Skip to main content

Section 7.2 Qualitative behavior of solutions to DEs

In earlier work, we have used the tangent line to the graph of a function \(f\) at a point \(a\) to approximate the values of \(f\) near \(a\text{.}\) The usefulness of this approximation is that we need to know very little about the function; armed with only the value \(f(a)\) and the derivative \(f'(a)\text{,}\) we may find the equation of the tangent line and the approximation

\begin{equation*} f(x) \approx f(a) + f'(a)(x-a)\text{.} \end{equation*}

Remember that a first-order differential equation gives us information about the derivative of an unknown function. Since the derivative at a point tells us the slope of the tangent line at this point, a differential equation gives us crucial information about the tangent lines to the graph of a solution. We will use this information about the tangent lines to create a slope field for the differential equation, which enables us to sketch solutions to initial value problems. Our aim will be to understand the solutions qualitatively. That is, we would like to understand the basic nature of solutions, such as their long-range behavior, without precisely determining the value of a solution at a particular point.

Preview Activity 7.2.1.

Let's consider the initial value problem

\begin{equation*} \frac{dy}{dt} = t - 2, \ \ y(0) = 1\text{.} \end{equation*}

  1. Use the differential equation to find the slope of the tangent line to the solution \(y(t)\) at \(t=0\text{.}\) Then use the initial value to find the equation of the tangent line at \(t=0\text{.}\) Sketch this tangent line over the interval \(-0.25 \leq t \leq 0.25\) on the axes provided in Figure 7.2.1.

    Figure 7.2.1. Grid for plotting partial tangent lines.

  2. Also shown in Figure 7.2.1 are the tangent lines to the solution \(y(t)\) at the points \(t=1, 2\text{,}\) and \(3\) (we will see how to find these later). Use the graph to measure the slope of each tangent line and verify that each agrees with the value specified by the differential equation.

  3. Using these tangent lines as a guide, sketch a graph of the solution \(y(t)\) over the interval \(0\leq t\leq 3\) so that the lines are tangent to the graph of \(y(t)\text{.}\)

  4. Use the Fundamental Theorem of Calculus to find \(y(t)\text{,}\) the solution to this initial value problem.

  5. Graph the solution you found in (d) on the axes provided, and compare it to the sketch you made using the tangent lines.

Subsection 7.2.1 Slope fields

Preview Activity 7.2.1 shows that we can sketch the solution to an initial value problem if we know an appropriate collection of tangent lines. We can use the differential equation to find the slope of the tangent line at any point of interest, and hence plot such a collection.

Let's continue looking at the differential equation \(\frac{dy}{dt} = t-2\text{.}\) If \(t=0\text{,}\) this equation says that \(dy/dt = 0-2=-2\text{.}\) Note that this value holds regardless of the value of \(y\text{.}\) We will therefore sketch tangent lines for several values of \(y\) and \(t=0\) with a slope of \(-2\text{,}\) as shown in Figure 7.2.2.

Figure 7.2.2. Tangent lines at points with \(t=0\text{.}\)
Figure 7.2.3. Adding tangent lines at points with \(t=1\text{.}\)

Let's continue in the same way: if \(t=1\text{,}\) the differential equation tells us that \(dy/dt = 1-2=-1\text{,}\) and this holds regardless of the value of \(y\text{.}\) We now sketch tangent lines for several values of \(y\) and \(t=1\) with a slope of \(-1\) in Figure 7.2.3.

Similarly, we see that when \(t=2\text{,}\) \(dy/dt = 0\) and when \(t=3\text{,}\) \(dy/dt=1\text{.}\) We may therefore add to our growing collection of tangent line plots to achieve Figure 7.2.4.

Figure 7.2.4. Adding tangent lines at points with \(t=2\) and \(t=3\text{.}\)
Figure 7.2.5. A completed slope field.

In Figure 7.2.4, we begin to see the solutions to the differential equation emerge. For the sake of even greater clarity, we add more tangent lines to provide the more complete picture shown at right in Figure 7.2.5.

