Let P(t) be the number of individuals in a particular population at time t. With unlimited resources, suppose the change in a population is affected only by the rate k of the difference in births and deaths per unit time. Then, over a change in time Dt,

Taking limits gives: dP/dt = kP

Equilibrium solution:  values where P' = 0, that is where there is no rate of change.  k P = 0 implies P=0 is the only equilibrium solution.
Initial condition:  P(0) is the value at time zero...generally assumed as a parameter...p₀
Initial Value Problem:  y' = f(t,y), y(0) = y₀

Here the rate of growth changes as the population increases.

Equilibrium: P'=0 implies P=0 or P=M.  Draw graph...notice P=0 is unstable and P=M is stable

Let A(t) be the amount of money in a savings account at time t, growing at a yearly rate of r% compounded continuously.  Then, over a change in time Dt, A(t + Dt) - A(t) = DA = r A(t) Dt. Taking limits gives: dA/dt = rA which yields an exponential growth model with solution A(t) = A(0)e^rt.

Now, suppose after T years, we want to start withdrawing M dollars at the beginning of each year.
Then, our differential equation model becomes
 

	={	rA, for 0 < t < T
dA/dt
		rA - M, for t < T

Notice, we have to solve two differential equations, the first starting with some given initial deposit A(0) and then use this model to create a value for A(T) which becomes the initial value for the second differential equation starting at time T.

Note: The use of graphical differential equations software is very useful in doing direction fields for complicated functions since the amount of numerical calculation and then graphical rendering is quite tedious if not downright impossible.

Special Cases:

y' = f(t) implies isoclines are vertical
y' = f(y) implies isoclines are horizontal

Numerical Techniques:

Given an IVP, a numerical solution is a collection of points (t_k, w_k) where w_k ~ y(t_k).

Euler's method: Use definition of derivative and drop the limit with h = t_k+1 - t_k. This suggests the iteration:

t_k = h k
w₀ = y₀
w_k+1 = w_k + h f(t_k,w_k)

w₀ = y₀
w_k+1 = w_k + h k w_k = (1 + h k)w_k

Remark: Rest points can be classified according to how solution curves act near these points. As the independent variable tends to get large, if solution curves stay close the equlibrum is called stable. Else, it is called unstable.

Classification of equilibrium solutions:

Stable = sink
Unstable = source
Neither = node

If f'(y₀)<0, then f is decreasing at y₀.  However, f(y₀)=0 implies y'<0 above y₀ and y'>0 below y₀.  Therefore, y₀ is a sink.
If f'(y₀)>0, then f is increasing at y₀.  However, f(y₀)=0 implies y'>0 above y₀ and y'<0 below y₀.  Therefore, y₀ is a source.

Modified Logistic Model

where N = maximum population/carrying capacity,
M = minimum sustainable population/sparsity constant.

Homework:

• Consider the differential equation y' = 2t+y.
• Sketch the slope field.
• Using this slope field, discuss the nature of solutions over time.
• Use Euler's method with h=1/4 to determine a numerical solution on [0,2] starting with y(0)=2.
• Consider the autonomous DE y' = y³ - 2y² + y.
• Determine all equilibria and classify each as a sink, source or node.
• Sketch the slope field.
• Using this slope field, sketch the graphs of solutions with the initial conditions y(0)=0, y(0)=1/2, y(1)=1/2, y(0)=3/2 and y(0) = -1/2.

For an autonomous DE, a bifurcation value of the parameter is a value for which the structure of the phase lines change. Collecting the phase lines for various values of the parameter give a bifurcation diagram.

Bifurcation Result: Consider the autonomous DE y'=f_m(y), where the partials of f_m(y) are continuous. For parameter m = m₀, consider the equilibrium y=y₀.

Source: If f '_m0(y₀)>0, then the DE y'=f_m(y) has a source for all m near m₀. Hence, no bifurcation.
Sink: If f '_m0(y₀)<0, then the DE y'=f_m(y) has a sink for all m near m₀. Hence, no bifurcation.
Bifurcation:  If the family of solutions has a bifurcation, then f_m0(y₀)=0 and f '_m0(y₀)=0, for some m = m₀ and y=y₀.

where C is Harvested from the Population each time period. Let C be the parameter of interest, with C>0.

So, our model can be written P' = f_C(P)

Case 1: C=0...leads to standard Logistic Model (no Harvesting) and solutions approach P=N.
Case 2: C>0...note f_C(P) is a concave downward quadratic which shifts downward as C increases.

For 0 < C < k N/4, the sink continues to exist but at a smaller location.
For C > k N/4, the sink ceases to exist and the Population moves toward extinction....that is, we harvested too much.

• Consider the autonomous differential equation y' = y⁶ - 2y³ + m.
• Identify the bifurcation values of m.  (Hint:  let x = y³ and use the quadratic formula.)
• Describe the bifurcations that take place as m increases.
• Create a bifurcation diagram.
• Consider the autonomous differential equation y' = y³ + m y².
• How many equilibrium points are near y=0 is m is just slightly greater than zero.
• How many equilibrium points are near y=0 is m is just slightly less than zero.
• Discuss the sensitivity of solutions to small changes in m.

