It is convenient to think that solutions of differential equations consist of a family of functions (just like indefinite integrals ).

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1 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?

2 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 2 nd Law: F = ma = m(dv/dt) Force of gravity: F = mg net force with m = mass in kg downward force g = acceleration of gravity, g = 9.8 m/s 2 Here F has units kg m/s 2. 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/s 2. The sum of forces = F = mg - v and F = ma = m(dv/dt), so we have equation is gamma

3 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/sec 2 in English units (Do not use 32.2 for this course!) g = 9.8 m/sec 2 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; dv 0 dt

4 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 2 nd Law: F = ma = m(dv/dt) Force of gravity: F = -mg net force with m = mass in kg downward force g = acceleration of gravity, g = 9.8 m/s 2 Here F has units kg m/s 2. 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/s 2. The sum of forces = F = -mg - v and F = ma = m(dv/dt), so we have equation dv m mg v dt

5 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 e kt is ke kt so y(t) = e kt. Lets consider a special case; take k = 1, then the ODE is y'(t) = y(t). One solution is y(t) = e t, but there is a family of solutions of the form y(t) = Ce t 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) = Ce t to the ODE y'(t) = y(t) is called the general solution of the ODE.

6 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.

7 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.

8 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. 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. Slope field.

9 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 (t 0, y 0 ) then we say we have an initial value problem which is designated by y'(t) = 3y(t) ODE y(t 0 ) = y 0 initial condition for our example.

10 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.

11 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

12 The direction field for over time interval 0 to 30. Equilibrium solution This was generated from an applet that sketches direction fields.

13 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.

14 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 What is this? 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

15 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).

16 Investigate our models for a pattern: Velocity of a falling object. m dv dt mg v For example let g = 9.8 m/s 2, 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 selfgoverning, 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 3 rd ed.

17 y'(t) = 3y(t), Definition: Any constant solutions of an ODE are called equilibrium solutions. The derivative = 0 at an equilibrium solution; the graph of such solutions is horizontal. 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. Set and solve for v. Equilibrium solution for y'(t) = 3y(t) is y = 0. Equilibrium solution for is p = 900. Set and solve for p. 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 2 nd Ed.

18 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.

19 The direction field for over time interval 0 to 30. Equilibrium solution

20 Example: Determine the equilibrium solution of the DE dy = 3y - 6 dt and 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. First note that the second derivative is y'' = 3y' = 9y 18. Case y = 1 y'= 3 6 < 0 y'' = 9 18 < 0 decreasing concave down Case y = 3 y'= 9 6 > 0 y'' = > 0 increasing concave up

21 Example: Determine a DE of the form dy = ay +b dt an equilibrium solution y = 4 as t. so that all solutions converge to 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. The equilibrium solution of So b/a must equal 4. Try a = 1 and b = -4 Try a = -1 and b = 4 dy = ay +b dt dy = y - 4 dt dy = 4 - y dt is y = -b/a y > 4 has y' > 0, so this choice doesn t work. y > 4 has y' < 0, and y < 4 has y' > 0, so this works.

22 Example of direction field for a non-autonomous ODE. This is the slope field of the DE y over a square of 6 units on a side centered at the origin. t From the line segments we can see where a particular solution curve is increasing, decreasing, concave up, or concave down

23 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. y t