Figure 7.2.5 is called a slope field for the differential equation. It allows us to sketch solutions just as we did in the preview activity. We can begin with the initial value \(y(0) = 1\) and start sketching the solution by following the tangent line. Whenever the solution passes through a point at which a tangent line is drawn, that line is tangent to the solution. This principle leads us to the sequence of images in Figure 7.2.6.

Figure 7.2.6. A sequence of images that show how to sketch the IVP solution that satisfies \(y(0)=1\text{.}\)

In fact, we can draw solutions for any initial value. Figure 7.2.7 shows solutions for several different initial values for \(y(0)\text{.}\)

Figure 7.2.7. Different solutions to \(\frac{dy}{dt} = t-2\) that correspond to different initial values.

Just as we did for the equation \(\frac{dy}{dt} = t-2\text{,}\) we can construct a slope field for any differential equation of interest. The slope field provides us with visual information about how we expect solutions to the differential equation to behave.

Activity 7.2.2.

Consider the autonomous differential equation

\begin{equation*} \frac{dy}{dt} = -\frac{1}{2}(y - 4)\text{.} \end{equation*}

  1. Make a plot of \(\frac{dy}{dt}\) versus \(y\) on the axes provided in Figure 7.2.8. Looking at the graph, for what values of \(y\) does \(y\) increase and for what values of \(y\) does \(y\) decrease?

    Figure 7.2.8. Axes for plotting \(\frac{dy}{dt}\) versus \(y\text{.}\)
    Figure 7.2.9. Axes for plotting the slope field for \(\frac{dy}{dt} = -\frac 12( y - 4)\text{.}\)

  2. Next, sketch the slope field for this differential equation on the axes provided in Figure 7.2.9.

  3. Use your work in (b) to sketch (on the same axes in Figure 7.2.9.) solutions that satisfy \(y(0) = 0\text{,}\) \(y(0) = 2\text{,}\) \(y(0) = 4\) and \(y(0) = 6\text{.}\)

  4. Verify that \(y(t) = 4 + 2e^{-t/2}\) is a solution to the given differential equation with the initial value \(y(0) = 6\text{.}\) Compare its graph to the one you sketched in (c).

  5. What is special about the solution where \(y(0) = 4\text{?}\)

Answer

  1. When \(y \lt 4\text{,}\) \(y\) is an increasing function of \(t\text{.}\) When \(y \gt 4\text{,}\) \(y\) is a decreasing function of \(t\text{.}\)

  4. \begin{equation*} \frac{dy}{dt} = 2 \left( -\frac{1}{2} e^{-t/2} \right) = -e^{-t/2} \end{equation*}

    and

    \begin{equation*} -\frac{1}{2}( y - 4 ) = -\frac{1}{2} \left( 4 + 2e^{-t/2} \right) = -e^{-t/2} \end{equation*}

    In addition, \(y(0) = 4 + 2e^0 = 6\text{.}\)

  5. A constant function.

Solution

  1. The graph below is a plot of \(\frac{dy}{dt}\) versus \(y\text{.}\) We see that when \(y \lt 4\text{,}\) \(\frac{dy}{dt} \gt 0\) and hence \(y\) is an increasing function of \(t\text{.}\) We also see that when \(y \gt 4\text{,}\) \(\frac{dy}{dt} \lt 0\) and hence \(y\) is a decreasing function of \(t\text{.}\)

  2. Next, by sampling points \((t,y)\) in the plane and plotting short lines whose slope is \(\left. \frac{dy}{dt}\right|_{(t,y)}\text{,}\) we create the following slope field for the differential equation \(\frac{dy}{dt} = -\frac{1}{2}(y - 4)\text{.}\)

  3. Finally, by choosing starting points and following the direction of the slope field, we generate plots of solutions that satisfy \(y(0) = 0\text{,}\) \(y(0) = 2\text{,}\) \(y(0) = 4\text{,}\) and \(y(0) = 6\text{,}\) as shown in the following figure.

