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Section1.3Geometric and Quantitative Analysis

Objectives
  • To understand that direction fields are a useful way of analyzing a differential equation from a geometric point of view, especially since not all differential equations can be solved analytically.

  • To understand that an autonomous equation is a differential equation of the form \(y' = f(y)\) and that phase lines can be used to analyze autonomous differential equations.

  • To understand equilibrium solutions to a differential equation \(y' = f(y)\text{.}\) are those solutions given by \(f(y) = 0\) for all \(y\text{.}\) In particular, an equilibrium solution is either a sink, source, or node.

If we view the differential equation \(y' = f(t,y)\) as a formula for the slope of a tangent line to a solution curve, we can approximate the graph of a solution curve. For example, if we consider the equation \(y' = t + y\text{,}\) then a solution curve will have a slope of \(2\) at the point \((1, 1)\text{.}\) We can use this information to obtain a geometric description of the solutions to the equation.

Subsection1.3.1RC Circuits

Suppose that we wish to analyze how an electric current flows through a circuit. An RC circuit is a very simple circuit might contain a voltage source, a capacitor, and a resistor (Figure1.3.1). A battery or generator is an example of a voltage source. The glowing red heating element in a toaster or an electric stove is an example of something that provides resistance in a circuit. A capacitor stores an electrical charge and can be made by separating two metal plates with an insulating material. Capacitors are used to power the electronic flashes for cameras. Current, \(I(t)\text{,}\) is the rate at which a charge flows through this circuit and is measured in amperes or amps (A). We assign a direction to the current. A current flowing in the opposite direction will be given negative values.

an electrical circuit containing a voltage source and  a resistor and a capacitor
Figure1.3.1An RC circuit

The source voltage, \(E(t)\text{,}\) is measured in volts (V). Kirchhoff's Second Law tells us that the impressed voltage in a closed circuit is equal to the sum of the voltage drops in the rest of the circuit. Thus, we need only compute the voltage drop across the resistor, \(E_R\text{,}\) and the voltage drop across the capacitor, \(E_C\text{.}\) According to Kirchhoff's Law, this is

\begin{equation*} E_R + E_C = E. \end{equation*}

Resistance, \(R\text{,}\) to the current is measured in ohms (\(\Omega\)). Ohm's Law tells us that the voltage drop across a resistor is given by

\begin{equation} E_R = IR.\label{firstlook03-equation-ohms-law}\tag{1.3.1} \end{equation}

Finally, capacitance, \(C\text{,}\) is measured in farads (F). Coulomb's Law tells us how current flows across a capacitor,

\begin{equation} I = C \frac{d E_C}{dt}. \label{firstlook03-equation-coulombs-law}\tag{1.3.2} \end{equation}

Thus, if we combine the equations(1.3.1) and(1.3.2), our equation \(E_R + E_C = E\) becomes

\begin{equation} RC \frac{d E_C}{dt} + E_C = E(t).\label{firstlook03-equation-rc-circuit}\tag{1.3.3} \end{equation}

We will now investigate how our circuit reacts under different voltage sources. For example, we might have a zero voltage source (the capacitor could still hold a charge). We could also have a constant nonzero source of voltage such as a battery or a fluctuating source of voltage such as a generator. We might even have a series of pulses of voltage where the current is periodically turned on and off. We would like to be able to understand the solutions to the equation(1.3.3) for different voltage sources \(E(t)\text{.}\) If we view the differential equation(1.3.3) as an expression for computing how fast current is flowing across the capacitor, we can analyze our circuit from a geometric point of view and can actually say a great deal about circuits without solving a differential equation.

Subsection1.3.2Direction Fields

Any differential equation

\begin{equation*} \frac{dy}{dt} = f(t, y) \end{equation*}

can be viewed as a formula for the slope of a function \(y = y(t)\text{.}\) Geometrically, the equation tells us that, at any point \((t_0, y_0)\text{,}\) the slope of a solution curve is given by \(f(t_0, y_0)\text{.}\) Suppose that our differential equation is defined on the rectangle \(R = [a, b] \times [c, d]\text{,}\) and let \(y(t)\) be a solution curve for \(y' = f(t, y)\) that passes through the point \((t_0, y_0)\text{.}\) Then the differential equation tells us the slope of this solution curve at \((t_0, y_0)\text{.}\) We can indicate this on the \((t, y)\)-plane by drawing a short line segment at the point \((t, y)\) with slope \(f(t_0, y_0)\text{.}\) Thus, we can obtain a direction field or slope field for the differential equation. A solution curve must be tangent to its direction field at every point.

