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Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. $$ \frac{d x}{d t}=x e^{-x} $$

Short Answer

Expert verified
There is one equilibrium at \( x = 0 \), and it is unstable with an eigenvalue of 1.

Step by step solution

01

Find The Equilibria

An equilibrium point occurs when \( \frac{dx}{dt} = 0 \). For the given differential equation \( \frac{dx}{dt} = x e^{-x} \), set this equation to 0 to find the equilibria. This gives:\[x e^{-x} = 0\]The only solution for this is when \( x = 0 \), since \( e^{-x} \) is never zero.
02

Calculate the Derivative to Find the Eigenvalue

To determine stability, calculate the derivative of \( \frac{dx}{dt} = x e^{-x} \) with respect to \( x \) to find the eigenvalue. Differentiate using the product rule:\[\frac{d}{dx}(x e^{-x}) = e^{-x} + (-e^{-x})x = e^{-x} (1-x)\]The eigenvalue at the equilibrium point \( x = 0 \) is \( e^{-x} (1-x) \) evaluated at \( x = 0 \). This gives: \[e^{0} (1-0) = 1\]
03

Determine Stability of the Equilibrium

The equilibrium point \( x = 0 \) has an eigenvalue of 1, which is positive. Therefore, the equilibrium point \( x = 0 \) is unstable since any small perturbation will cause the solution to move away from this point as time progresses.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Points
Equilibrium points in differential equations are essential for understanding the behavior of a system over time. These are the points where the system remains unchanged if it starts there. In simpler terms, if the system is at an equilibrium point, it will stay there and not change unless disturbed.

For the differential equation \( \frac{dx}{dt} = x e^{-x} \), finding equilibrium points means setting the derivative to zero: \( x e^{-x} = 0 \). This might look a bit tricky, but the solution is quite straightforward:
  • Since \( e^{-x} \) is never zero for any real \( x \), the only way the equation can be zero is if \( x = 0 \).
So, the equilibrium point for our differential equation is \( x = 0 \). This is where the system's behavior will stabilize if undisturbed. By identifying the equilibrium, we set the stage for further analysis on whether this point is stable or unstable.
Stability Analysis
Once we've identified equilibrium points, it's important to determine their stability. Stability tells us whether an equilibrium will maintain its state or deviate when slightly disturbed. In our example, we found the equilibrium point at \( x = 0 \).

To determine if this point is stable or unstable, we consider small movements away from it. If such perturbations cause further deviation, the equilibrium is considered unstable. Conversely, if the system returns to the equilibrium after a slight disturbance, it's stable.

For our differential equation, performing stability analysis involves calculating the derivative of \( \frac{dx}{dt} = x e^{-x} \). By differentiating, we find the expression \( e^{-x} (1-x) \). This expression, evaluated at the equilibrium point \( x = 0 \), gives us insight into the behavior of the system. Since the derived expression evaluates to 1 (a positive value), the equilibrium is unstable. This implies any small deviation will likely cause the system to move away from \( x = 0 \).
Eigenvalues
Eigenvalues are crucial in the study of differential equations and stability analysis as they help determine the nature of equilibrium points. They are essentially numbers that provide us with insights regarding the growth or decay of solutions near an equilibrium.
  • If an eigenvalue is positive, it often signifies an unstable equilibrium, as solutions tend to move away from the equilibrium point.
  • If negative, the eigenvalue indicates stability, suggesting that solutions will return to the equilibrium after small disturbances.
To compute the eigenvalue, we revisited our differentiated expression: \( e^{-x} (1-x) \), which results from applying the derivative to \( \frac{dx}{dt} = x e^{-x} \). Evaluating this at the equilibrium point \( x = 0 \), the eigenvalue is 1.

The positive eigenvalue confirms the instability of our equilibrium point at \( x = 0 \). This means that, in practice, if the system is at equilibrium, any slight change will cause it to quickly diverge rather than settle back to equilibrium.

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Most popular questions from this chapter

Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d y}{d t}=y^{2}-1\) (a) \(y(0)=-1\), (b) \(y(0)=-1 / 2\), (c) \(y(0)=1 / 2\), (d) \(y(0)=2\).

In Problems 13 and 14 we assumed that the per colony extinction rate was proportional to \(p .\) This means that the per colony extinction rate goes to 0 for small \(p .\) This may not be realisticsubpopulations may still go extinct even if they are not competing among themselves. One way to model this is to say that the per colony extinction rate is a function \(m(p)\) of \(p\). In Problems 15 and 16 we will assume that \(m(p)=a+b p\) for some constants \(a, b>0 .\) That is, the extinction rate increases with \(p\) because of competition between subpopulations, but \(m(p)\) does not vanish as \(p \rightarrow 0\). Then our model for proportion of occupied sites must be modified to: $$ \frac{d p}{d t}=c p(1-p)-(a+b p) p $$ where \(c, a, b\) are all positive constants. Assuming that the subpopulations obey the differential Equation (8.60) and the coefficients are \(a=1, b=2\), but \(c\) is allowed to take any value: (a) Find the equilibrium values of \(p\) (your answer will depend on the unknown coefficient \(c\) ). (b) What are the conditions on \(c\) for \(p\) to have a nontrivial equilibrium, that is, an equilibrium in which \(p \in(0,1] ?\) (c) Show that if your condition from (b) is met, then the nontrivial equilibrium is also stable.

Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d y}{d x}=-y-y^{3}\) (a) \(y(0)=0\), (b) \(y(0)=1\), (c) \(y(0)=2\), (d) \(y(0)=3\).

Interactions between different subpopulations need not be competitive. In fact, different subpopulations may share resources, and the presence of many subpopulations may provide a pool of genetic diversity that helps the population of organisms to react to changing conditions. We will model cooperation between subpopulations by again assuming that the extinction rate depends on \(p\), but now \(m(p)=a-b p\), where a and \(b\) are both positive constants. So \(m(p)\) decreases as \(p\) increases. Ourmodel for the number of subpopulations then becomes: $$ \frac{d p}{d t}=c p(1-p)-(a-b p) p $$ We will analyze this model in Problems 17 and \(18 .\) (a) Find the equilibrium values of \(p\) (your answer will depend on the constant \(c\) ). You may assume \(c>1\). (b) What is the condition on \(c\) for \(p\) to have a nontrivial equilibrium (i.e., an equilibrium in which \(\hat{p} \in(0,1])\) ? (c) Show that if your condition from (b) is met, then the nontrivial equilibrium is also stable.Assume that the number of subpopulations obeys (8.61) with \(a=2, b=1\), and \(c\) some unknown (positive) constant.

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d y}{d t}=y^{2}-y $$

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