  4. For \(y(t) = 4 + 2e^{-t/2}\text{,}\) we see that

    \begin{equation*} \frac{dy}{dt} = 2 \left( -\frac{1}{2} e^{-t/2} \right) = -e^{-t/2} \end{equation*}

    and

    \begin{equation*} -\frac{1}{2}( y - 4 ) = -\frac{1}{2} \left( 4 + 2e^{-t/2} \right) = -e^{-t/2} \end{equation*}

    In addition, \(y(0) = 4 + 2e^0 = 6\text{.}\) So \(y(t) = 4 + 2e^{-t/2}\) is a solution of the differential equation \(\frac{dy}{dt} = -\frac{1}{2}(y - 4)\) with \(y(0) = 6\text{.}\) The graph of this function is identical to the top graph in part (c).

  5. The solution where \(y(0) = 4\) is a constant function.

Subsection 7.2.2 Equilibrium solutions and stability

As our work in Activity 7.2.2 demonstrates, first-order autonomous equations may have solutions that are constant. These are simple to detect by inspecting the differential equation \(dy/dt = f(y)\text{:}\) constant solutions necessarily have a zero derivative, so \(dy/dt = 0 = f(y)\text{.}\)

For example, in Activity 7.2.2, we considered the equation \(\frac{dy}{dt} = f(y)=-\frac12(y-4)\text{.}\) Constant solutions are found by setting \(f(y) = -\frac12(y-4) = 0\text{,}\) which we immediately see implies that \(y = 4\text{.}\)

Values of \(y\) for which \(f(y) = 0\) in an autonomous differential equation \(\frac{dy}{dt} = f(y)\) are called equilibrium solutions of the differential equation.

Activity 7.2.3.

Consider the autonomous differential equation

\begin{equation*} \frac{dy}{dt} = -\frac{1}{2}y(y-4)\text{.} \end{equation*}

  1. Make a plot of \(\frac{dy}{dt}\) versus \(y\) on the axes provided in Figure 7.2.10. Looking at the graph, for what values of \(y\) does \(y\) increase and for what values of \(y\) does \(y\) decrease?

    Figure 7.2.10. Axes for plotting \(dy/dt\) vs \(y\) for \(\frac{dy}{dt} = -\frac 12 y(y-4)\text{.}\)
    Figure 7.2.11. Axes for plotting the slope field for \(\frac{dy}{dt} = -\frac 12 y(y-4)\text{.}\)

  2. Identify any equilibrium solutions of the given differential equation.

  3. Now sketch the slope field for the given differential equation on the axes provided in Figure 7.2.11.

  4. Sketch the solutions to the given differential equation that correspond to initial values \(y(0)=-1, 0, 1, \ldots, 5\text{.}\)

  5. An equilibrium solution \(\overline{y}\) is called stable if nearby solutions converge to \(\overline{y}\text{.}\) This means that if the initial condition varies slightly from \(\overline{y}\text{,}\) then \(\lim_{t\to\infty}y(t) = \overline{y}\text{.}\) Conversely, an equilibrium solution \(\overline{y}\) is called unstable if nearby solutions are pushed away from \(\overline{y}\text{.}\) Using your work above, classify the equilibrium solutions you found in (b) as either stable or unstable.

  6. Suppose that \(y(t)\) describes the population of a species of living organisms and that the initial value \(y(0)\) is positive. What can you say about the eventual fate of this population?

  7. Now consider a general autonomous differential equation of the form \(dy/dt = f(y)\text{.}\) Remember that an equilibrium solution \(\overline{y}\) satisfies \(f(\overline{y}) = 0\text{.}\) If we graph \(dy/dt = f(y)\) as a function of \(y\text{,}\) for which of the differential equations represented in Figure 7.2.12 and Figure 7.2.13 is \(\overline{y}\) a stable equilibrium and for which is \(\overline{y}\) unstable? Why?

Figure 7.2.12. Plot of \(\frac{dy}{dt}\) as a function of \(y\text{.}\)
Figure 7.2.13. Plot of \(\frac{dy}{dt}\) as a different function of \(y\text{.}\)
Answer

  1. When \(y \lt 0\) and when \(y \gt 4\text{,}\) \(y\) is a decreasing function of \(t\text{.}\) When \(0 \lt y \lt 4\text{,}\) \(y\) is a increasing function of \(t\text{.}\)

  2. \(y = 0\) and \(y = 4\text{.}\)

  5. \(y = 4\) is stable; \(y = 0\) is unstable.

