Modeling with Differential Equations

Transcription

1 Modeling with Differential Equations 1. Exponential Growth and Decay models. Definition. A quantity y(t) is said to have an exponential growth model if it increases at a rate proportional to the amount present. It is said to have an exponential decay model if it decreases at a rate that is proportional to the amount present. In the first case, y satisfies the differential equation dy ky ( k 0) dt = > and in the second case it satisfies dy ky ( k 0) dt = >

2 Suppose that y satisfies an exponential growth model, and we know that y(0) = y 0 (that is we know the value of y at some time when we begin keeping track - set to 0). Then y is the solution of the initial value problem dy kyy ; (0) y dt = = 0 This equation is both separable and linear. By separability we have: dy kdt y = Thus, ln y = kt+ C or y= Cekt 1 Substituting the initial conditions into the result gives y 0 = C, so that y= ye 0 kt

3 In exactly the same way we can show that the solution to the exponential decay initial value problem is dy ky; y(0) dt = = y= ye 0 kt y 0 Note in both cases that the proportionality constant k is still in the final formula. Thus to get a numerical answer we will also need to know k. This is usually accomplished by giving one additional piece of information that lets us solve for k.

4 Half life and doubling time. Suppose that a quantity y(t) is subject to exponential growth. How long will it take to double in size? This is called the doubling time, and it appears that this could depend on time or on the size of y, but we will see that neither of these is true. If the value of y is y 0 at time 0, then we want to know when the value will be 2y 0. Thus we want to solve the equation We get 2y = yekt 0 0 ln2 = kt or t= ln2 k

5 Thus the doubling time depends on the rate of growth k, but not on the initial amount present or on the time when we began to keep track of the process. In particular, if the quantity y begins with 10 units, the time it takes to get to 20 units is the same as the time it takes to get from 20 to 40 or the time it takes to get from 200 to 400.

6 Problem. A cell of E. coli divides into two cells every 20 minutes when placed in a nutrient culture. Let y(t) be the number of cells present at time t and suppose that the growth of bacteria is approximated by a continuous exponential growth model. (a) Find an initial value problem whose solution is y(t). (b) Find a formula for y(t). (b) How many cells are present after 2 hrs.? (c) How long does it take to for the number of cells to reach a million? (a) Clearly the initial value problem is dy kyy ; (0) 1 dt = =

7 Now the quantity of cells doubles every twenty minute. Using the formula for doubling time we get 20 = ln2 or k = ln2 k 20 so (b) y satisfies the equation t t t ln2 20 y= e20 = eln[2 ] = 2 20 (c) After 2 hours, we have 120 y= 2 20 = 26= 64

8 (d) To see when y will be , we solve t = 2 20 and get ln = ln(2) t or t= 20 ln ln2 minutes.

9 Suppose that a quantity y(t) is subject to exponential decay. How long will it take to half in size? This is called the half life. If the value of y is y 0 at time 0, then we want to know when the value will be (1/2y 0. Thus we want to solve the equation y0 = ye kt 2 0 We get ln 1 = ln2 = kt or t= ln2 2 k This is exactly the same formula we got for doubling time.

10 Problem. Polonium 210 is a radioactive element with a half life of 140 days. Assume that 10 milligrams of the element are placed in a lead container and that y(t) is the number of milligrams present t days later. (a) Find an initial-value problem whose solution is y(t). (b) Find a formula for y(t). (c) How many milligrams will be present after 10 weeks? (d) How long will it take for 70% of the original sample to decay? (a) Clearly the initial value problem is dy kyy ; (0) 10 dt = =

11 The formula for half life is t= ln2 k so we have 140 = ln2 or k = ln k 140 and (b) y satisfies the equation y= 10e t (c) After 10 weeks or 70 days we have y= 10e milligrams.

12 (d) We want the time in days until the amount present is 3 mg. Thus we need to solve the equation 3= 10 e t or.3= e t The solution is ln.3= t or t= ln

13 Problem. Assume the town of Grayrock had a population of in 1987 and in Assuming an exponential rate of growth, in what year will the population reach 20000? Solution. We are assuming that the population P(t) is given by a formula y= ye 0 kt Let us start counting in 1987 and take that as time 0. Then we know that y 0 = 10000, and so the formula for the population is y= 10000ekt Where t is the amount of time after the year The second piece of information is that P = in 1997, that is when t = 10.

