local Stability
Sketch the Nullclines
Sketch the Phase Portrait
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
The given system of equations consists of two first-order differential equations:
- \(x' = y - x^3\)
- \(y' = x - y\)
Equilibrium Points
For the given system:
- \(x' = y - x^3 = 0 \Rightarrow y = x^3\)
- \(y' = x - y = 0 \Rightarrow y = x\)
Stability Analysis
We analyze stability by evaluating the Jacobian matrix of the system at each equilibrium point. The Jacobian matrix, given by:
\[J = \begin{pmatrix} -3x^2 & 1 \ 1 & -1 \end{pmatrix}\]
provides a linear approximation of the system near the equilibrium points.
For each point:
- \((0,0)\): Calculated eigenvalues reveal a saddle point, indicating instability.
- \((1,1)\): Eigenvalues are negative, showcasing a stable node.
- \((-1,-1)\): Similar negative eigenvalues, confirming stability like \((1,1)\).
Nullclines
In our system, the nullclines are given by:
- \(y = x^3\): The \(x'\) nullcline, where the differential equation for \(x\) remains zero.
- \(y = x\): The \(y'\) nullcline, indicating zero rate of change for \(y\).