Many of the equations that we have examined have a parameter, which means that we actually have a family of differential equations. For example, \begin{equation*} \frac{dx}{dt} = kx \end{equation*} has the growth rate parameter $k\text{.}$ The logistic equation \begin{equation*} \frac{dP}{dt} = kP\left( 1 - \frac{P}{N} \right) \end{equation*} has two parameters, $k$ and $N\text{.}$ In this section we will investigate how the solutions of a differential equation vary as we change the value of a parameter.

# Subsection1.7.1The Logistic Model with Harvesting Revisited¶ permalink

Recall how we modeled logistic growth in a trout pond in Example 1.24 with the equation \begin{equation*} \frac{dP}{dt} = P\left(1 - \frac{P}{200} \right). \end{equation*} If we allowed fishing in our pond at a rate of 32 fish per year, 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 $P_2 = 40$ (a source). If the population of the pond falls below 40, then the fish will die out unless the pond is restocked or fishing is banned (Figure 1.46).

Now let us see 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(1 - \frac{P}{200} \right) - 100. \end{equation*} To determine the equilibrium solutions, we must solve $$\frac{dP}{dt} = P\left(1 - \frac{P}{200} \right) - 100 = 0\label{equation-firstlook07-over-harvesting}\tag{1.29}$$ for $P\text{.}$ This last equation can be rewritten as $P^2 - 200P - 20000 = 0\text{.}$ Thus, \begin{equation*} P = \frac{200 \pm \sqrt{200^2 - 4 \cdot 20000}}{2} = 100 \pm \sqrt{-10000}, \end{equation*} which means that equation (1.29) has no real solutions and that we have no equilbrium solutions. Furthermore, $dP/dt \lt 0$ for all values of $P\text{.}$ This means that no matter how many fish are in the pond initially, the trout population will eventually die out due to over fishing (Figure 1.47).

Finally, we will let $H = 50\text{.}$ In this case, we must solve \begin{equation*} \frac{dP}{dt} = P\left(1 - \frac{P}{200} \right) - 50 = 0 \end{equation*} in order to determine any equilibrium solutions. We 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 $P \lt 100\text{,}$ we have $dP/dt \lt 0\text{,}$ and the number of fish in the pond will decrease (Figure 1.48).

To better understand what is happening, we will generalize our model. Suppose that a population with a limited carrying capacity $N$ is modeled with the logistic equation \begin{equation*} \frac{dP}{dt} = kP\left(1 - \frac{P}{N}\right). \end{equation*} If we allow harvesting at a constant rate $H\text{,}$ our model now becomes \begin{equation*} \frac{dP}{dt} = kP\left(1 - \frac{P}{N}\right) - H. \end{equation*} To analyze our model, we will first find the equilibrium solutions. If we will let \begin{equation*} f_H(P) = kP\left(1 - \frac{P}{N}\right) - H, \end{equation*} each equilibrium solution must satisfy $f_H(P) = 0$ or \begin{equation*} -k P^2 + kNP - H N = 0. \end{equation*} Therefore, our equilibrium solutions are given by \begin{equation*} P = \frac{-kN \pm \sqrt{k^2N^2 - 4kHN}}{-2k} = \frac{N}{2} \pm \sqrt{\frac{N^2}{4} - \frac{HN}{k}}. \end{equation*}

The explanation of how our model behaves lies in the discriminant, \begin{equation*} \frac{N^2}{4} - \frac{HN}{k}. \end{equation*} If \begin{equation*} \frac{N^2}{4} - \frac{HN}{k} \lt 0 \end{equation*} or, equivalently if $H \gt kN/4\text{,}$ 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 as $t \to \infty\text{.}$ In this case, the population is doomed to extinction no matter how large the initial population is. Since negative populations do not make 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}{4} - \frac{HN}{k}} \end{equation*} and \begin{equation*} P_2 = \frac{N}{2} - \sqrt{\frac{N^2}{4} - \frac{HN}{k}}. \end{equation*} The first equilibrium solution, $P_1$ is a sink, while the second, $P_2$ is a source.

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

In our example, we have a family of differential equations—one for each value of $H\text{,}$ $$\frac{dP}{dt} = P\left( 1 - \frac{P}{200}\right) - H.\label{equation-firstlook07-fishing}\tag{1.30}$$ A small change in $H$ can have a dramatic effect on how the solutions of the differential equation behave. Changing 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 two to one and then to none. This change occurs exactly at $H = 50\text{.}$ We say that a bifurcation occurs at $H = 50$ for equation (1.30).

Let us consider the equation $$\frac{dx}{dt} = x^2 - 4x + \lambda\label{equation-firstlook07-one-parameter-bifurcation}\tag{1.31}$$ as a family of differential equations indexed by the parameter $\lambda\text{.}$ If we let $f_\lambda(x) = x^2 - 4x + \lambda\text{,}$ then \begin{equation*} \frac{dx}{dt} = f_\lambda(x) \end{equation*} is a called one-parameter family 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 different phase line to examine.