14 Thus we have12000= 10000e10k orln(1.2) = 10 k. This leads to the result that k = ln(1.2) 10 so ln(1.2) t y= 10000e 10 The population will reach when20000= 10000e or We have ln(2) = ln(1.2) 10 t t = 10ln(2) 38 ln(1.2) This means that the population will reach in about ln(1.2) 10 t

15 2. The logistic growth model. Normally, populations do not grow exponentially forever, since they must live in a system that can support no more than a maximum of L individuals, called the carrying capacity of the system. The population tends to increase when it is below L, decrease when it is above L, and stay stable when it is in equilibrium with the system, that is when the population is L. Logically, when the population is significantly below L, the growth should be close to exponential. One model that meets all these conditions is called the logistic model or the inhibited growth model.

16 In this model, the population satisfies the Logistic Differential Equation. dy y = k 1 y ; y (0) y (1) dt = L 0 Note that (assuming k to be positive) the rate of growth is positive when y/l < 1, negative when y/l > 1, and 0 when y = L. Moreover it is nearly exponential when y is much smaller than L. We can rewrite the differential equation in (1) as dy = k ( L yy ) = k yl ( y) dt L L This equation is separable and we separate the variables, getting Ldy yl ( y ) =kdt

17 We can use the method of partial fractions on the left side, and we will see that Thus so y L y We can then get dy= kdt y L y y ln y ln L y= kt+ C or ln = kt+ C L y = ekt+ C or y kt L y = Ae or L =+ 1 Ae kt or y= L y 1+ Ae kt L y = Ae kt y

18 Now the initial condition is that y(0) = y 0. This leads to L L y y =, so that A= A y0 This leads to the final form of the solution which is y= y0l y0+ ( L y0) e kt This function has a number of forms depending on the relationship between y 0 and L.

19 Problem. Suppose that a population y(t) grows in accordance with the logistic model dy 50y 0.001y 2 dt = (a) What is the carrying capacity? (b) What is the value of k? (c) For what value of y is the population growing most rapidly? Solution. The standard form of the logistic equation is dy = k yl ( y) = ky k y2 dt L L Comparing this DE with the given one, we see that k = 50 and k/l =.001. Thus means that L = 1000k = Thus we have answered (a) and (b).

20 For (c), we see that the population grows most rapidly when the derivative is a maximum. Since dy 50y 0.001y 2 dt = we must maximize 50y 0.001y 2. This maximum occurs when 0= d 50y 0.001y 2 = y dt or y =(50)(500)=25000.

21 If we substitute these numbers into the solution for the logistic equation, we get 50000y y= 0 y (50000 ) 50t 0+ y0 e y 0 = y 0 = 75000

22 3. Blocks and Springs A block is attached to a vertical spring and allowed to settle into an equilibrium position as shown. It is then set into motion by pulling or pushing. up l l - y(t)

23 We know by Hooke s law that if the spring is stretched or compressed d units from its natural position, it resists with a force F = kd, where k is the spring constant. If m is the mass of the block, then the weight provides a downward force of - mg at all times. Thus we see that if the equilibrium position stretches the spring by an amount l, then we must have kl = mg. When the block is moving and is in position y(t), the spring provides a force of k(l y(t)). Thus the total force on the mass is k(l y(t)) mg. By Newton s second law F = ma, so we have the equation my () t = kl ( yt ()) mg= kl ky() t mg= ky() t

24 or y () t + k yt () = 0 m Although this is a second order DE, it is shown in courses on differential equations that the general solution of this equation is k k y= c cos sin 1 t + c t m 2 m This is called simple harmonic motion. If the block is positioned at y 0 initially and released from rest, the solution is easily determined by the conditions y(0) = y 0 and y (0) = 0. It is 0 cos k y= y t m

25 Two possible graphs are shown below, depending on whether the spring is initially stretched or compressed. y 0 y 0 positive T y 0 y 0 negative The quantity T= 2π = 2π km mk is called the period of vibration. It is the time required to complete one cycle. The frequency is 1/T or f = km y 0 is the amplitude. 2π

26 Problem. Suppose a block weighs w pounds and vibrates with a period of 3 s when it is pulled below the equilibrium position and released. Suppose that when the process is repeated with an additional 4 lbs of weight, then the period is 5 s. (a) Find the spring constant (b) Find w. Solution. We know the weight of the block is mg and the period of the motion is 3= 2π mk= 2π wkg The additional weight brings the total to w + 4, and so produces a motion of period 5= 2π w+ 4 kg

27 These equations tell us that 2 3 w 4 2 or k π w 2π = = kg 9g 2 The other equation tells us similarly that k = 4 π ( w+ 4) 25g 2 2 Thus 4π w = 4 π ( w+ 4) or 25 w= 9( w+ 4) 9g 25g so w=, k= π w = π w = π 4 9g