Chapter 10: Problem 26
In Problems \(23-26\) solve the given nonlinear plane autonomous system by changing to polar coordinates. Describe the geometric behavior of the solution that satisfies the given initial condition(s). \(x^{\prime}=y-\frac{x}{\sqrt{x^{2}+y^{2}}}\left(4-x^{2}-y^{2}\right)\) \(y^{\prime}=-x-\frac{y}{\sqrt{x^{2}+y^{2}}}\left(4-x^{2}-y^{2}\right), \quad \mathbf{X}(0)=(1,0), \mathbf{X}(0)=(2,0)\)
Short Answer
Step by step solution
Change to Polar Coordinates
Substitute Polar Coordinates into the System
Simplify the System
Analyze the System
Determine the Geometric Behavior
Draw Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nonlinear Autonomous System
For the given problem, the system is expressed as:
- \(x' = y - \frac{x}{\sqrt{x^2+y^2}}(4-x^2-y^2)\)
- \(y' = -x - \frac{y}{\sqrt{x^2+y^2}}(4-x^2-y^2)\)
Critical Points
For the exercise at hand, we determine the critical points by setting the radial component of our polar-transformed system to zero:
- \(r' = r(1-r^2) = 0\)
- \(r = 0\) or \(r = 1\)
- \(r = 0\) corresponds to the origin, a stable equilibrium.
- \(r = 1\) represents a unit circle, indicating a cyclical behavior around the origin.
Geometric Behavior
For the exercise, the given polar system:
- \(r' = r(1-r^2)\)
- \(\theta' = -1\)
- When \(r = 1\), the solution traces a unit circle, denoting a stable orbit.
- If \(r > 1\), the trajectories will spiral inward towards the unit circle due to the term \(1-r^2\), which attains negative values, thereby reducing \(r\).
- The angle \(\theta\) decreases steadily for all \(r\), indicating a continuous counterclockwise rotation.
Coordinate Transformation
- \(x = r \cos(\theta)\)
- \(y = r \sin(\theta)\)
By substituting these into the original equations, derivatives transform as:
- \(x' = r' \cos(\theta) - r \sin(\theta) \theta'\)
- \(y' = r' \sin(\theta) + r \cos(\theta) \theta'\)