  6. Tend to 4.

  7. Figure 7.2.12 is for an ustable equilibrium; Figure 7.2.13 is for a stable equilibrium.

Solution

  1. The graph below is a plot of \(\frac{dy}{dt}\) versus \(y\text{.}\) We see that when \(y \lt 0\) and when \(y \gt 4\text{,}\) \(\frac{dy}{dt} \lt 0\) and hence \(y\) is a decreasing function of \(t\text{.}\) We also see that when \(0 \lt y \lt 4\text{,}\) \(\frac{dy}{dt} \gt 0\) and hence \(y\) is a increasing function of \(t\text{.}\)

  2. The equilibrium solutions occur when \(\frac{dy}{dt} = 0\) and so the equilibrium solutions are \(y = 0\) and \(y = 4\text{.}\)

  3. Following is a slope field for the differential equation \(\frac{dy}{dt} = -\frac{1}{2}y(y - 4)\text{.}\)

  4. Below is the slope field for the differential equation \(\frac{dy}{dt} = -\frac{1}{2}y(y - 4)\) along with solutions that satisfy \(y(0) = 1\text{,}\) \(y(0) = 2\text{,}\) \(y(0) = 3\text{,}\) \(y(0) = 4\text{,}\) and \(y(0) = 5\text{.}\)

  5. The equilibrium solution \(y = 4\) is stable because trajectories nearby approach this constant solution, while the equilibrium solution \(y = 0\) is unstable because trajectories nearby veer away from this constant solution.

  6. The population will eventually tend to 4.

  7. Figure 7.2.12 is for an ustable equilibrium. This is because if \(y\) is slightly less than \(\bar{y}\text{,}\) then \(\frac{dy}{dt} \lt 0\) and the function \(y(t)\) will be decreasing. In addition, if \(y\) is slightly greater than \(\bar{y}\text{,}\) then \(\frac{dy}{dt} \gt 0\) and the function \(y(t)\) will be increasing. In both situations, \(y(t)\) is “moving away from” \(\bar{y}\text{.}\) Figure 7.2.13 is for a stable equilibrium. This is because if \(y\) is slightly less than \(\bar{y}\text{,}\) then \(\frac{dy}{dt} \gt 0\) and the function \(y(t)\) will be increasing. In addition, if \(y\) is slightly greater than \(\bar{y}\text{,}\) then \(\frac{dy}{dt} \lt 0\) and the function \(y(t)\) will be decreasing. In both situations, \(y(t)\) is “moving toward” \(\bar{y}\text{.}\)

Subsection 7.2.3 Summary

  • A slope field is a plot created by graphing the tangent lines of many different solutions to a differential equation.

  • Once we have a slope field, we may sketch the graph of solutions by drawing a curve that is always tangent to the lines in the slope field.

  • Autonomous differential equations sometimes have constant solutions that we call equilibrium solutions. These may be classified as stable or unstable, depending on the behavior of nearby solutions.

Exercises 7.2.4 Exercises

1. Graphing equilibrium solutions.
2. Sketching solution curves.
3. Matching equations with direction fields.
4. Describing equilibrium solutions.
5.

Consider the differential equation

\begin{equation*} \frac{dy}{dt} = t-y\text{.} \end{equation*}

  1. Sketch a slope field on the axes at right.

  2. Sketch the solutions whose initial values are \(y(0)= -4, -3, \ldots, 4\text{.}\)

  3. What do your sketches suggest is the solution whose initial value is \(y(0) = -1\text{?}\) Verify that this is indeed the solution to this initial value problem.

  4. By considering the differential equation and the graphs you have sketched, what is the relationship between \(t\) and \(y\) at a point where a solution has a local minimum?

Answer

  2. Sketch curves through appropriate points in the slope field above.

  3. \(y(t) = t-1\text{.}\)

  4. \(t\) and \(y\) are equal.

Solution

  1. Sketching the slopes at individual points by computing \(\frac{dy}{dt}\vert_{(t,y)}\) at various values of \((t,y)\text{,}\) we see the following slope field.

