What Are the Bifurcation Values of the One-parameter Family of Differential Equations Dy/dt = a + 4?

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Many of the equations that nosotros have examined have a parameter, which means that we actually accept a family of differential equations. For example,

\begin{equation*} \frac{dx}{dt} = kx \end{equation*}

has the growth charge per unit parameter \(yard\text{.}\) The logistic equation

\begin{equation*} \frac{dP}{dt} = kP\left( 1 - \frac{P}{North} \correct) \terminate{equation*}

has two parameters, \(k\) and \(North\text{.}\) In this department nosotros will investigate how the solutions of a differential equation vary every bit we change the value of a parameter.

Subsection 1.7.ane The Logistic Model with Harvesting Revisited

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Recall how nosotros modeled logistic growth in a trout pond in Example one.3.eight with the equation

\begin{equation*} \frac{dP}{dt} = P\left(1 - \frac{P}{200} \right). \finish{equation*}

If we allowed line-fishing in our pond at a rate of 32 fish per twelvemonth, then the equation became

\begin{equation*} \frac{dP}{dt} = P\left(1 - \frac{P}{200} \right) - 32. \end{equation*}

There are two equilibrium solutions for this equation, \(P_1 = 160\) (a sink) and \(P_2 = forty\) (a source). If the population of the pond falls below xl, and so the fish will die out unless the pond is restocked or fishing is banned (Effigy ane.7.1).

Figure ane.seven.1 Harvesting with \(H = 32\)

Now permit us meet what happens when we allow more fishing in our pond, say \(H = 100\text{.}\) Our differential equation now becomes

\begin{equation*} \frac{dP}{dt} = P\left(one - \frac{P}{200} \correct) - 100. \cease{equation*}

To determine the equilibrium solutions, we must solve

\begin{equation} \frac{dP}{dt} = P\left(1 - \frac{P}{200} \right) - 100 = 0\characterization{equation-firstlook07-over-harvesting}\tag{1.7.1} \end{equation}

for \(P\text{.}\) This last equation can be rewritten as \(P^two - 200P - 20000 = 0\text{.}\) Thus,

\begin{equation*} P = \frac{200 \pm \sqrt{200^2 - four \cdot 20000}}{2} = 100 \pm \sqrt{-10000}, \end{equation*}

which means that equation (1.vii.1) has no real solutions and that we have no equilibrium solutions. Furthermore, \(dP/dt \lt 0\) for all values of \(P\text{.}\) This ways that no matter how many fish are in the swimming initially, the trout population will eventually die out due to overfishing (Figure 1.seven.2).

Figure 1.7.two Harvesting with \(H = 100\)

Finally, we volition let \(H = l\text{.}\) In this case, we must solve

\begin{equation*} \frac{dP}{dt} = P\left(1 - \frac{P}{200} \right) - fifty = 0 \end{equation*}

in gild to determine whatsoever equilibrium solutions. Nosotros now obtain a single equilibrium solution at \(P = 50\text{.}\) In fact, \(P = 100\) will be a node. For values of \(P \gt 100\) as well as values of \(P \lt 100\text{,}\) nosotros have \(dP/dt \lt 0\text{,}\) and the number of fish in the pond will decrease (Figure one.seven.three).

Figure 1.seven.three Harvesting with \(H = l\)

To better understand what is happening, we will generalize our model. Suppose that a population with a limited conveying capacity \(N\) is modeled with the logistic equation

\brainstorm{equation*} \frac{dP}{dt} = kP\left(1 - \frac{P}{N}\right). \stop{equation*}

If we allow harvesting at a constant charge per unit \(H\text{,}\) our model now becomes

\begin{equation*} \frac{dP}{dt} = kP\left(i - \frac{P}{Northward}\right) - H. \end{equation*}

To analyze our model, nosotros will first find the equilibrium solutions. If nosotros volition allow

\begin{equation*} f_H(P) = kP\left(ane - \frac{P}{N}\right) - H, \end{equation*}

each equilibrium solution must satisfy \(f_H(P) = 0\) or

\brainstorm{equation*} -k P^two + kNP - H N = 0. \finish{equation*}

Therefore, our equilibrium solutions are given by

\begin{equation*} P = \frac{-kN \pm \sqrt{k^2N^two - 4kHN}}{-2k} = \frac{N}{2} \pm \sqrt{\frac{N^2}{four} - \frac{HN}{k}}. \end{equation*}

