Section 1.1 Direction Fields

Key Terms/Ideas:

• Mathematical model

• Geometric behavior of solutions without solving the

model using calculus

• Graphical description using direction fields

• Equilibrium solution

• General Solution

• Initial Value Problem

• Autonomous DE

It is convenient to think that solutions of differential equations consist of a

family of functions (just like indefinite integrals ).

We are interested in the long term behavior of the solution Q(t); that is,

how does behave?

Model: Velocity of a falling object.

Formulate a differential equation describing motion of an object falling under the force

of gravity in the atmosphere near sea level.

Variables: time t, velocity v; time is in sec. and velocity in meter/s (m/s)

Newton’s 2nd Law: F = ma = m(dv/dt) net force with m = mass in kg

Force of gravity: F = mg downward force

g = acceleration of gravity, g = 9.8 m/s2

Here F has units kg·m/s2 .

The sum of forces = F = mg - v and F = ma = m(dv/dt),

so we have equation

If we also consider the force of air resistance which is proportional to velocity then we

have an additional upward force - v since velocity is going downward it is positive.

is called the drag coefficient and has units kg/s so the drag force has units kg·m/s2.

is gamma

This is a mathematical model of an object falling in the atmosphere

near sea level.

Note that the model contains the three constants m, g, and .

The constants m and depend very much on the particular object

that is falling, and they are usually different for different objects. It

is common to refer to them as parameters, since they may take on

a range of values during the course of experiments.

On the other hand, g the acceleration of gravity is a physical constant,

whose value is the same for all objects.

g = 32 ft/sec2 in English units (Do not use 32.2 for this course!)

g = 9.8 m/sec2 in metric units.

Limiting or terminal velocity is the speed when a falling object is no longer getting

faster. That is, when the rate of change of the velocity is zero;

0

dv

dt

Model: Velocity of an object thrown upward.

Formulate a differential equation describing motion of an object thrown upward using

the force of gravity in the atmosphere near sea level.

Variables: time t, velocity v; time is in sec. and velocity in meter/s (m/s)

Newton’s 2nd Law: F = ma = m(dv/dt) net force with m = mass in kg

Force of gravity: F = -mg downward force

g = acceleration of gravity, g = 9.8 m/s2

Here F has units kg·m/s2 .

If we also consider the force of air resistance which is proportional to velocity then we

have an additional downward force - v since velocity is going upward it is negative.

is called the drag coefficient and has units kg/s so the drag force has units kg·m/s2.

The sum of forces = F = -mg - v and F = ma = m(dv/dt), so

we have equation

dvm mg v

dt

Model Unlimited Population Growth.

If the population at time t is denoted by y(t), this model says the rate of change

of the population is directly proportional to the current population. This is

expressed as the y'(t) = ky(t), where k is a constant that depends on the

species under consideration.

So we want a function y(t) whose derivative is a constant times y(t). From

calculus we know that the derivative of ekt is kekt so y(t) = ekt .

Lets consider a special case; take k = 1, then the ODE is y'(t) = y(t). One

solution is y(t) = et , but there is a family of solutions of the form y(t) = Cet

where C is an arbitrary constant.

If C = 0, then y(t) = 0 (the zero function; uninteresting for populations).

This is a constant solution to the ODE y'(t) = y(t). Any constant solution to

an ODE is called an equilibrium solution, which is an important type of

solution.

The family of solutions y(t) = Cet to the ODE y'(t) = y(t) is called the

general solution of the ODE.

Let’s take a geometric look at the

solutions of the unlimited population

growth model. The ODE is y'(t) = ky(t).

So for a particular value of t the ODE

supplies the following information: at

time = t and population y, which is

considered a point in the ty-plane, the

slope of the solution is given by the

expression ky(t).

In order to get a picture

of this we need to

specify a value of k.

Suppose k = 3, then

the ODE is y'(t) = 3y(t).

Here we show a set of

points in the t-y plane.

At each point we can

draw a short line

segment to indicate the

slope of the solution to

the ODE at that point.

