Qualitative analysis of differential equations: Part I

Transcription

1 Qualitative analysis of differential equations: Part I Math 12 Section 16 November 7, 216

2 Hi, I m Kelly. Cole is away. Office hours are cancelled. Cole is available by zmurchok@math.ubc.ca.

3 Today Steady-states 2. Phase-line analysis 3. Sketching solutions 4. Slope fields

4 From last time... Q1. The fuel load carried by a hummingbird is F (t). Suppose that the intake takes place at a constant rate a and fuel consumption is proportional to the amount of fuel being carried with constant b. Which differential equation describes how F (t) changes? A. df = a bt B. df = (a b)t C. df = (a b)f D. df = a bf

5 Hummingbird Example (Steady-states = new concept)3 Q2. What are the steady-states of the differential equation df = a bf A. F = a B. F = b C. F = b a D. F = a b Set df = a bf = F = a b Steady-states are constant solutions!

6 Phase line What do the solutions to df = a bf look like? When F a b, df. This means that F is increasing or decreasing! df a b F

7 Phase line What do the solutions to df = a bf look like? When F > b a, df <, so F is decreasing When F < b a, df >, so F is increasing df a b F

8 Phase line to solutions df F(t) a b a b F t The steady-state a solution! It is stable. Arrows show the flow magnitude & direction

9 Phase line to solutions Some comments: Pay attention to the axes! The steady-state is a constant solution to the differential equation. A steady-state is stable if states that are initially close enough will get closer with time. A steady-state is unstable if states that are initially very close will eventually move away from that steady-state.

10 The Logistic Equation ( dn = rn 1 N ) K What effect does changing the parameters r and K on the population size?

11 The Logistic Equation dn = rn ( 1 N K The steady-states are N = and N = K. N = is unstable (open dot) and N = K is stable (filled in dot) dn N K ) K N t

12 The Logistic Equation dn = rn ( 1 N K The steady-states are N = and N = K. N = is unstable and N = K is stable dn N K ) K N t

13 The Logistic Equation dn N K K N t N = is unstable and N = K is stable. As t, N(t) K (provided N() > ). The population size grows or shrinks to the carrying capacity, K.

14 How can we more accurately sketch the solutions to a differential equation?

15 Slope Fields Consider the differential equation dy = f(y) Compute some of the slopes for various y values and use this to sketch a slope field for this differential equation. Idea: dy is the slope of the tangent line to the solution y(t). The tangent line thus has slope equal to f(y) since dy = f(y).

16 Slope Fields: Document Camera/Board Example Example: dy = 2y Idea: dy is the slope of the tangent line to the solution y(t). The tangent line thus has slope equal to 2y since dy = 2y.

17 its tangent line at that point has a slope 2y, regardlessofthevalueoft. Thisexampleis Slope simple enough Fields: that we can state Document the following: for positive Camera values of y, theslopeispositive, for negative values of y, theslopeisnegative, andfory =the slope is zero. We provide some tabulated values of y indicating the values of the slope f(y), itssign,andwhatthis implies about the local behaviour of the solution and its direction. Then, in Figure 13.1 we combine this information to generate the direction field and the corresponding solution curves. Note that the direction of the arrows (rather than their absolute magnitude) provides the most important qualitative tendency for the slope field sketch. y f(y) =2y slope of tangent line behaviour of y direction of arrow ve decreasing ve decreasing no change in y 1 2 +ve increasing 2 4 +ve increasing Table Table of derivatives and slopes for the differential equation (13.4) in Example y 1.5 y 1.5

18 Slope Fields: Document Camera y y t -2

19 Slope Fields: Document Camera y t t

20 t Your turn... Sketch the slope field for df = a bf on the following set of axes. Then sketch some solution curves. F(t) a b

21 The Logistic Equation ( dn = rn 1 N ) K What do the solutions to the logistic equation look like?

22 The Logistic Equation: Slope field y() = y.thesuccessivevaluesofy were calculated according to 256For r = 1 and Chapter K = 1: 13. Qualitative methods for differential equations population 1.2 (a) population 1.25 (b) time time 1 Figure (a) Slope field and (b) solution curves for the logistic equation (13.9).

23 The Logistic Equation 256For r = 1 and Chapter K = 1: 13. Qualitative methods for differential equations population 1.2 (a) population 1.25 (b) time time 1 Next class: Do the solutions have an inflection point? Do the solutions y 1 = y +.5y ever cross? (1 y )h Figure (a) Slope field and (b) solution curves for the logistic equation (13.9). y() = y.thesuccessivevaluesofy were calculated according to

24 Qualitative analysis for dy = f(y): Phase-line analysis: 1. Sketch dy vs. f(y). 2. Identify steady-states on y axis, and draw arrows. Identify stability. 3. Sketch solutions. Slope field: 1. Figure out slope of tangent line for various y values. 2. Determine behaviour of y(t) (increasing vs. decreasing) 3. Plot slope field on y vs. t. 4. Sketch solutions.

25 Answers 1. B 2. C

26 Related Exam Problems 1. Consider the differential equation dy = y y3. a. Sketch the slope field corresponding to his equation in the range 2 y 2, and sketch on the same figure the graphs of the solutions that also satisfy y() = 1 and y() =.5. b. Find all steady state (i.e., constant or equilibrium) solutions of this different equation and classify their stability.