Review of Lecture 5. F = GMm r 2. = m dv dt Expressed in terms of altitude x = r R, we have. mv dv dx = GMm. (R + x) 2. Max altitude. 2GM v 2 0 R.

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1 Review of Lecture 5 Models could involve just one or two equations (e.g. orbit calculation), or hundreds of equations (as in climate modeling). To model a vertical cannon shot: F = GMm r 2 = m dv dt Expressed in terms of altitude x = r R, we have Max altitude Escape velocit mv dv dx = GMm (R + x) 2 ξ = v 2 0 R2 2GM v 2 0 R. v escape = 2GM R

2 Autonomous equations The general first-order equation is d dt = F (, t) If F (t, ) actuall depends onl on we have d dt = f (). Such an equation is called autonomous

3 Autonomous equations The general first-order equation is d dt = F (, t) If F (t, ) actuall depends onl on we have d dt = f (). Such an equation is called autonomous For example the simplest equations for growth or deca: = k Growth with k > 0, deca with k < 0, since = Ae kt

4 Modeling population growth With a constant birth rate and constant death rate, we get for the population, = r. Here r is the difference of the birth rate and the death rate. With r > 0 the model predicts exponential growth. In a finite environment, that can t be right for long.

5 The logistic equation We need a revised model that allows to be bounded when grows large. One such model is the logistic equation: ( = r 1 ) K For small it s almost = r. For large will be negative, so will decrease.

6 The logistic equation We need a revised model that allows to be bounded when grows large. One such model is the logistic equation: ( = r 1 ) K For small it s almost = r. For large will be negative, so will decrease. Is there an equilibrium solution?

7 Is there an equilibrium solution? ( = r 1 ) K To find an equilibrium solution, set = 0: ( 0 = r 1 ) K There are two equilibrium solutions: = 0 and = K.

8 Is there an equilibrium solution? ( = r 1 ) K To find an equilibrium solution, set = 0: ( 0 = r 1 ) K There are two equilibrium solutions: = 0 and = K. K is called the carring capacit of the environment.

9 Solutions can t cross or touch an equilibrium solution, because through each point (t, ) there is a unique solution. So if another solution touched or crossed an equilibrium solution, it would have to coincide with that solution. As engineers ou ll probabl take that on authorit or faith. The curious or skeptical can read section 2.8 of the text

10 Graphical investigation In class we will graph the solutions and see that all solutions approach the equilibrium solution = K for large t. Solutions with 0 < K increase Solutions with > 0 decrease If ou are viewing these slides at another time, refer to pp , and use MathXpert Grapher or other software to graph the solutions.

11 Concavit Recall from calculus that the second derivative determines the concavit (upwards or downwards), and the inflection points (where = 0). We can do some of this analsis for an autonomous equation = f (). = d dt

12 Concavit Recall from calculus that the second derivative determines the concavit (upwards or downwards), and the inflection points (where = 0). We can do some of this analsis for an autonomous equation = f (). = d dt = d dt f ()

13 Concavit Recall from calculus that the second derivative determines the concavit (upwards or downwards), and the inflection points (where = 0). We can do some of this analsis for an autonomous equation = f (). = d dt = d dt f () = f () d dt

14 After-the-lecture viewers, see pages to replace graphs done in class. Concavit Recall from calculus that the second derivative determines the concavit (upwards or downwards), and the inflection points (where = 0). We can do some of this analsis for an autonomous equation = f (). = d dt = d dt f () = f () d dt = f ()f () Thus the signs of f and f tell us about concavit

15 The phase line If we think of t as time and onl use one space dimension, then our graph becomes a moving point on a line. It is customar to draw that line verticall. The direction field from the 2-d graph now becomes just an arrow up or down at ever point on the line. This line, with arrows, is the phase line. In our case, the arrows point up at points below = K, and down at points above = K. at = K the arrow has zero length. That is a critical point. There is another critical point at = 0, so technicall the arrow doesn t point up there, but it does point up at points just below or above there At the critical point = K the arrows do change direction. After-lecture viewers, see Fig

16 Determining qualitative behavior We can determine the qualitative behavior of the solutions from the graph of f alone. We did not need a computer or an explicit solution, although in class a computer was used first. We could have started with f, found the critical points and the phase line, sketched the direction field and solutions using the concavit information. This procedure was illustrated at the whiteboard.