For example, consider the differential equation \(y' = y^2/2 - t\text{.}\) The direction field for this equation is given in Figure1.3.2 along with several solution curves.

a direction field of slope arrows and three curves following the arrows
Figure1.3.2The direction field for \(y' = y^2/2 - t\)

Direction fields can be tedious to compute using pencil and paper, and it is often more efficient to use a computer program. The Desmos module below plots a direction field corresponding to a given first order differential equation \(\frac{dy}{dt} = f(t,y)\text{.}\) Try it yourself!

Activity1.3.1Plotting Direction Fields

Plot the direction field and the solution curves for each of the following initial conditions.

(a)

\(y' = y - 2x\text{,}\) \(y(0) = 4\text{,}\) \(y(0) = 1\text{.}\)

(b)

\(y' = y(1 + x)\text{,}\) \(y(0) = 1\text{,}\) \(y(0) = 0\text{,}\) \(y(0) = -1\text{.}\)

(c)

\(\dfrac{dx}{dt} + 2tx = x\text{,}\) \(x(0) = 5\text{.}\)

(d)

\(\dfrac{dy}{dt} = 2y(1 - y/4)\text{,}\) \(y(0) = 1\text{,}\) \(y(0) = 4\text{,}\) \(y(0) = 5\text{.}\)

(e)

\(y' = -1 - y^4\text{,}\) \(y(0) = 4\text{,}\) \(y(0) = 0\text{,}\) \(y(0) = -4\text{.}\)

In each case comment on anything that you notice about the direction field and the solutions.

Subsection1.3.3RC Circuits Revisited

Now let us return to our RC circuit and consider different functions \(E(t)\) for the differential equation

\begin{equation*} \frac{d E_C}{dt} = \frac{E(t) - E_C}{RC}, \end{equation*}

with \(R = 1\) and \(C = 1\text{.}\) First suppose that there is no voltage source in the circuit. If we let \(E(t) = 0\) for all \(t \ge 0\text{,}\) we will get the direction field of given in Figure1.3.3. The direction field agrees with our analytic solution

\begin{equation*} E_C(t) = v_0 e^{-t}, \end{equation*}

where \(v_0 = E_C(0)\text{.}\)

a direction field of slope arrows and three curves following the arrows with the curves approaching zero
Figure1.3.3The direction field for no current

If we assume that we have a nonzero constant source of voltage, \(E(t) = K\text{,}\) in our circuit such as a battery, then we obtain the separable differential equation

\begin{equation*} \frac{d E_C}{dt} = K - E_C. \end{equation*}

The direction field for this differential equation for \(K = 10\) is given in Figure1.3.4.

a direction field of slope arrows and three curves following the arrows with the curves approaching two
Figure1.3.4The direction field for a constant current

If we attach a battery to our circuit at time \(t = 0\) and then disconnect the battery at \(t = 4\text{,}\) then we obtain a different solution. For example, if

\begin{equation*} E(t) = \begin{cases} 10 & 0 \leq t \leq 4, \\ 0 & t \gt 4, \end{cases} \end{equation*}

we will obtain a different direction field (Figure1.3.5).

a direction field of slope arrows and three curves following the arrows with the curves approaching ten and then to zero once the time is four
Figure1.3.5The direction field for a single pulse

If our voltage source emits a series of pulses, say

\begin{equation*} E(t) = \begin{cases} 10 & 0 \leq t \lt 4, \\ 0 & 4 \leq t \lt 8, \\ 10 & 8 \leq t \lt 12, \\ & \vdots \end{cases} \end{equation*}

then the direction field for our differential equation is given in Figure1.3.6.

a direction field of slope arrows and a curve following the arrows with the curve decreasing to ten and then to zero and then increasing to ten and then decreasing
Figure1.3.6The direction field for a series of pulses

Finally, if we use a generator for a voltage source, the voltage source might be given by a function such as \(E(t) = \sin(\pi t /2)\text{.}\) The direction field for this circuit is given in Figure1.3.7.

a direction field of slope arrows and three curves following the arrows and eventually converging to the same oscillating curve
Figure1.3.7The direction field for an oscillating voltage

Subsection1.3.4Autonomous Differential Equations

An autonomous differential equation is one of the form

\begin{equation*} \frac{dx}{dt} = f(x). \end{equation*}

In other words, a differential equation is autonomous if the variable \(t\) does not appear on the righthand side of the equation. Since an autonomous differential equation \(dx/dt = f(x)\) only depends on the variable \(x\text{,}\) its direction field is particularly easy to graph. The slope only depends on \(x\) and is the same for all values of \(t\text{.}\)

Let us consider a trout pond that has a carrying capacity of 200 fish. Suppose that the trout population can be modeled according to the logistic equation

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

where \(t\) is the time in years. If we allow the fish to be harvested at a constant rate of 32 per year, our equation becomes

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

The direction field for this equation is given in Figure1.3.9.

a direction field of slope arrows with two  curves approaching 160 and one curve aproaching zero
Figure1.3.9Logistic growth with harvesting

One of the basic questions that we can ask of our model is whether or not we have a sustainable population in our trout pond given this harvest rate. If so, under what conditions for sustainablility?