Note the behavior on opposite sides of the

equilibrium solution y = 0.

Diagrams like this are called slope

fields or direction fields.

At each point we can

draw a short line

segment to indicate the

slope of the solution to

the ODE at that point.

Note the behavior on opposite sides

of the equilibrium solution y = 0.

Slope field.

Here we show an approximation

to the way the slope changes at

successive points to simulate a

curve that would be a particular

solution to the DE.

Each curve is an

approximation to a particular

member of the family of

solutions of the ODE.

A particular member of the family

of solution curves is obtained by

specifying a point in the t-y plane

that is to lie on that particular

solution curve. If the point is

designated as (t0, y0) then we say

we have an initial value problem

which is designated by

y'(t) = 3y(t) ODE

y(t0) = y0 initial condition

for our example.

Model: Field Mice and Owls.

Consider a population of field mice who inhabit a certain rural area. In the absence

of predators we assume that the mouse population increases at a rate proportional

to the current population. (That is, we have unlimited growth.)

If we denote time by t and the mouse population by p(t), then the assumption

about population growth can be expressed by the equation

where the proportionality factor r is called the rate constant or growth rate. To be

specific, suppose that time is measured in months and that the rate constant r has

the value 0.5/month. Then each term in the equation has the units of mice/month.

Now let us add to the problem by supposing that several owls live in the same

neighborhood and that they kill 15 field mice per day. To incorporate this

information into the model, we must add another term to the differential equation ,

so that it becomes

Observe that the predation term is -450 rather than -15 because time is measured

in months and the monthly predation rate is needed.

Let’s find equilibrium solutions for ODE

Set 0.5p – 450 = 0 and solve for p: p = 900

p = 900 is an equilibrium solution since when we set p = 900 both the left and

right side of the ODE is zero and p = 900 is a constant solution.

Next we inspect the slope field (direction field) of ODE

The direction field for over time interval 0 to 30.

Equilibrium

solution

This was generated from an applet that sketches direction fields.

The direction field for over time interval 0 to 30.

If p < 900, then p' < 0

so solutions are

decreasing in this

region of the tp-plane.

If p > 900, then p' > 0

so solutions are

increasing in this

region of the tp-plane.

So we can say that as t →∞ solutions from both sides are diverging from the

equilibrium solution.

Another model:

Newton’s Law of Cooling/Heating

The temperature of an object changes at a rate proportional to the difference

between the temperature of the object itself and the temperature of its

surroundings (the ambient temperature).

Let Q(t) be the temperature of the object at time t. Suppose the ambient

temperature is 80°F and the rate constant is 0.03 per minute. We get

differential equation dQ= 0.03(Q - 80)

dt

Tank Problem Example

A 500 gallon tank initially contains 200 gallons of a salt solution which contains

25 pounds of salt. Water containing 2lbs of salt per gallon flows in at the rate of

5 gallons a minute (the mixture is kept well stirred) and flows out of the tank at

the same rate. Let Q(t) be the amount of salt in the tank at time t. Then the rate

of change of the salt in the tank is modeled by the DE

dQ= Input Rate of Salt - Output Rate of Salt

dt

dQ= 2lb / gal * 5gal / min - Qlb / 200gal * 5gal / min = 10 - Q / 40

dt

Q(0) = 25

What is this?

Investigating behavior of solutions to ODEs.

Here we investigate our math models without solving the differential equations.

1. We look at the equations from a geometrical point of view and use calculus to

describe the behavior of solutions. This particularly effective if we are interested

in the long term behavior; that is, the behavior of solutions as t →∞.

2. Graphically we can generate a direction field (also called a slope field).

A slope field (or direction field) is a graphical representation of the solutions of

a first-order differential equation. It is achieved without solving the differential

equation analytically, and thus it is useful. The representation may be used to

qualitatively visualize solutions, or to numerically approximate them.

(From Wikipedia.)

