CHAPTER 3. Mathematical Models and Numerical Methods Involving First-Order Equations

Transcription

1 CHAPTER 3 Mathematical Models and Numerical Methods Involving First-Order Equations 2. Compartmental Analysis The asic one-compartment system consists of a function x(t) that represents the amount of a sustance in the compartment at time t, an input rate at which the sustance enters the compartment, and an output rate at which the sustance leaves the compartment. Interpreting dx as the rate of change in the amount of the sustance in the compartment at time t, we get dx = input rate output rate. 59

2 60 3. MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS Mixing Prolems Example (A one compartment system orange juice prolem). There are 50,000 gal of orange juice in a vat with 10% concentrate, the rest water. A solution of 40% concentrate is sent to the vat at 100 gal/min. Similarly, 100 gal/min is drained from the vat, where there is constant mixing. We want to (1) express the percentage of concentrate in the vat as a function of time and (2) predict when the concentrate will reach 12%. 50,000 gal juice 10% concentrate 100 gal/min 40% concentrate 100 gal/min?% concentrate Let x(t) = the amount of concentrate in the vat. Let c(t) = the percentage of concentrate in the vat. Let V (t) = the volume of liquid in the vat. c(t) = x(t) and x(t) = c(t) V (t). V (t) ( ) Assume good mixing so that the percentage of concentrate leaving the vat is the same as that in the vat. The rate of change of the amount of concentrate in the vat changes with time and is given y dx. Thus we can find x(t) as a solution to a DE modeling the rate of change. Note. (1) amount of concentrate=(# of gal) (percentage of concentrate/gal) (2) # of gal flowing in or out=(rate of flow) (time)

3 We have, using ( ), 2. COMPARTMENTAL ANALYSIS 61 inflow rate = (100 gal/min)(.40 gal conc./gal) = 40 gal conc./min x(t) gal conc. outlow rate = (100 gal/min) = x(t) gal conc./min gal 500 Thus the IVP modeling the process is dx = 40 x(t), x(0) = dx +.002x = 40, x(0) = 5000 µ = e.002t e.002t x e.002t x = 40e.002t =) d h i e.002t x = 40e.002t =) Z e.002t x = 40 e.002t + C =) 1 e.002t x = 40 e.002t + C =).002 x(t) = Ce.002t Now x(0) = C = 5000 =) C = =) x(t) = e.002t =) c(t) =.4.3e.002t so.12 =.4.3e.002t =) t = 34.5 min. Note that lim x(t) = 0.4. t!1 Maple. See orangejuice.mw or orangejuice.pdf

4 62 3. MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS Prolem. A 200 gal tank initially contains 50 gal salt water from dissolving 5 l salt in 50 gal water. Salt water with concentration of.25 l/gal of salt is pumped into the tank at 6 gal/min. Salt water is pumped out at 4 gal/min. Find a formula for the concentration of salt in the tank up to the point of overflowing. 6 gal/min 200 gal tank 4 gal/min.25 l/gal 50 gal conc.=.1 l/gal? l/gal Let x(t)=l of salt in tank, V(t)=gal in tank, c(t)=concentration of salt in tank in l/gal Thus c(t) = x(t) and x(t) = c(t)v (t). V (t) inflow rate = (6 gal/min)(.25 l/gal) = 1.5 l/min x(t) l outlow rate = (4 gal/min) = 4x(t) V (t) gal V (t) l/min Since V (t) = t, the model is dx ( ) = 1.5 4x, s(0) = t Alternately, since x(t) = c(t)(50 + 2t), we could also use d h i (#) (50 + 2t)c(t) = 1.5 4c(t) =)

5 Solving ( ), 2. COMPARTMENTAL ANALYSIS 63 2c(t) + (50 + 2t) dc = 1.5 4c(t) =) dc + 6c t = 1.5, t c(0) =.1 x 0 + 2x 25 + t = 1.5, x(0) = 5 µ = e R 2 25+t = e 2 R 25+t = e 2 ln t+25 = e ln(t+25)2 = (t + 25) 2 (t + 25) 2 x 0 + 2(t + 25)x = 1.5(t + 25) 2 =) d h i (t + 25) 2 x = 1.5(t + 25) 2 =) Z (t + 25) 2 x = 1.5 (t + 25) 2 + C =) (t + 25) 2 (t + 25)3 x = x(t) = t C =) C (t + 25) = (t + 25)3 + 2C. 2 2(t + 25) 2 x(0) = C = 5 =) C = =) Then 2C = = 25 2 (10 25) =) C = c(t) = (t + 25) (t + 25) 2 x(t) = (t + 25) (t + 25) 2 and Maple. See extrapro.mw or extrapro.pdf 1 2(t + 25) = (t + 25) (t + 25) 3