For $\lambda = 0\text{,}$ the differential equation \begin{equation*} \frac{dx}{dt} = f_0(x) = x^2 - 4x = x(x-4), \end{equation*} there is a sink at $x = 0$ and a source at $x = 4$ (Figure 1.49).

For $\lambda = 4\text{,}$ the differential equation \begin{equation*} \frac{dx}{dt} = f_4(x) = x^2 - 2x + 4 = (x - 2)^2, \end{equation*} we have exactly one equilibrium solution, a node at $x = 2$ (Figure 1.50).

If $\lambda = 8\text{,}$ then the differential equation \begin{equation*} \frac{dx}{dt} = f_8(x) = x^2 - 2x + 8 \end{equation*} has no equilibrium solutions (Figure 1.51).

In fact, the number of equilibrium solutions for (1.31) changes at $\lambda = 4\text{.}$ We say that $\lambda = 4$ is a bifurcation value for the differential equation $$\frac{dx}{dt} = f_\lambda(x) = x^2 - 4x + \lambda.\label{equation-firstlook07-bifurcation-1}\tag{1.32}$$ For $\lambda \lt 4\text{,}$ we have two equilibrium solutions. \begin{equation*} x = 2 \pm \sqrt{4 - \lambda}. \end{equation*} For values of $\lambda \gt 1\text{,}$ there are no equilibrium solutions. We 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 different values of $\lambda\text{.}$ The bifurcation diagram for equation (1.32) is a parabola (Figure 1.52). We have a phase line for each value of $\lambda\text{.}$

Bifurcations for a one-parameter family of differential equations $dx/dt = f_\lambda(x)$ are, in fact, rare. Let us 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 $x$ 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 $x$ near $x_0$ with $f'_{\lambda_0}(x_0) \lt 0\text{,}$ then for all $\lambda_1$ sufficiently close to $\lambda_0\text{,}$ the differential equation \begin{equation*} \frac{dx}{dt} = f_{\lambda_1}(x) \end{equation*} must have sink at a point $x = x_1$ very close 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 can only occur when $f_{\lambda_0}(x_0) = 0$ and $f'_{\lambda_0}(x_0) = 0\text{.}$

##### Example1.53

Now consider the one-parameter family \begin{equation*} \frac{dy}{dt} = y^3 - \alpha y = y (y^2 - \alpha). \end{equation*} We will have an equilibrium solution at zero for all values of $\alpha$ and two additional equilibrium solutions at $\pm \sqrt{\alpha}$ for $\alpha \gt 0\text{.}$ This type of bifurcation is a pitch fork bifurcation (Figure 1.54).

##### Example1.55

Let us find the bifurcation values of the one-parameter family $$\frac{dy}{dt} = y(y - 2)^2 + \lambda.\label{equation-firstlook07-bifurcation-2}\tag{1.33}$$ If $g_\lambda(y) = y(y - 2)^2 + \lambda\text{,}$ then $g'_\lambda(y) = 3y^2 - 8y + 4\text{.}$ The roots of $g'_\lambda(y) = 0$ are $y = 2$ and $y = 2/3\text{.}$ In order for $\lambda$ to be a bifurcation value, we must have $g_\lambda(2) = \lambda = 0$ or \begin{equation*} g_\lambda(2/3) = \frac{32}{27} + \lambda = 0 \end{equation*} Thus, equation (1.33) has two bifurcation values, $\lambda = -32/27$ and $\lambda = 0\text{.}$ The bifurcation diagram for this one-parameter family is given in Figure 1.56.

• A one-parameter family 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 $f'_{\lambda_0}(x_0) = 0\text{.}$

##### 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. $y' = \lambda y^2 - 1$ for $\lambda \in {\mathbb R}\text{.}$
##### 2

Outbreaks of the spruce budworm have been responsible for some major deforestations in Canada and the United States. The equation \begin{equation*} x' = r \left( 1 - \frac{x}{K} \right) x - c \frac{x^2}{a + x^2} \end{equation*} has been used to describe the dynamics of spruce budworm populations, where the variable $x$ denotes the population or density of the insect [16]. One explanation that has been given for the occurrence of outbreaks is based on the multiple bifurcations that occur with this differential equation.

1. If $a = 0.01\text{,}$ $c = 1\text{,}$ and $K = 1\text{,}$ we have a family of differential equations parameterized by $r\text{,}$ \begin{equation*} x' = rx(1-x) - \frac{x^2}{0.01 + x^2}. \end{equation*} Solve the equation equation \begin{equation*} rx(1-x) - \frac{x^2}{0.01 + x^2} = 0 \end{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^2 - 8yt + 16 t^2 + 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 form \begin{equation*} \frac{du}{dt}= f(u) \end{equation*} by letting $u = y - 4t\text{.}$
2. Sketch the phase 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 $ty$-plane. Also, 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.