Let the population of Prey be denoted by R(t) (for rabbits) and the population of Predators by F(t) (for foxes)

Consider the following assumptions:

In the absence of Predators, the Prey will grow unrestricted.  Hence, R' must have a term of the form a R

Predators eat Prey at a rate proportional to how often they interact.  This can be modeled with a term of the form -b R F.

In the absence of Prey, Predators will die off.  Hence, F' must have a term of the form - c F.

The growth rate of Predators is proportional to the number of Prey eaten by the Predator.  As in (2), this yields a term d R F.

The system is coupled because the rates R' and F' both depend upon R and F.

The system is called autonomous since the independent variable t does not appear.

The Phase Plane:  Analogous the the phase line for single autonomous differential equations, the phase plane demonstrates the nature of solutions for a system of two autonomous differential equations.  Indeed, given the autonomous system x' = f(x,y) and y' = g(x,y), for each time t, graph the point (x(t), y(t)).  Graphing several solution curves with differing starting values (x(0), y(0)) gives a phase portrait.  Equilibrium points are locations where the solutions are constant.

Application: Harmonic Oscillator

Newton's Law of Motion: F = ma = m y''

Hooke's Law: The force exerted by a spring is proportional to the distance the spring is stretched. F = k y

Shock Absorber: The force exerted by a shock absorber (dashpot) is proportional to its velocity. F = c y'

Automobile Suspension: Assume a wheel on an automobile is suspended (ie, off the ground) and the equilibrium position is given the value y=0.
At the equilibrium, the spring is not moving and so mg=kl, where l is the distance gravity pulls the spring from its natural length. The forces acting on the suspended suspension are: gravity=mg pulling down and opposed by the force in the spring=k(y+l) and (if the wheel is moving) by the force in the shock absorber = cy'. So,

my'' = mg - k(y+l) - cy'

Hence, we have

my'' = kl - k(y+l) - cy' = ky - cy', or

my'' + cy' + ky = 0.

If we also apply an external force f(t) to the suspension (ie, we pull on it), we obtain

my'' + cy' + ky = f(t),

where m = mass of object pulling string, k = spring constant, c = coefficient of damping.

• y' = z
• z' = -cy - bz + f(t)

Euler's Method for Autonomous Systems: Consider the vector field. Starting at some point Y₀ (IC) in this field, follow the direction specified by the corresponding vector F(Y₀) for a distance t. Take this point as a new starting point and do it again. This generates an approximate solution curve for the system x' = f(x,y), y'=g(x,y) using the iteration:

Van der Pol Equation:

Model for Swaying Skyscraper:

Application:  Pendulums
Let a mass m be suspended by a (nonflexible) rod of length r.  Let j be the angle the rod is from vertical and let x be the arc length from vertical, x=x(j)=r*j.  Notice, if j is known, then the precise location of the mass is known.  The amplitude is the maximum j and the period is the time required for the pendulum to go through a complete cycle.  Gravity will not affect the motion of the pendulum if it is in the middle (at equilibrium) but will affect it otherwise.  However, for our purposes, we only are concerned with the motion  of the mass in the direction of its tangent, where the force in the tangent due to gravity is (-mg*sin(j)) where the negative sign indicates a restoring force always directed toward the equilibrium.

- Assume the rod is massless and the mass m to be concentrated at one point.  Neglect buoyancy, friction in the hinge and fluid resistance (ie. the pendulum is in a vacuum).  So, we have   m x'' = ma = F = -mg sin(j), or   x'' + g sin(j) = 0.
But, arc length x =r*j implies x'' = r j'' which yields the nonlinear model  r j'' + g sin(j) = 0.  We can "linearize" this by using the Maclaurin expansion of sin(j) = j - (j)3/3! + ...  or approximately we replace the sin(j) term with j, provided the mass moves through a reasonably small angle j.  This yields the linear approximate model r j'' + g j = 0.

- Assume now that the pendulum is damped (air or fluid resistance).
This yields mx'' = -mg sin(j) - bx', or as above we obtain mr*j'' + brj' + mg*j = 0.

- Assume we add an external force f(t).
This yields mr*j'' + brj' + mg*j = f(t).

Application:  Automobile Suspension Systems Revisited
- Now assume the mass m is the portion of the car's mass supported by one tire.  The same equation developed earlier still holds.
- Determine the spring and shock absorber constants to give a good, smooth, safe ride.
- Model 1:  No spring or shock.  No good.
- Model 2:  Using a spring but no shock.  Oscillatory results or Simple Harmonic motion.
- Model 3:  Both spring and shock used.  Damped results.
- Model 4:  Add an external force.  Since we have not yet solved (LNC) yet, we must wait till later

Homework:  page 394 #3, 6; page 418 #1, 5