The explanation of how our model behaves lies in the discriminant,

\begin{equation*} \frac{N^2}{iv} - \frac{HN}{k}. \end{equation*}

If

\begin{equation*} \frac{N^2}{four} - \frac{HN}{k} \lt 0 \end{equation*}

or, equivalently if \(H \gt kN/iv\text{,}\) there are no equilibrium solutions and

\begin{equation*} \frac{dP}{dt} = f_H(P) \lt 0 \end{equation*}

for all values of \(P\text{.}\) In particular, all solutions of \(dP/dt = f_H(P)\) tend towards negative infinity equally \(t \to \infty\text{.}\) In this case, the population is doomed to extinction no matter how large the initial population is. Since negative populations do non brand sense, we say that the population is extinct when \(P = 0\text{.}\)

On the other hand, if \(H \lt kN/4\text{,}\) we have equilibrium solutions at

\begin{equation*} P_1 =\frac{N}{2} + \sqrt{\frac{N^2}{iv} - \frac{HN}{k}} \finish{equation*}

and

\begin{equation*} P_2 = \frac{Due north}{2} - \sqrt{\frac{N^2}{4} - \frac{HN}{k}}. \stop{equation*}

The first equilibrium solution, \(P_1\) is a sink, while the second, \(P_2\) is a source.

Finally, if \(H = kN/4\text{,}\) and then we will have exactly one equilibrium solution at \(P = N/2\text{.}\) Although \(dP/dt \lt 0\) for all \(P \neq North/2\text{,}\) we see that \(P \to N/2\) as \(t \to \infty\) for all initial values of \(P\) greater than \(N/ii\text{.}\) For initial values of \(P\) less than \(Northward/2\text{,}\) solutions tend towards \(- \infty\) as \(t \to \infty\text{.}\) Thus, the initial population of fish must exist at least \(kN/4\text{;}\) otherwise, the fish will go extinct.

In our example, we have a family of differential equations—1 for each value of \(H\text{,}\)

\begin{equation} \frac{dP}{dt} = P\left( 1 - \frac{P}{200}\right) - H.\characterization{equation-firstlook07-angling}\tag{i.7.2} \end{equation}

A small change in \(H\) can accept a dramatic effect on how the solutions of the differential equation comport. Irresolute the value of \(H\) from \(50\) to \(50.1\) will doom the population of fish to extinction no matter what the initial population is. As we increase the value of \(H\text{,}\) the number of equilibrium solutions changes from 2 to one then to none. This change occurs exactly at \(H = fifty\text{.}\) Nosotros say that a bifurcation occurs at \(H = 50\) for equation (i.vii.2).

Subsection 1.7.2 One-Parameter Families

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Let us consider the equation

\brainstorm{equation} \frac{dx}{dt} = x^2 - 4x + \lambda\label{equation-firstlook07-ane-parameter-bifurcation}\tag{1.7.3} \end{equation}

as a family of differential equations indexed past the parameter \(\lambda\text{.}\) If nosotros let \(f_\lambda(x) = ten^2 - 4x + \lambda\text{,}\) then

\brainstorm{equation*} \frac{dx}{dt} = f_\lambda(10) \end{equation*}

is a chosen one-parameter family unit of differential equations. For each value of \(\lambda\text{,}\) we obtain an autonomous differential equation, and for each value of \(\lambda\text{,}\) we have a unlike stage line to examine.

For \(\lambda = 0\text{,}\) the differential equation

\brainstorm{equation*} \frac{dx}{dt} = f_0(ten) = ten^two - 4x = x(x-4), \finish{equation*}

at that place is a sink at \(ten = 0\) and a source at \(ten = four\) (Effigy i.seven.4).

Figure 1.7.4 \(x' = x^2 - 4x + \lambda\) for \(\lambda = 0\)

For \(\lambda = 4\text{,}\) the differential equation

\begin{equation*} \frac{dx}{dt} = f_4(x) = x^2 - 2x + iv = (10 - 2)^2, \end{equation*}

we have exactly i equilibrium solution, a node at \(x = 2\) (Figure ane.7.5).