  2. By following the slope field so that we pass through the points \((0,-4)\text{,}\) \((0,-3)\text{,}\) \(\ldots\text{,}\) \((0,4)\text{,}\) we see the collection of curves in the figure above.

  3. It looks like the solution that passes through \((0,-1)\) is a straight line with slope \(m = 1\text{,}\) and thus has the formula \(y(t) = t-1\text{.}\) To verify this function is a solution, we compute both \(\frac{dy}{dt}\) and \(t-y\text{.}\) Doing so, we see

    \begin{equation*} \frac{dy}{dt} = 1 \end{equation*}

    and

    \begin{equation*} t-y = t - (t-1) = 1\text{.} \end{equation*}

    Thus, \(y(t) = t-1\) is indeed a solution to the differential equation.

  4. From both the differential equation and the graphs we have sketched, it appears that \(t\) and \(y\) are equal at any point where a solution has a local minimum. This is because when \(t = y\text{,}\) \(\frac{dy}{dt} = t - y = 0\text{,}\) and having the derivative be zero is required for any local minimum on a differentiable function.

6.

Consider the situation from problem 2 of Section 7.1: Suppose that the population of a particular species is described by the function \(P(t)\text{,}\) where \(P\) is expressed in millions. Suppose further that the population's rate of change is governed by the differential equation

\begin{equation*} \frac{dP}{dt} = f(P) \end{equation*}

where \(f(P)\) is the function graphed below.

  1. Sketch a slope field for this differential equation. You do not have enough information to determine the actual slopes, but you should have enough information to determine where slopes are positive, negative, zero, large, or small, and hence determine the qualitative behavior of solutions.

  2. Sketch some solutions to this differential equation when the initial population \(P(0) \gt 0\text{.}\)

  3. Identify any equilibrium solutions to the differential equation and classify them as stable or unstable.

  4. If \(P(0) \gt 1\text{,}\) what is the eventual fate of the species? if \(P(0) \lt 1\text{?}\)

  5. Remember that we referred to this model for population growth as “growth with a threshold.” Explain why this characterization makes sense by considering solutions whose inital value is close to 1.

Answer

  2. Any solution curve that starts with \(P(0) \gt 3\) will decrease to \(P(t) = 3\) as \(t \to \infty\text{;}\) any curve that starts with \(1 \lt P(0) \lt 3\) will increase to \(P(t) = 3\text{;}\) any curve that starts with \(0 \lt P(0) \lt 1\) will decrease to \(P(t) = 0\text{.}\)

  3. \(P = 0\text{,}\) \(P = 1\text{,}\) and \(P = 3\text{.}\) \(P = 1\) is unstable; \(P = 0\) and \(P = 3\) are stable.

  4. The population will stabilize either at the value \(P = 3\) or at \(P = 0\text{.}\)

  5. \(P(t) = 1\) is the threshold.

Solution

  1. We first observe from the given plot of \(\frac{dP}{dt}\) versus \(P\) that \(\frac{dP}{dt} = 0\) when \(P = 0\text{,}\) \(P = 1\text{,}\) and \(P = 3\text{.}\) Moreover, we see that \(\frac{dP}{dt} \gt 0\) when \(P \lt 0\) or \(1 \lt P \lt 3\text{,}\) and \(\frac{dP}{dt} \lt 0\) otherwise. The closer \(P\) is to \(0\text{,}\) \(1\text{,}\) or \(3\text{,}\) the closer \(\frac{dP}{dt}\) is to \(0\text{.}\) This leads us to sketch a direction field like the following.

  2. Following the slope field plotted above, we see that any solution curve that starts with \(P(0) \gt 3\) will decrease to \(P(t) = 3\) as \(t \to \infty\text{.}\) Any curve that starts with \(1 \lt P(0) \lt 3\) will increase to \(P(t) = 3\text{.}\) And any curve that starts with \(0 \lt P(0) \lt 1\) will decrease to \(P(t) = 0\text{.}\)

  3. The equilibrium solutions are the values of \(P\) where \(\frac{dP}{dt} = 0\text{.}\) These are \(P = 0\text{,}\) \(P = 1\text{,}\) and \(P = 3\text{.}\) The equilibrium solution \(P = 1\) is unstable, while the solutions \(P = 0\) and \(P = 3\) are stable.