6 64 3. MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS Radioactive Decay Suppose t is a su ciently small time interval (i.e., as small as we need) and P ( t) is the proaility that a given nucleus will decay over any time period of length t. We assume P ( t) / approx t =) P ( t) t, > 0 and that P ( t) << 1 (P ( t) is much smaller than 1). P ( t) Also assume that lim =. t!0 t Let N(t) = # of nuclei at time t. Then the numer of atoms decaying over a time period of t eginning at time t is N(t)P ( t) N(t) t =) N(t + t) = N(t) N(t)P ( t) N(t) N(t) t =) dn = lim N(t + t) N(t) t!0 t N(t) t = lim t!0 t lim t!0 = N(t) = N(t) Thus the model is dn = N(t), N(0) = N 0. However, this model is flawed due to the discreteness of N(t), whose values are integers, resulting in a discontinuous function.

7 2. COMPARTMENTAL ANALYSIS 65 We switch to mass m(t) in grams, a continuous quantity. Then the model is given y dm = m(t), m(0) = m 0 and is called the radioactive decay law. > 0 is the decay constant. m(t) outflow rate=λm(t) Malthusian population model We want to model the population p(t) at time t. Assume a virtual lack of death over the oserved time interval. This is asically the same situation as radioactive decay, except there is growth here. Instead of losing a nucleus, we gain a irth. Thus the model is = p, p(0) = p o, > 0 inflow rate=p p(t) is the growth rate and / p is the per capita growth rate.

8 66 3. MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS Now add in a death rate where d > 0 is the proportionality constant for the death rate (death y natural causes). irth rate p p(t) death rate Then, the Malthusian model is = p = ( d)p = kp, p(0) = p 0, where the per capita growth rate / p = k is constant. Example. In the 18th century, the population of the US douled every 25 years. Use the Malthusian model to estimate the per capita rate of growth. The Malthusian model is = kp or kp = 0, p(0) = p 0 µ = e kt =) e kt p 0 ke kt p = 0 =) d h i e kt p = 0 =) e kt p = C =) p(t) = Ce kt Now p(0) = C = p 0, so ( ) p(t) = p 0 e kt is the solution for the Malthusian model. Then

9 2. COMPARTMENTAL ANALYSIS 67 2p 0 = p 0 e 25k =) 2 = e 25k =) 25k = ln 2 =) k = ln 2 25 =.0277 Then k =.0277 is the per capita growth rate. Prolem (Page 100 # 14). In 1980 the population of alligators on the Kennedy Center Space grounds was was estimated to e In 2006 the population had grown to an estimated Using the Mathusian model for population growth, estimate the alligator population of the alligators on the Kennedy Center Space grounds in the year Solution. Counting time from the year 1980 with p(t) = the population of the alligators, we have p(0) = Then, from ( ), For 2006, t = 26, and thus p(t) = 1500e kt. p(26) = 1500e k(26) = 6000 =) e k(26) = 4 =) e k = 4 1/26 = 2 1/13. Then, for 2020, t = 40, and so y the Malthusian model, there will e p(40) = 1500e k(40) = / alligators on the Kennedy Center Space grounds

10 68 3. MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS Logistic Population Model Now add in the death rate due to the rate of competition (e.g., limited resources, disease, etc.). Then = kp rate of competition]. This rate of competition is proportional to the numer of possile 2-party interactions p p(p 1) pc 2 = C p,2 = =. 2 2 Thus and = kp k p(p 1) 1 2 = k + k 1 2 p k 1 p 2 2, = Ap(p p 1), p(0) = p 0 is the logistic model where A = k 1 2 and p 1 = 2k +1. The equation is separale. k 1 First, Thus are the constant solutions. = 0 =) Ap(p p 1) = 0 =) p = 0 or p = p 1. p = 0 and p = p 1

11 If Ap(p p 1 ) 6= 0, Z p(p p 1 ) = A p p 1 p Since p(0) = p 0, 1 p 1 Z 1 p 1 2. COMPARTMENTAL ANALYSIS 69 Z p + Z + c 1 =) Z 1 p 1 (using partial fractions) = At + c 1 =) p p Z 1 = At + c 1 =) p p 1 p + 1 p 1 1 ln p + 1 ln p p 1 p 1 p 1 = At + c 1 =) ln p p 1 p = Ap 1 t + p 1 c 1 =) = e Ap 1t+p 1 c 1 = e Ap 1t e p 1c 1 = c 2 e Ap 1t p p 1 p = Ce Ap 1t p 0 p 1 p 0 = C =) (C 6= 0) p p 1 = p 0 p 1 p p 0 e =) Ap 1t p(p 0 p 1 ) = p 0 (p p 1 )e Ap1t =) pp 0 pp 1 pp 0 e Ap1t = p 0 p 1 e Ap1t =) p 0 p 1 p(t) = p 0 + (p 1 p 0 )e. Ap 1t (c 2 > 0) =) Note: lim t!1 p(t) = p 1, the carrying capacity of the environment.

