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Magazine Chapter 11: Problem 14
analyze the phase plane of the differential equations for \(x, y \geq 0 .\) Show the nullclines and equilibrium points, and sketch the direction of the trajectories in each region. $$\begin{aligned} &\frac{d x}{d t}=x(2-x-3 y)\\\ &\frac{d y}{d t}=y(1-2 x) \end{aligned}$$
Short Answer
Expert verified
Identify nullclines and equilibria, analyze direction, and sketch phase plane.
Step by step solution
01
Identify the nullclines
For the given system of differential equations, nullclines are found by setting each derivative to zero.\\1. For \(dx/dt = x(2-x-3y) = 0\), the nullclines are:\ - \(x = 0\) or \(2-x-3y = 0\). This gives:\ - \(x = 0\) and \(y = \frac{2-x}{3}\).\ \ 2. For \(dy/dt = y(1-2x) = 0\), the nullclines are:\ - \(y = 0\) or \(1-2x = 0\). This gives: \ - \(y = 0\) and \(x = \frac{1}{2}\).
02
Find the equilibrium points
Equilibrium points occur where both \\(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\).\From Step 1 we have:\- Using \(x=0\) and \(y=0\), equilibrium at \((0,0)\).\- Plugging \(x=0\) into \(y = \frac{2-x}{3}\), equilibrium at \((0,\frac{2}{3})\).\- Setting \(x = \frac{1}{2}\) and \(y = 0\), equilibrium at \((\frac{1}{2}, 0)\).\- Solve \(2-x-3y = 0\) with \(1-2x=0\), giving equilibrium at \(\left(\frac{1}{2}, \frac{1}{2}\right)\).
03
Analyze phase plane and direction
For each region divided by the nullclines, determine the sign of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) to sketch trajectory directions:\- **Region 1**: Above \(y=\frac{2-x}{3}\) and on the left of \(x=\frac{1}{2}\), \(\frac{dx}{dt} < 0\) and \(\frac{dy}{dt} < 0\).\- **Region 2**: Between \(x=0\) and \(x=\frac{1}{2}\), below \(y=\frac{2-x}{3}\), \(\frac{dx}{dt} > 0\) and \(\frac{dy}{dt} < 0\).\- **Region 3**: Right of \(x=\frac{1}{2}\) and below \(y=\frac{2-x}{3}\), \(\frac{dx}{dt} < 0\) and \(\frac{dy}{dt} > 0\).
04
Sketch the phase plane
Draw the nullclines: "\(x=0\)", "\(y=0\)", "\(x=\frac{1}{2}\)", and "\(y=\frac{2-x}{3}\)."\Plot the equilibrium points: \((0,0)\), \((0,\frac{2}{3})\), \((\frac{1}{2},0)\), and \((\frac{1}{2}, \frac{1}{2})\).\In regions defined in Step 3, use arrows to show directions of trajectories based on the sign of derivatives, indicating where \(x\) and \(y\) are increasing or decreasing.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Nullclines
Nullclines are an essential part of analyzing the phase plane of a system of differential equations. To find the nullclines, we set each derivative equal to zero. This helps us identify significant lines in the phase plane that describe the behavior of the system. For the equations \(\frac{d x}{d t}=x(2-x-3 y)\) and \(\frac{d y}{d t}=y(1-2 x)\), two sets of nullclines are discovered:
- For \(\frac{dx}{dt}=0\): The solutions are \(x=0\) or \(2-x-3y=0\). This translates to nullclines at \(x=0\) and \(y=\frac{2-x}{3}\).
- For \(\frac{dy}{dt}=0\): The solutions are \(y=0\) or \(1-2x=0\). This gives us nullclines at \(y=0\) and \(x=\frac{1}{2}\).
Equilibrium Points
Equilibrium points occur at the intersection of nullclines, where both derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are zero. It's at these points that the system doesn鈥檛 move, meaning \(x\) and \(y\) remain constant over time. For our system, there are several equilibrium points:
- \((0, 0)\): Both derivatives are automatically zero when \(x\) and \(y\) are zero.
- \((0, \frac{2}{3})\): When \(x = 0\) plugged into \(y=\frac{2-x}{3}\), \(y\) equals \(\frac{2}{3}\).
- \((\frac{1}{2}, 0)\): Substituting \(x = \frac{1}{2}\) into \(1-2x=0\) sets \(y = 0\).
- \((\frac{1}{2}, \frac{1}{2})\): Solving \(2-x-3y=0\) with \(1-2x=0\) results in this point.
Trajectory Directions
In each region of the phase plane divided by the nullclines, direction of movement helps analyze how the system behaves. By determining the sign of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), you can identify trajectory directions鈥攕howing whether \(x\) and \(y\) increase or decrease. For the given system, consider:
- **Region 1**: Above \(y=\frac{2-x}{3}\) and left of \(x=\frac{1}{2}\), both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are negative, meaning both \(x\) and \(y\) decrease.
- **Region 2**: Below \(y=\frac{2-x}{3}\) between \(x=0\) and \(x=\frac{1}{2}\), \(\frac{dx}{dt}\) is positive and \(\frac{dy}{dt}\) is negative, showing \(x\) increases while \(y\) decreases.
- **Region 3**: Right of \(x=\frac{1}{2}\) and below \(y=\frac{2-x}{3}\), \(\frac{dx}{dt}\) is negative while \(\frac{dy}{dt}\) is positive, indicating \(x\) decreases and \(y\) increases.
Phase Plane Sketch
The phase plane sketch is a visual representation of the system's dynamics. It combines nullclines, equilibrium points, and trajectory directions to create a comprehensive picture of the behavior. When sketching a phase plane:
- Start by plotting the nullclines like lines where the derivatives are zero. This includes \(x=0\), \(y=0\), \(x=\frac{1}{2}\), and \(y=\frac{2-x}{3}\).
- Mark equilibrium points: \((0,0)\), \((0,\frac{2}{3})\), \((\frac{1}{2},0)\), and \((\frac{1}{2}, \frac{1}{2})\).
- In the spaces between these lines, draw arrows representing trajectory directions from each region analysis, showing flow based on the signs of derivatives.