  4. If \(P(0) \gt 1\text{,}\) the population's value will tend to \(P(t) = 3\) as \(t \to \infty\text{;}\) if \(P(0) \lt 1\text{,}\) the population's value will tend to \(P(t) = 0\text{.}\) Thus, the population will stabilize either way, but either at the value \(P = 3\) or at a level that corresponds to the population going extinct.

  5. We can view \(P(t) = 1\) as the threshold for this population since if there aren't at least \(P(t) = 1\) members of the population, the population cannot grow. This aligns well with the definition of the word threshold: “the magnitude or intensity that must be exceeded for a certain reaction, phenomenon, result, or condition to occur or be manifested”.

7.

The population of a species of fish in a lake is \(P(t)\) where \(P\) is measured in thousands of fish and \(t\) is measured in months. The growth of the population is described by the differential equation

\begin{equation*} \frac{dP}{dt} = f(P) = P(6-P)\text{.} \end{equation*}

  1. Sketch a graph of \(f(P) = P(6-P)\) and use it to determine the equilibrium solutions and whether they are stable or unstable. Write a complete sentence that describes the long-term behavior of the fish population.

  2. Suppose now that the owners of the lake allow fishers to remove 1000 fish from the lake every month (remember that \(P(t)\) is measured in thousands of fish). Modify the differential equation to take this into account. Sketch the new graph of \(dP/dt\) versus \(P\text{.}\) Determine the new equilibrium solutions and decide whether they are stable or unstable.

  3. Given the situation in part (b), give a description of the long-term behavior of the fish population.

  4. Suppose that fishermen remove \(h\) thousand fish per month. How is the differential equation modified?

  5. What is the largest number of fish that can be removed per month without eliminating the fish population? If fish are removed at this maximum rate, what is the eventual population of fish?

Answer

  1. A graph of \(f\) against \(P\) is given in blue in the figure below. The equilibrium solutions are \(P=0\) (unstable) and \(P=6\) (stable).

  2. \(\frac{dP}{dt} = g(P) = P(6-P)-1\) ; the equilibrium at \(P\approx 0.172\) is unstable; the equilibrium at \(P \approx 5.83\) is stable.

  3. If \(P \lt \frac{6-\sqrt{32}}{2}\text{,}\) then the fish population will die out. If \(\frac{6-\sqrt{32}}{2} \lt \text{,}\) then the fish population will approach \(\frac{6+\sqrt{32}}{2}\) thousand fish.

  4. \(\frac{dP}{dt} = g(P) = P(6-P)-h \text{;}\) equilibrium solutions \(P = \frac{6+\sqrt{36-4h}}{2}, \ \frac{6-\sqrt{36-4h}}{2} \text{.}\)

  5. \(9000\) fish; harvesting at that rate will maintain the number of fish we start with, provided it's at least \(3000\text{.}\)

Solution

  1. A graph of \(f\) against \(P\) is given in blue in the figure below. This graph and the differential equation show that the equilibrium solutions occur when \(f(P)=0\text{,}\) or when \(P=0\) or \(P=6\text{.}\) Since \(P' \lt 0\) for \(P\) slightly less than 0, the solutions with \(P \lt 0\) decrease away from the equilibrium \(P=0\text{.}\) Also, \(P'>0\) for \(P\) slightly larger than 0 means that the solutions for \(P \gt \) increase away from the equilibrium \(P=0\text{.}\) So \(P=0\) is an unstable equilibrium. If \(P\) is slightly less than \(6\) we see that \(P \gt 0\text{,}\) so these solutions increase toward the equilibrium \(P=0\text{.}\) When \(P\) is slightly larger than \(6\text{,}\) we have \(P' \lt 0\) and so these solutions decrease toward the equilibrium \(P = 6\text{.}\) We conclude that the equilibrium solution \(P=6\) is stable.