Figure 1.7.5 \(10' = x^ii - 4x + \lambda\) for \(\lambda = four\)

If \(\lambda = eight\text{,}\) then the differential equation

\begin{equation*} \frac{dx}{dt} = f_8(ten) = x^2 - 2x + eight \terminate{equation*}

has no equilibrium solutions (Figure one.7.6).

Figure 1.vii.6 \(x' = ten^2 - 2x + \lambda\) for \(\lambda = 8\)

In fact, the number of equilibrium solutions for (one.7.3) changes at \(\lambda = 4\text{.}\) We say that \(\lambda = four\) is a bifurcation value for the differential equation

\begin{equation} \frac{dx}{dt} = f_\lambda(x) = x^2 - 4x + \lambda.\label{equation-firstlook07-bifurcation-1}\tag{1.7.4} \end{equation}

For \(\lambda \lt 4\text{,}\) nosotros have two equilibrium solutions.

\begin{equation*} ten = 2 \pm \sqrt{4 - \lambda}. \end{equation*}

For values of \(\lambda \gt four\text{,}\) there are no equilibrium solutions. We tin can record all of the information for the various values in a graph called the bifurcation diagram. The horizontal axis is \(\lambda\) and the vertical axis is \(x\text{.}\) Over each value of \(\lambda\text{,}\) we will plot the corresponding phase line. The curve in the graph represents the various equilibrium solutions for the unlike values of \(\lambda\text{.}\) The bifurcation diagram for equation (1.7.4) is a parabola (Effigy 1.7.vii). We take a phase line for each value of \(\lambda\text{.}\)

Figure one.7.seven Bifurcation diagram for \(x' = x^2 - 4x + \lambda\)

Bifurcations for a one-parameter family of differential equations \(dx/dt = f_\lambda(ten)\) are, in fact, rare. Allow the states consider a bifurcation where a sink changes to a source as we vary the parameter \(\lambda\text{.}\) Suppose that for \(\lambda = \lambda_0\text{,}\) we have a sink at \(x_0\text{.}\) Then

\begin{equation*} \frac{dx}{dt} = f_{\lambda_0}(x_0) = 0. \end{equation*}

Furthermore, the graph of \(f_{\lambda_0}(x)\) must be decreasing for \(ten\) near \(x_0\text{,}\) since \(f_{\lambda_0}(x)\) must be postive for values of \(x \lt x_0\) and negative for values of \(x \gt x_0\text{.}\) In other words, \(f'_{\lambda_0}(x) \lt 0\) for \(ten\) nigh \(x_0\) with \(f'_{\lambda_0}(x_0) \lt 0\text{,}\) and so for all \(\lambda_1\) sufficiently shut to \(\lambda_0\text{,}\) the differential equation

\brainstorm{equation*} \frac{dx}{dt} = f_{\lambda_1}(10) \end{equation*}

must accept sink at a point \(x = x_1\) very shut to \(x_0\text{.}\) A similar situation hold if \(x_0\) is a source and \(f'_{\lambda_0}(x_0) \gt 0\text{.}\) Thus, bifurcations tin can only occur when \(f_{\lambda_0}(x_0) = 0\) and \(f'_{\lambda_0}(x_0) = 0\text{.}\)

Case ane.7.viii

Now consider the one-parameter family

\begin{equation*} \frac{dy}{dt} = y^iii - \alpha y = y (y^2 - \alpha). \end{equation*}

Nosotros volition have an equilibrium solution at zilch for all values of \(\blastoff\) and two additional equilibrium solutions at \(\pm \sqrt{\alpha}\) for \(\alpha \gt 0\text{.}\) This type of bifurcation is a pitch fork bifurcation (Figure one.7.ix).