17 Solving the logistic equation Since solutions can t cross equilibrium solutions, if we have a non-equilibrium solution then is never equal to 0 or K, for an value of t. Therefore we won t get a zero denominator when we separate the variables: rdt = d (1 /K)

18 Solving the logistic equation rdt = rt = d (1 /K) d (1 /K)

19 Solving the logistic equation rdt = rt = rt = d (1 /K) d (1 /K) 1 + 1/K 1 /K d

20 Solving the logistic equation d rdt = (1 /K) d rt = (1 /K) 1 rt = + 1/K 1 /K d rt + c = ln ln 1 K Can we drop the absolute value signs?

21 Solving the logistic equation d rdt = (1 /K) d rt = (1 /K) 1 rt = + 1/K 1 /K d rt + c = ln ln 1 K Can we drop the absolute value signs? Yes, if 0 < 0 < K, since then stas in this interval. ( rt + c = ln ln 1 ) K

22 So we need to solve this equation for : rt + c = ln ln = ln 1 /K ( 1 K )

23 So we need to solve this equation for : rt + c = ( ln ln 1 ) K = ln 1 /K e rt+c = 1 /K since e ln u = u

24 So we need to solve this equation for : rt + c = ( ln ln 1 ) K = ln 1 /K e rt+c = 1 /K since e ln u = u 1 /K = ec e rt = Ce r t with C = e c

25 So we need to solve this equation for : rt + c = ( ln ln 1 ) K = ln 1 /K e rt+c = 1 /K since e ln u = u 1 /K = ec e rt = Ce r t with C = e c It works if > K too, but then we have a negative sign on the left.

26 Initial condition determines C When t = 0, we have = 0. So we put those values into and we get 1 /K = ec e rt = Ce r t with C = e c /K = C B putting the absolute value signs in, the equation works for all 0 > 0.

27 Solving for So put that value for C in, and we get 1 /K = Ce rt = 0 e rt 1 0 /K

28 Solving for So put that value for C in, and we get 1 /K = Ce rt = 0 e rt 1 0 /K But since 1 /K has the same sign as 1 0 /K we can take the absolute value signs off again. 1 /K = 0 e rt 1 0 /K

29 Solving for So put that value for C in, and we get 1 /K = Ce rt = 0 e rt 1 0 /K But since 1 /K has the same sign as 1 0 /K we can take the absolute value signs off again. 1 /K = 0 e rt 1 0 /K (1 0 /K) = 0 e rt (1 /K)

30 Solving for So put that value for C in, and we get 1 /K = Ce rt = 0 e rt 1 0 /K But since 1 /K has the same sign as 1 0 /K we can take the absolute value signs off again. 1 /K = 0 e rt 1 0 /K (1 0 /K) = 0 e rt (1 /K) ( 1 0 K + 0e rt ) = 0 e rt K

31 ( 1 0 K + 0e rt ) K = 0 e rt = 0 e rt 1 0 K + 0e rt K

32 ( 1 0 K + 0e rt ) K = 0 e rt = = = 0 e rt 1 0 K + 0e rt K 0 e rt 1 0 K + 0e rt K K 0 e rt K e rt

33 = K 0 e rt K e rt Divide numerator and denominator both b e rt : = K (K 0 )e rt

34 Qualitative conclusions check out? Now we can confirm our qualitative conclusions: The limit as t goes to is K. Solutions starting in some interval around K converge to K. Solutions starting near 0 diverge from 0 (well, at least positive ones)

35 Qualitative conclusions check out? Now we can confirm our qualitative conclusions: The limit as t goes to is K. Solutions starting in some interval around K converge to K. Solutions starting near 0 diverge from 0 (well, at least positive ones) We sa = K is a stable equilibrium and = 0 is an unstable equilibrium.

36 Growth with a threshold ( = r 1 ) = f () T

37 Growth with a threshold ( = r 1 ) = f () T Practice what ou learned in the first part of the lecture: Draw the graph of f. Draw the phase line. Sketch the direction field and some solutions.