12 70 3. MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS We look at the graph of the function p(t) for two cases elow. Note that p 0 is the p-intercept. These two curves are called logistic curves Maple. See logistic.mw or logistic.pdf

13 4. NEWTONIAN MECHANICS Newtonian Mechanics Mechanics is the study of the motion of ojects and the e ect of forces acting on them. Newtonian, or classical mechanics deals with the motion of ordinary ojects that is, ojects that are large compared to an atom and slow-moving compared with the speed of light. A model for Newtonian mechanics can e ased on Newton s laws of motion: (1) When a ody is sujected to no resultant external force, it moves with constant velocity. (2) When a ody is sujected to one or more external forces, the time rate of change of the ody s momentum is equal to the vector sum of the external forces acting on it. (3) When one ody interacts with a second ody, the force of the first ody on the second is equal in magnitude, ut in opposite direction, to the force of the second ody on the first. We can express Newton s second law y = F (t, x, v) where F (t, x, v) is the resultant force on the ody at time t, location x, and velocity v, and p(t) is the momentum of the ody at time t. We have p(t) = mv(t) =) m dv = ma = F (t, x, v), where a = dv dx is the acceleration of the ody at time t. With v =, we have a second order DE in x(t). For F independent of x, we have (?) m dv = F (t, v), a first order equation in v(t).

14 72 3. MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS Procedure for Newtonian Models (1) Determine all relevant forces acting on the oject eing studied. It is helpful to draw a simple diagram of the oject that depicts these forces. (2) Choose an appropriate axis or coordinate system in which to represent the motion of the oject and the forces acting on it. Keep in mind that this coordinate system must e an inertial reference frame (a reference frame in which an undistured ody moves with a constant velocity). (3) Apply Newton s second law as expressed in equation (?) to determine the equations of motion for the oject. We have the following choices for units:

15 4. NEWTONIAN MECHANICS 73 Vertical motion with a velocity-dependent drag force An oject of mass m is given an initial downward velocity v 0 and allowed to fall under the influence of gravity. Assume the gravitational force is constant and the force due to air resistance is proportional to the velocity of the oject. We use a vertical axis and let x(t) denote the distance fallen at time t. The force due to gravity is F 1 = mg where g is the acceleration due to gravity at the Earth s surface. The force due to air resistance isn F 2 = v(t) where > 0 is the proportionality constant and the sign is present since the air resistance is opposite in direction to gravity. We have F = F 1 + F 2 = mg v(t) =) (#) m dv = mg v, v(0) = v 0 is the IVP for this model with m > 0 and > 0. (#) is separale and first order linear and can e solved to give (##) v(t) = mg Integrating v = dx + v 0 with respect to t, we get Z x(t) = v(t) = mg t m v 0 mg e t/m mg e t/m + c.

16 74 3. MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS Since x(0) = 0, we have 0 = m mg v 0 + c =) c = m and thus the equation of motion is (###) x(t) = mg t + m v 0 We note at this time that mg lim v(t) = and lim x(t) = mg t!1 t!1 t v 0 mg, mg (1 e t/m ). m 2 g 2 + m v 0. The value mg of the horizontal asymptote for v(t) is called the limiting or terminal velocity of the oject and is a constant solution of (#). Since the two forces are in alance here, this is known as an equilirium solution. Note also that the terminal velocity is independent of the initial velocity. We also have x(t) asymptotically approaching the line mg t m 2 g 2 + m v 0.