Since

\begin{equation*} \frac{dP}{dt} = P\left(1 - \frac{P}{200} \right) - 32 = -\frac{1}{200}(P - 40)(P - 160) \end{equation*}

is an autonomous differential equation, the direction field does not depend on \(t\text{.}\) Consequently, we need only keep track of what happens on the vertical axis. We can do this with a phase line. Instead of drawing the entire direction field, we can draw a single line containing the same information (Figure1.3.10).

a vertical line with arrows approaching a sink at 160 and diverging from a source at 40
Figure1.3.10Phase line diagram

Notice that \(dP/dt = 0\) when \(P= 40\) or \(P = 160\text{.}\) Thus, the two constant solutions \(P(t) = 40\) and \(P(t) = 160\) are the same for all values of the independent variable \(t\text{.}\) We say that such a solution is an equilibrium solution. Equilibrium solutions graph as horizontal lines on the direction field. We can identify equilibrium solutions by setting the derivative of the function equal to zero. On our phase line we will represent these solutions as equilibrium points. For values of \(P\) between 40 and 160, we know that \(dP/dt \gt 0\text{.}\) Thus, any solution curve must be increasing. We denote this property on the phase line by drawing an upward pointing arrow. On the other hand, we know that \(dP/dt \lt 0\) when \(P \lt 40\) or \(P \gt 160\text{.}\) In this case any solution curve will be decreasing, and we will indicate this by a downward pointing arrow.

Let \(y' = f(y)\) and suppose that \(y = y_0\) is an equilibrium solution. We say this solution is a sink if for any solution \(y(t)\) with initial condition sufficiently close to \(y_0\text{,}\) we have

\begin{equation*} \lim_{t \to \infty} y(t) = y_0. \end{equation*}

We say that an equilibrium point is a source if all solutions that start sufficiently close to \(y_0\) tend toward \(y_0\) as \(t \to - \infty\text{.}\) An equilibrium solution that is neither a sink or a source is called a node (Figure1.3.10). When \(P= 40\text{,}\) we have a source, and when \(P = 160\text{,}\) we have a sink.

An equilibrium solution is stable if a small change in the initial conditions gives a solution which tends toward the equilibrium as the independent variable tends towards positive infinity. An equilibrium solution is unstable if a small change in the initial conditions gives a solution which veers away from the equilibrium as the independent variable tends towards positive infinity.

Consider the differential equation

\begin{equation} \frac{dy}{dt} = y^4 - 4y^2 = y^2(y + 2)(y - 2).\label{firstlook03-equation-sinks-sources-nodes}\tag{1.3.4} \end{equation}

The graph of \(f(y) = y^4 - 4y^2\) is given in Figure1.3.11. If \(y = -2\text{,}\) we have a sink. If \(y = 2\text{,}\) we have a source. Finally, if \(y = 0\text{,}\) we have a node.

a fourth degree polynomial curve with two local minimums and one local maximum
Figure1.3.11Sinks, sources, and nodes

It is easy to generate a phase line diagram for equation(1.3.4) from the graph of \(f(y) = y^2(y + 2)(y - 2)\) (Figure1.3.11). If the graph is above the \(y\)-axis, then \(y\) is increasing. If the graph is below the \(y\)-axis, then \(y\) is decreasing. Therefore, the phase line is easy to sketch (Figure1.3.12).

a vertical line with arrows diverging from a source at two and approaching a sink at negative two with zero being a node
Figure1.3.12Phase line diagram for \(y' = y^2(y + 2)(y - 2)\)
Activity1.3.2Autonomous Equations and Phase Lines

For each of the differential equations below, draw the phase line and classify each equilibrium solution as a sink, a source, or a node.

(a)

\(y' = y(y - 2)(y + 3)\text{.}\)

(b)

\(y' = y^2(y - 2)(y + 3)\text{.}\)

(c)

\(y' = \cos y\text{.}\)

(d)

\(y' = \cos^2 y\text{.}\)

In each case comment on anything that you notice about the phase line and the equilibrium solutions.