  2. Since \(P\) will decrease by \(1\) thousand fish every month through harvesting, the new differential equation is

    \begin{equation*} \frac{dP}{dt} = g(P) = P(6-P)-1\text{.} \end{equation*}

    A sketch of \(g\) against \(P\) is given in green in the figure in (a). The equilibrium solutions now occur when \(g(P) = P(6-P)-1 = 0\text{.}\) This happens when \(P^2-6P+1 = 0\text{.}\) The quadratic formula tells us that the solutions to this quadratic equation are

    \begin{equation*} P = \frac{6+\sqrt{32}}{2} \approx 5.83 \ \ \ \text{ and } \ \ \ P = \frac{6-\sqrt{32}}{2} \approx 0.172\text{.} \end{equation*}

    The same argument as in (a) shows that the equilibrium at \(P\approx 0.172\) is unstable and the equilibrium at \(P \approx 5.83\) is stable.

  3. Assuming that there are fish in the lake, so that \(P > 0\text{,}\) there are two cases to consider. If \(P \lt \frac{6-\sqrt{32}}{2}\text{,}\) then \(P' \lt 0\) and the fish population will die out. If \(\frac{6-\sqrt{32}}{2} \lt P \lt \frac{6+\sqrt{32}}{2}\text{,}\) then \(P' \gt 0\) and the fish population will increase and approach a stable equilibrium population of \(\frac{6+\sqrt{32}}{2}\) thousand fish. If \(P \gt \frac{6+\sqrt{32}}{2}\text{,}\) then \(P' \lt 0\) and the fish population will decrease and approach a stable equilibrium population of \(\frac{6+\sqrt{32}}{2}\) thousand fish.

  4. In this case the differential equation becomes

    \begin{equation*} \frac{dP}{dt} = g(P) = P(6-P)-h\text{.} \end{equation*}

    We will have equilibrium solutions when \(P^2 - 6P + h = 0\) or when

    \begin{equation*} P = \frac{6+\sqrt{36-4h}}{2} \ \ \ \text{ and } \ \ \ P = \frac{6-\sqrt{36-4h}}{2}\text{.} \end{equation*}

  5. We need a positive stable equilibrium. This will will happen as long as \(36-4h \ge 0\) or when \(h \le 9\text{.}\) So the maximum harvest is \(9000\) fish. Note that \(P'\) is negative, so we need to start with more than \(3000\) fish. Since \(P(6-P)\) has a maximum value of \(9\text{,}\) the fish will reproduce quickly enough so that harvesting \(9000\) fish will maintain a stable population.

8.

Let \(y(t)\) be the number of thousands of mice that live on a farm; assume time \(t\) is measured in years. 1 

  1. The population of the mice grows at a yearly rate that is twenty times the number of mice. Express this as a differential equation.

  2. At some point, the farmer brings \(C\) cats to the farm. The number of mice that the cats can eat in a year is

    \begin{equation*} M(y) = C\frac{y}{2+y} \end{equation*}

    thousand mice per year. Explain how this modifies the differential equation that you found in part a).

  3. Sketch a graph of the function \(M(y)\) for a single cat \(C=1\) and explain its features by looking, for instance, at the behavior of \(M(y)\) when \(y\) is small and when \(y\) is large.

  4. Suppose that \(C=1\text{.}\) Find the equilibrium solutions and determine whether they are stable or unstable. Use this to explain the long-term behavior of the mice population depending on the initial population of the mice.

  5. Suppose that \(C=60\text{.}\) Find the equilibrium solutions and determine whether they are stable or unstable. Use this to explain the long-term behavior of the mice population depending on the initial population of the mice.

  6. What is the smallest number of cats you would need to keep the mice population from growing arbitrarily large?

This problem is based on an ecological analysis presented in a research paper by C.S. Hollings: The Components of Predation as Revealed by a Study of Small Mammal Predation of the European Pine Sawfly, Canadian Entomology 91: 283-320.
Answer

  1. \(\frac{dy}{dt} = 20y\text{.}\)

  2. \(\frac{dy}{dt} = 20y - C\frac{y}{2+y}\)

  3. For positive \(y\) near \(0\text{,}\) \(M(y) = \frac{y}{2+y} \approx 0\text{;}\) for large values of \(y\text{,}\) \(M(y) = \frac{y}{2+y} \approx 1\text{.}\)