Figure 1.7.9 The bifurcation diagram for \(y' = y^3 - \alpha y\)

Example 1.seven.10

Let us find the bifurcation values of the one-parameter family

\brainstorm{equation} \frac{dy}{dt} = y(y - two)^2 + \lambda.\label{equation-firstlook07-bifurcation-2}\tag{1.7.5} \finish{equation}

If \(g_\lambda(y) = y(y - two)^ii + \lambda\text{,}\) then \(k'_\lambda(y) = 3y^2 - 8y + 4\text{.}\) The roots of \(thou'_\lambda(y) = 0\) are \(y = 2\) and \(y = 2/3\text{.}\) In order for \(\lambda\) to be a bifurcation value, we must have \(g_\lambda(two) = \lambda = 0\) or

\begin{equation*} g_\lambda(2/3) = \frac{32}{27} + \lambda = 0 \end{equation*}

Thus, equation (1.7.5) has two bifurcation values, \(\lambda = -32/27\) and \(\lambda = 0\text{.}\) The bifurcation diagram for this one-parameter family is given in Figure i.vii.11.

Effigy ane.7.11 The bifurcation diagram for \(y' = y(1 - y)^2 + \lambda\)

Subsection 1.seven.three Important Lessons

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• A 1-parameter family unit of differential equations

  \begin{equation*} \frac{dx}{dt} = f_\lambda(x) \end{equation*}

  has a bifurcation at \(\lambda = \lambda_0\) if a change in the number of equilibrium solutions occurs.
• Bifurcation diagrams are an effective way of representing the nature of the solutions of a one-parameter family of differential equations.
• Bifurcations for a one-parameter family of differential equations \(dx/dt = f_\lambda(x)\) are rare. Bifurcations occur when \(f_{\lambda_0}(x_0) = 0\) and \(f'_{\lambda_0}(x_0) = 0\text{.}\)

Subsection Exercises

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1

Describe the phase line portraits for each of the following equations and how they depend on the parameter \(\lambda\text{.}\) Draw the bifurcation diagram for each equation.

1. \(y' = \lambda y - \sin y\) for \(\lambda > 2 / \pi\text{.}\)
2. \(y' = \lambda y^2 - ane\) for \(\lambda \in {\mathbb R}\text{.}\)

2

Outbreaks of the spruce budworm have been responsible for some major deforestations in Canada and the U.s.a.. The equation

\begin{equation*} x' = r \left( 1 - \frac{x}{K} \right) x - c \frac{10^2}{a + x^2} \end{equation*}

has been used to describe the dynamics of spruce budworm populations, where the variable \(ten\) denotes the population or density of the insect [sixteen]. I explanation that has been given for the occurrence of outbreaks is based on the multiple bifurcations that occur with this differential equation.

1. If \(a = 0.01\text{,}\) \(c = ane\text{,}\) and \(K = 1\text{,}\) we accept a family of differential equations parameterized past \(r\text{,}\)

   \begin{equation*} x' = rx(1-x) - \frac{x^two}{0.01 + x^2}. \end{equation*}

   Solve the equation equation

   \begin{equation*} rx(ane-10) - \frac{x^2}{0.01 + 10^2} = 0 \finish{equation*}

   and plot the result in the \(xr\)-plane for \(0 \leq r \lt 1\text{.}\)
2. To find the bifurcation diagram for the spruce budworm equation, reflect the graph obtained in part (1) about the line \(r= x\) line.
3. Estimate the two bifurcation values from your graph. Explain what happens to the population as \(r\) increases. That is, when does an outbreak occur? What happens after an outbreak?

3

The differential equation

\begin{equation*} \frac{dy}{dt} = y - 4t + y^ii - 8yt + sixteen t^two + 4. \end{equation*}

is not autonomous, separable, or linear; however, we can solve this equation with a change of variable.

1. Transform this equation into a new differential equation of the class

   \begin{equation*} \frac{du}{dt}= f(u) \stop{equation*}

   past letting \(u = y - 4t\text{.}\)
2. Sketch the stage line for this new equation, \(u' = f(u)\text{,}\) and sketch several solutions.
3. Find the solutions of the original differential equation that correspond to the equilibrium solutions of \(u' = f(u)\text{.}\) Graph these solutions in \(ty\)-aeroplane. Too, sketch the graphs of the solutions that you plotted in part (b).
4. Solve the differential new equation and use this information to solve the original differential equation.

Subsection 1.7.4 Project—Bifurcations

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Source: http://faculty.sfasu.edu/judsontw/ode/html-20180819/firstlook07.html

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