One of the reasons why autonomous equations are so important is Taylor's theorem, which tells us that any function \(f(x)\) can be approximated near a point \(x_0\) by an \(n\)th degree polynomial,

\begin{equation*} f(x) \approx f(x_0) + f'(x_0)(x - x_0) + \frac{ f''(x_0)}{2!} (x - x_0)^2 + \cdots + \frac{ f^{(n)}(x_0)}{n!} (x - x_0)^n \end{equation*}

near \(x_0\text{.}\) For example, if

\begin{equation*} \frac{dx}{dt} = f(x) = \cos(x^2 + \pi) \end{equation*}

with \(x(0) = x_0\text{,}\) then we may approximate this initial value problem near \(x_0\) with

\begin{align*} \frac{dx}{dt} & = f(x_0) + f'(x_0)(x - x_0) = \cos(x_0^2 + \pi) + 2 x _0 \sin(x_0^2 + \pi)(x - x_0)\\ x(0) & = x_0. \end{align*}

Of course, this strategy might not work very well if \(f(x_0) = \cos(x_0^2 + \pi) = 0\) or \(f'(x_0) = 2 x _0 \sin(x_0^2 + \pi) = 0\text{.}\)

Subsection1.3.5Important Lessons

  • Direction fields and phase lines are a useful way of analyzing a differential equation from a geometric point of view, especially since not all differential equations can be solved analytically.

  • An autonomous equation is a differential equation of the form \(y' = f(y)\text{.}\) We can use a phase line to analyze autonomous differential equations.

  • Equilibrium solutions to a differential equation \(y' = f(y)\) are those solutions given by \(f(y) = 0\) for all \(y\text{.}\) In this case, any solution must be constant. We can classify equilibrium solutions according to whether they are stable or unstable. In particular, an equilibrium solution is either a sink, source, or node.

Exercise1.3.13

Explain why solution curves to a differential equation cannot intersect.

Exercise1.3.14

Explain in your own words what an autonomous differential equation is.

Subsection1.3.6Exercises

For each of the differential equations in Exercise Group1.3.6.16, plot the direction field on the integer coordinates \((t,x)\) of the rectangle \(-2 \lt t \lt 2\) and \(-2 \lt x \lt 2\) by drawing a short line of the appropriate slope.

\(x' = x + t\)

\(x' = xt\)

\(x' = x^2 + t^2\)

\(x' = t + \tan(x)\)

\(x' = (x + t)/(x^2 + t^2)\)

\(x' = x - t + 1\)

Find the equilibrium solutions for each of the differential equations in Exercise Group1.3.6.712. Draw the phase line for each equation and classify each equilibrium solution as a sink, a source, or a node.

\(y' = 2y - 5\)

\(\dfrac{dx}{dt} = (x - 1)(x + 2)\)

\(\dfrac{dx}{dt} = (x^2 - 1)(x - 2)\)

\(\dfrac{dy}{dx} = \sin 2y\)

\(x' = (x^2 + 1)(x - 1)\)

\(x' = x^2 + x + 1\)

Each of the differential equations in Exercise Group1.3.6.1318 has several initial conditions specified. Sketch the solution curves that satisfy the initial conditions. Sketch your solutions for each equation on the same pair of axes.

\(y' = 2y - 5\text{,}\) \(y(0) = 4\text{,}\) \(y(0) = 2\text{,}\) \(y(0) = 0\text{.}\)

\(\dfrac{dx}{dt} = (x - 1)(x + 2)\text{,}\) \(x(0) = 2\text{,}\) \(x(0) = 1\text{,}\) \(x(0) = 0\text{.}\)

\(\dfrac{dx}{dt} = (x^2 - 1)(x - 2)\text{,}\) \(x(0) = 1\text{,}\) \(x(0) = 0\text{,}\) \(x(0) = -1\text{.}\)

\(\dfrac{dy}{dx} = \sin 2y\text{,}\) \(y(0) = 2\text{,}\) \(y(0) = 1\text{,}\) \(y(0) = 0\text{.}\)

\(x' = (x^2 + 1)(x - 1)\text{,}\) \(x(0) = 2\text{,}\) \(x(0) = 1\text{,}\) \(x(0) = 0\text{.}\)

\(x' = x^2 + x + 1\text{,}\) \(x(0) = 1\text{,}\) \(x(0) = 0\text{,}\) \(x(0) = -1\text{.}\)

Consider the differential equation \(y' = f(y)\text{,}\) where the graph of \(f(y)\) is given in Exercise Group1.3.6.1922. Draw the phase line for each equation and classify each equilibrium solution as a sink, a source, or a node.

What happens if we increase the harvest rate to 100 in Example1.3.8? What should be our strategy to maintain a viable population in the trout pond and still permit fishing?