Graphically a direction field consists of an array of short line segments drawn at

various points in the ty-plane showing the slope of the solution curve for the

DE. Because the direction field gives the “flow of solutions”, it assists in the

drawing of any particular solution (once an initial condition is specified).

Investigate our models for a pattern:

Velocity of a falling object.

vmg

dt

dvm

For example let g = 9.8 m/s2, m = 10 kg, and the drag constant = 2 kg/s

then after a bit of algebra the ODE is

Unlimited Population Growth. y'(t) = ky(t), for example y'(t) = 3y(t),

The Mice and Owl model.

Each of the ODEs has the expression for the derivative that depends only on the

dependent variables, v, y, and p respectively. The independent variable t does not

appear explicitly in the formula for the derivative. So the ODEs have general form

ODEs of this form are called autonomous DEs.

If t is interpreted as time, this name reflects the fact that such equations are self-

governing, in the sense that the derivative y' is steered by a function f determined

solely by the current state (or value) y, and not by any external controller watching

a clock. (That is, yꞌ does not depend on t.) Ref: Nagel, Saff, & Snider 3rd ed.

Definition: Any constant solutions of an ODE are called equilibrium solutions.

The equilibrium solutions of an autonomous DE are the zeros of f(y). (There can

be more than one equilibrium solution.)

Equilibrium solution for is v = 49.

Equilibrium solution for is p = 900.

Equilibrium solutions are often important in applications. In addition, when a DE

has an equilibrium solution it is helpful in describing the behavior of other

solutions.

Ref: Campbell 2nd Ed.

Set and solve for v.

Set

and solve for p.

The derivative = 0 at an equilibrium solution; the graph of such solutions is horizontal.

y'(t) = 3y(t),

Equilibrium solution for y'(t) = 3y(t) is y = 0.

Case: falling body model. Describe the solutions of

Recall that this DE has equilibrium solution v = 49.

We investigate the solutions on either side of the equilibrium solution.

If v < 49, then v' > 0 so solutions are increasing in this region of the tv-plane.

If v > 49, then v' < 0 so solutions are decreasing in this region of the tv-plane.

So we can say that as t →∞

solutions from both sides are

converging to the equilibrium

solution v = 49.

The direction field for over time interval 0 to 30.

Equilibrium

solution

Example: Determine the equilibrium solution of the DE

dy= 3y - 6

dtand the behavior of solutions on either side of the equilibrium solution.

Setting 3y – 6 = 0 we find that the equilibrium solution is y = 2.

Consider behavior of the solutions when y = 1 and when y = 3.

Case y = 1

First note that the second derivative is y'' = 3y' = 9y – 18.

y'= 3 – 6 < 0 y'' = 9 – 18 < 0

decreasing concave down

Case y = 3 y'= 9 – 6 > 0 y'' = 27 – 18 > 0

increasing concave up

Determine a DE of the form so that all solutions converge to

an equilibrium solution y = 4 as t →∞.

dy= ay +b

dt

We need to choose a and b so that y = 4 is an equilibrium solution and that

for y > 4, dy/dt < 0 and for y < 4, dy/dt > 0.

Try a = 1 and b = -4

dy= ay +b

dtThe equilibrium solution of is y = -b/a

dy= y - 4

dt

y > 4 has y' > 0, so

this choice doesn’t

work.

Try a = -1 and b = 4 dy

= 4 - ydt

y > 4 has y' < 0, and

y < 4 has y' > 0, so

this works.

Example:

So –b/a must equal 4.

This is the slope field of

the DE

over a square of 6

units on a side

centered at the

origin.

From the line segments we

can see where a particular

solution curve is increasing,

decreasing, concave up, or

concave down

t

y

Example of direction field for a non-autonomous ODE.

Here are what several solution

curves look like for this DE. The

curves were approximated by

numerical methods, not

analytical (or calculus based)

methods.

To generate the solution

curve an initial condition

(that is, a point on the curve

was specified) and the

numerical method then

generated a sequence of

(approximate) points that

would lie close to the true

particular solution curve.

t

y