Edwards-Penney Chapter 2, sections 2.1, 2.2, 2.3 Topics, Definitions, Theorems, Examples 2.1 Population models, Malthus and Verhulst ==== MALTHUS MODEL P'(t) = (beta - delta) P(t) P'(t) = k P(t) LOGISTIC MODEL P'(t) = (a-b P(t)) P(t) EXAMPLE. Alligators and population explosion This model uses delta=0 and beta=0.0005 P(t) to obtain P'(t) = -0.0005 P(t)^2, P(0)=100 P(t)=2000/(20-t) explosion at t=20 EXAMPLE. Population equilibrium This model uses delta=0.06 and delta=0.0004 P(t) as the natural Mathusian birth rate and the death rate or inhibition factor. P'(t) = 0.06 P(t) - 0.0004 P(t)^2 P(t) = 150 P0/(P0+(150-P0)exp(-0.06 t)) Limiting population is limit P(t) at t=infinity, which is 150, regardless of the size of P0>0. DEF. Carrying capacity = a/b in the logistic model P'=(a-bP)P EXAMPLE. Find the constants a, b in the logistic model given the following information: P=50 million in 1885 Growth rate = 0.75 million per year in 1885 P=100 million in 1940 Growth rate = 1 million per year in 1940 ANSWER: b=0.0001 and a=0.02 are obtained by solving 2 linear algebraic equations in 2 unknowns a,b: P' = aP - bP^2 Start with the logistic model 0.75 = 50a - 2500b Substitute P'=0.75 and P=50 1.00 = 100a - 10000b Substitute P'=1.00 and P=100 MAPLE ANSWER CHECK. eq:= 0.75=50*a-2500*b,1.00=100*a-10000*b; solve({eq},{a,b}); # or solve([eq],[a,b]); EXAMPLE. Malthus model for the US population. Assume model P'=rP and solution P(t)=P0 exp(rt). At t=0, corresponding to 1800, the population was 5.308 million. At t=100 corresponding to 1900, the population was 76.212. Find P0 and r in the Malthus model P(t)=P0 exp(rt). ANSWER: Substitute t=0, P=5.308 and t=100, P=76.212 into the Mathus model P(t)=P0 exp(rt) to obtain 2 equations in two unknowns P0, r. Then solve for P0=5.308, r=0.026643. EXPONENTIAL EQUATION. The problem involves solving for r in the equation 76.212=5.308 exp(100 r). Take logs to do it, then use a calculator to get r=0.02664303814. MAPLE ANSWER CHECK. eq:=76.212=5.308*exp(100*r); solve(eq,r); EXAMPLE. Logistic model for the US population. Assume logistic model P'=kP(M-P) REMARK. The equation P'=(kM - kP)P is exactly P'=(a-bP)P. Assume three data items t=1800, P=5.308 million t=1850, P=23.192 million t=1900, P=76.212 million Find k, M. Then report the values of a=kM and b=k. ANSWER: k=0.00167716, M=188.121; a=0.3155090164, b=0.00167716 MAPLE ANSWER CHECK. # First, develop three equations in 3 unknowns P0,k,M from the # logistic model solution formula P=MP0/U, U=P0+(M-P0)exp(-kMt). # Take t=0 to be the year 1800. U:=(P0,k,M,t)->P0+(M-P0)*exp(-k*M*t); P:=(P0,k,M,t)->M*P0/U(P0,k,M,t); eq1:=P(P0,k,M,0)=5.308; eq2:=P(P0,k,M,50)=23.192; eq3:=P(P0,k,M,100)=76.212; solve([eq1,eq2,eq3],[P0,k,M]); # ans P0 = 5.308000000, k = 0.001677157274, M = 188.1208275 LOGISTIC APPLICATIONS Limited environment Competition, cannibalistic population Spread of a disease, infective and susceptible EXAMPLE. Spread of a rumor. At time t=0, 10 thousand people in a population of 100 thousand have heard a rumor about a politician's tax return. After 1 week, the number who have heard the rumor increased to 20 thousand. Assume logistic model P'=(a-bP)P and compute the time t when 80 percent of the population has heard the rumor. ANSWER: 4.42 weeks. The carrying capacity is a/b=M=100 thousand. This implicit information is needed to solve the two equations obtained from data pairs t=0, P=10 and t=1, P=20. Then a/b=100, exp(-100b)=4/9 and the time t satisfies exp(-100bt)=1/36 or (9/4)^t = 36. MAPLE ANSWER CHECK solve((9/4)^t=36,t); # 4.419022582 EXAMPLE. Explosion and Extinction P' = 0.0004 P^2 - 0.06 P Model gives extinction if P0 < M and explosion if P0 > M. The ideal or threshold population M is an unstable equilibrium of the DE, found where P'=0, then P=0 or P= 150. Case 1. P0<150 Then P(t) is a fraction with denominator having limit infinity. Extinction. Case 2. P0>150 Then P(t) is a fraction with denominator having value zero at some future time t=T. Then the fraction becomes infinite at t=T. Explosion. 