  4. The only equilibrium solution is \(y = 0\text{,}\) which is unstable.

  5. The equilibrium solutions are \(y = 0\) (stable) and \(y = 1\) (unstable).

  6. At least \(41\) cats.

Solution

  1. If we let \(y(t)\) be the population of the mice at year \(t\) in thousands and it grows at a yearly rate that is twenty times the number of mice, the governing differential equation is \(\frac{dy}{dt} = 20y\text{.}\)

  2. On their own, the mice add \(20y\) thousand mice per year to their total population. With \(C\) cats present, the cats now remove \(M(y) = C\frac{y}{2+y}\) thousand mice per year. By taking both the rates at which mice are added and removed from the population into account, we find the updated differential equation

    \begin{equation*} \frac{dy}{dt} = 20y - C\frac{y}{2+y} \end{equation*}

  3. For positive \(y\) near \(0\text{,}\) \(M(y) = \frac{y}{2+y} \approx 0\text{;}\) for large values of \(y\text{,}\) \(M(y) = \frac{y}{2+y} \approx 1\text{.}\) This means that a single cat can remove about \(1\) thousand mice per year from the population when the mice population is large.

  4. Letting \(C=1\text{,}\) we are working with the differential equation

    \begin{equation*} \frac{dy}{dt} = 20y - \frac{y}{2+y}\text{.} \end{equation*}

    To find the equilibrium solutions, we first set \(\frac{dy}{dt} = 0\text{.}\) Doing so, we find that \(20y = \frac{y}{2+y}\text{,}\) so \(40y + 20y^2 = y\text{,}\) or \(39y + 20y^2 = 0\text{.}\) Factoring, we see \(y(39y + 20) = 0\) and therefore \(y = 0\) or \(y = -\frac{20}{39}\text{.}\) Since a negative equilibrium solution doesn't make sense in the context of the problem, the only equilibrium solution to study is \(y = 0\text{.}\) If we plot \(dy/dt\) versus \(y\) for \(y \gt 0\text{,}\) we see that \(dy/dt\) is always positive, and thus the population will grow away from \(y = 0\text{,}\) making this equilibrium unstable. And this makes sense: the population of mice will grow without bound, because one cat is insufficient to keep up with a mice population growing at a rate of \(20\) times its size per year.

  5. Letting \(C=60\text{,}\) we are working with the differential equation

    \begin{equation*} \frac{dy}{dt} = 20y - \frac{60y}{2+y}\text{.} \end{equation*}

    Similar work as in (d) shows that the equilibrium solutions are \(y = 0\) or \(y = 1\text{,}\) and if we plot \(dy/dt\) versus \(y\text{,}\) we see that for \(0 \lt y \lt 1\text{,}\) \(dy/dt \lt 0\text{,}\) while for \(y \gt 1\text{,}\) \(dt/dt \gt 0\text{.}\) Thus, with \(60\) cats present, as long as there are initially fewer than \(1000\) mice, the cats will drive the mice population to \(0\text{.}\) If there are more than \(1000\) mice initially present, the mice population will grow without bound. And if there are \(1000\) mice initially present, this will be an (unstable) equilibrium.

  6. In order to keep the mice population from growing arbitrarily large, we need to bring enough cats that for certain initial mice populations, we can make \(dy/dt \le 0\text{.}\) In other words, we need enough cats that there's a positive equilibrium solution. Considering the general differential equation,

    \begin{equation*} \frac{dy}{dt} = 20y - C\frac{y}{2+y}\text{,} \end{equation*}

    we thus set \(dy/dt = 0\) and solve for \(C\text{.}\) Doing so, we find that \(0 = 20y - C\frac{y}{2+y}\text{,}\) so \(20y(y+2) = Cy\) or \(20y^2 + 40y - Cy = 0\text{.}\) Factoring, we find that

    \begin{equation*} y(20y + (40-C)) = 0\text{.} \end{equation*}

    To have a positive equilibrium solution, we require that \(40-C \lt 0\text{,}\) and therefore the farmer must bring at least \(41\) cats to the farm. Even in that circumstance, whether the cats can keep the number of mice from growing without bound will depend on the initial number of mice present.

login