17 4. NEWTONIAN MECHANICS 75 Prolem (Page 115 # 10). An oject of mass 2 kg is released from rest from a platform 30 m aove the water and allowed to fall under the influence of gravity. After the oject strikes the water, it egins to sink with gravity pulling down and a ouyancy force pushing up. Assume that the force of gravity is constant, no change of momentum occurs on impact with the water, the ouyancy force is 1/2 the weight (weight = mg), and the force due to air resistance or water resistance is proportional to the velocity, with proportionality constant 1 = 10 N-sec/m in the air and 2 = 100 N-sec/m in the water. Find the equation of motion of the oject. What is the velocity of the oject 1 min after it is released? Solution. We have motion equations for oth air and water. For motion in the air, let x 1 (t) e the distance from the oject to the platform and v 1 (t) = x 0 (t) its velocity at time t. We use (##) and (###) from aove with m = 2, = 1 = 10, v 0 = v 1 (0) = 0, and g = 9.81 to get v(t) = mg + mg v 0 e t/m = 1.962(1 e 5t ) and x(t) = mg t + m mg v 0 (1 e t/m ) = 1.962t 0.392(1 e 5t ). Solving x 1 (t) = 1.962t 0.392(1 e 5t ) = 30 for t gives t 15.5 sec for the time the oject hits the water. The velocity at this moment was v 1 (15.5) = 1.962(1 e 5(15.5) ) For motion in the water, we reset t and let x 2 (t) e the distance passed y the oject from the water s surface and y v 2 (t) it velocity at (reset) time t, we have IC at the water s surface of v 2 (0) = 1.962, x 2 (0) = 0. With the gravity force F g = mg, the resistance force F r = 100v, and the ouyancy force F = (1/2)mg, Newton s second law gives (adding ouyancy to (#))

18 76 3. MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS m dv 2 = mg 100v 2 dv 2 (?) = g 2 Solving (?) with the aove IC, 1 2 mg = 1 2 mg 100v 2 =) 100 m v 2 = v 2. v 2 (t) = Ce 50t, v 2 (0) = C = =) C = =) x 2 (t) = Z t 0 v 2 (t) = e 50t =) v 2 (s) ds = 0.098t 0.037e 50t Taking into account the time shift made, the distance from the oject to the platform is ( 1.962t 0.392(1 e 5t ), t apple 15.5 x(t) = 0.098(t 15.5) 0.037e 50(t 15.5) , t > min after the oject was released, it traveled in the water for = 44.5 sec, so it had velocity v 2 (44.5) = e 50(44.5) m/sec. Resistive drag forces proportional to v r, r 6= 1. The signum function, which we use here is defind y 8 >< +1, if x > 0 sgn(x) = 0, if x = 0 >: 1, if x < 0 Note that for x 6= 0, sgn(x) = x and indicates the sign of x. Your text uses x sign(x) for this function.

19 4. NEWTONIAN MECHANICS 77 Using Newton s second law of motion, the DE that models vertical motion of a ody through a medium that exerts a drag force on the ody that is approximately proportional to a power r of its velocity is m dv = mg v r sgn(v) Find a constant solution of this model if r = 2. dv = 0 =) g m v 2 sgn(v) = 0 =) v 2 sgn(v) = mg Thus sgn(v) = +1 =) v > 0, and so v 2 = mg =) r mg v = since the velocity is positive. Maple. See quadratic drag.mw or quadratic drag.pdf.

20 78 3. MATHEMATICAL MODELS AND NUMERICAL METHODS INVOLVING FIRST-ORDER EQUATIONS Prolem (Page 116 # 22). A sailoat of mass 50 kg has een running (on a straight course) under a light wind at 1 m/sec. Suddenly the wind picks up, lowing hard enough to apply a constant force of 600 N to the sailoat. The only other force acting on the oat is water resistance that is proportional to the velocity of the oat. When the velocity of the sailoat reaches 5 m/sec, the oat egins to rise out of the water and plane. When this happens, the proportionality constant for the water resistance drops to 0 = 60 N-sec/m. Find the equation of motion of the sailoat. What is the limiting velocity of the sailoat under this wind as it is planing? Solution. There are two opposing forces here: a constant horizontal force due to the wind and an opposing force due to water resistance. All the motion occurs on a horizontal axis. Set t = 0 at the point where the oat egins to plane, and let x(t) and v(t) = x 0 (t) denote the distance traveled and velocity, respectively, at time t. The forces due to the wind and water resitance are F w = 600 N and F r = 60v N. Applying Newton s second law, we have 50 dv = v This gives us the IVP dv = 6 (10 v), v(0) = 5. 5 This is a first order linear equation whose solution is Then v(t) = 10 + Ce 6t/5. v(0) = 5 =) 10+C = 5 =) C = 5 =) v(t) = 10 5e 6t/5 =) lim t!1 v(t) = 10. Integrating v(t) with v(0) = 0 gives the following equation of motion: Z t x(t) = 10 5e 6s/5 ds = 10s + 25 t 6 e 6s/5 25 = 10t (e 6t/5 1). 0

21 4. NEWTONIAN MECHANICS 79