Problem 13
Question
For \(\mathbf{X}=\left(K_{1}, 0\right), \tau=-r_{1}+r_{2}\left[1-\left(K_{1} \alpha_{21} / K_{2}\right)\right]\) and \(\Delta=-r_{1} r_{2}\left[1-\left(K_{1} \alpha_{21} / K_{2}\right)\right] .\) If we let \(c=1-K_{1} \alpha_{21} / K_{2}\) \(\tau^{2}-4 \Delta=\left(c r_{2}+r_{1}\right)^{2}>0 .\) Now if \(k_{1} >K_{2} / \alpha_{21}, c< 0\) and so \(\tau<0, \Delta>0 .\) Therefore \(\left(K_{1}, 0\right)\) is a stable node. If \(K_{1}< K_{2} / \alpha_{21}, c>0\) and so \(\Delta< 0 .\) In this case \(\left(K_{1}, 0\right)\) is a saddle point.
Step-by-Step Solution
Verified Answer
\((K_1, 0)\) is a stable node if \( K_1 > K_2 / \alpha_{21} \), and a saddle point if \( K_1 < K_2 / \alpha_{21} \).
1Step 1: Understanding the Parameters
We have the point \( \mathbf{X} = (K_1, 0) \), and several parameters that define its nature: \( \tau \), \( \Delta \), and \( c \). These parameters are functions of \( K_1 \), \( r_1 \), \( r_2 \), \( \alpha_{21} \), and \( K_2 \). Our task is to determine the properties of the point \( (K_1, 0) \) based on inequalities involving these parameters.
2Step 2: Defining the Variables
1. \( \tau = -r_1 + r_2[1 - (K_1\alpha_{21} / K_2)] \)2. \( \Delta = -r_1 r_2[1 - (K_1\alpha_{21} / K_2)] \)3. Let \( c = 1 - (K_1\alpha_{21} / K_2) \)
3Step 3: Expression for Discriminant
We find \( \tau^2 - 4\Delta = (c r_2 + r_1)^2 \). This expression implies that the discriminant is always positive, \( \tau^2 - 4\Delta > 0 \). This suggests that our equilibrium points are real and distinct.
4Step 4: Analyze Case: \( k_1 > K_2/\alpha_{21} \)
If \( K_1 > K_2/\alpha_{21} \), then \( c < 0 \). Consequently, \( \tau < 0 \) and since \( \Delta = -r_1 r_2 c \), this implies \( \Delta > 0 \). This combination (\( \tau < 0 \) and \( \Delta > 0 \)) indicates that the equilibrium point \( (K_1, 0) \) is a stable node.
5Step 5: Analyze Case: \( K_1 < K_2/\alpha_{21} \)
For \( K_1 < K_2/\alpha_{21} \), \( c > 0 \) making \( \Delta < 0 \). Since \( \Delta < 0 \), \( (K_1, 0) \) is a saddle point under these conditions.
Key Concepts
Stability AnalysisDiscriminantSaddle PointStable Node
Stability Analysis
Understanding stability analysis is like trying to figure out whether an equilibrium point will remain steady or not over time. Equilibrium points are where systems do not change; they are like the calm center of a storm. In our example, we have analyzed the points using parameters such as \(\tau\) and \(\Delta\). These parameters help us to determine whether the point \((K_1, 0)\) will stay stable or become unstable.
Stability analysis allows us to predict the future behavior of the system around these points. It's like forecasting weather, but for mathematical systems.
- For the given system, when \(\tau < 0\) and \(\Delta > 0\), the equilibrium point is known as a stable node.
- If you find \(\Delta < 0\), then it indicates a saddle point.
Stability analysis allows us to predict the future behavior of the system around these points. It's like forecasting weather, but for mathematical systems.
Discriminant
The discriminant is a valuable tool in the stability analysis of equilibrium points. It can be thought of as a mathematical litmus test that helps us determine the nature of equilibrium solutions. In the context of our example, the formula \(\tau^2 - 4\Delta\) plays the role of the discriminant.
The concept of a discriminant simplifies complex stability questions, making it invaluable in analyzing how systems behave near equilibrium.
- If \(\tau^2 - 4\Delta > 0\), it indicates that the system's solutions are real and distinct, hinting at physical stability.
- This distinguishes whether we have two different types of roots, which then leads us to identify other behaviors like stable nodes or saddle points.
The concept of a discriminant simplifies complex stability questions, making it invaluable in analyzing how systems behave near equilibrium.
Saddle Point
A saddle point is one of the possible equilibrium states you might encounter in stability analysis. It is characterized by being unstable in one direction and stable in another – much like a saddle on a horse. In our example problem, when \(K_1 < K_2/\alpha_{21}\), we find \(c > 0\), which causes \(\Delta < 0\).
Recognizing a saddle point helps in predicting the specific instabilities that create new pathways for dynamical behavior.
- This \(\Delta < 0\) outcome confirms the presence of a saddle point at \((K_1, 0)\).
- Saddle points are important, as they highlight conditions under which small disturbances grow large in the system, leading to instability.
Recognizing a saddle point helps in predicting the specific instabilities that create new pathways for dynamical behavior.
Stable Node
A stable node is an equilibrium point where the system naturally settles over time, akin to a marble rolling to the bottom of a bowl. In stability analysis, this is identified when both conditions \(\tau < 0\) and \(\Delta > 0\) are satisfied. For the system described in the exercise, this occurs when \(K_1 > K_2/\alpha_{21}\).
Identifying stable nodes is crucial for understanding long-term behavior in dynamic systems, giving insights into how the system can be maintained or controlled.
- Here, \(c < 0\) results in \(\Delta > 0\), therefore confirming a stable node at \((K_1, 0)\).
- In practical terms, a stable node means that even if the system is slightly disturbed, it will return to the equilibrium point over time.
Identifying stable nodes is crucial for understanding long-term behavior in dynamic systems, giving insights into how the system can be maintained or controlled.
Other exercises in this chapter
Problem 11
From \(x\left(10-x-\frac{1}{2} y\right)=0\) we obtain \(x=0\) or \(x+\frac{1}{2} y=10 .\) Likewise \(y(16-y-x)=0\) implies that \(y=0\) or \(x+y=16 .\) We there
View solution Problem 12
Critical points are (1,0) and \((-1,0),\) and $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{rr} 2 x & -2 y \\ 0 & 2 \end{array}\right)$$ At \(\mathbf{X}
View solution Problem 13
We have $$\mathbf{X}(0)=c_{1}\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right)+c_{2}\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)+c_{3}\left(\begin{
View solution Problem 13
\(y^{\prime}=2 x y-y=y(2 x-1) .\) Therefore if \((x, y)\) is a critical point, either \(x=1 / 2\) or \(y=0 .\) The case \(x=1 / 2\) and \(y-x^{2}+2=0\) implies
View solution