2.2 Equilibrium solutions and stability ==== EXAMPLE. Newton's law of cooling x'(t) = -k(x(t)-A) The equilibrium solution x(t)=A is the temperature of the surrounding medium. Every solution limit to the equilibrium solution. DEF. The equaton y'=f(y) is called an AUTONOMOUS first order differential equation. DEF. The constant solutions of y'=f(y), which are solutions of the form y=c, are called EQUILIBRUM POINTS or CRITICAL POINTS. DEF. An EQUILIBRIUM SOLUTION of y'=f(y) is a constant solution y=c. EXAMPLE. Logistic equation y'=(a-by)y has equilibria y=0, y=a/b which are the extinction state and the carrying capacity, resp. DEF. STABLE and UNSTABLE. A constant solution y=c of y'=f(y) is called STABLE provided all other solutions with start point x=0, y near y=c, stay near the line y=c. A constant solution y=c that is not stable is called UNSTABLE. In examples, it means limit y(x) = c as x goes to infinity, for all solutions y(x) starting close enough to y=c. A SPEED GOVERNOR in an auto behaves this way - a small push on the accelerator pedal will cause the speed to change (y>c), but quickly the speed returns to the design speed (y=c). The speed in the auto actually oscillates a bit above and below the design speed y=c. Math examples are MONOTONIC, they limit from above or below to the line y=c. The GOVERNOR problem is a control problem, which we don't study. DEF. PHASE DIAGRAM. It is a graphic that displays a number of threaded solutions to the differential equation y'=f(y). It should contain if possible all equilibrium solutions y=c (horizontal line). While it can be drawn from Euler's theory of direction fields, often shortcuts are used, especially these two: 1. TRANSLATION. If y(x) is a solution of y'=f(y), then so is y(x-a). 2. The direction field along vertical line x=0 is the same as the direction field along any x=a. DEF. Funnel, spout, node. A node is neither a funnel or a spout. Funnel is the only stable figure, the others are unstable. A spout in reverse time is a funnel. A funnel y=c is characterized in examples by nearby solutions y(x) to line y=c limiting from above or below to the line y=c. EXAMPLE. Logistic equation x'=(a-bx)x Equilibria are x=0 and x=a/b. Small changes in a,b result in small changes in the stable equilibrium x=a/b (for a>0, b>0). Solutions depend continuously on a,b,t,x(0), as can be seen from the solution formula ax(0) x(t) = ---------------------- x(0)+(a-bx(0))exp(-at) EXAMPLE. Logistic explosion-extinction equation x'=kx(x-M) The constant solutions are x=0 and x=M, but x=0 is stable and x=M is unstable. HARVESTING x'=(a-bx)x-h with a,b,h positive Key example is x(t)=number of fish in a lake, h=fishing rate EXAMPLE. Analysis of havesting to find the threshold population. ANSWER: H=M/2 (half the carrying capacity in the unharvested setting) The meaning is that x(0)H there is population equilibrium at carrying capacity N=sqrt(M^2-4h/k)/2. This interpretation is valid only if the square root is real. EXAMPLE. Logistic harvesting equation x'=x(x-4)-h for a fish population in a lake. Variable x(t) is in hundreds of fish and t is in years. The threshold is always 1/2 of the carrying capacity when h=0, or 400/2 fish. For h=3, the harvested capacity is 300. But H=1. Stocking the lake with less than 100 (=100H)0 fish results in no fish, but stocking more than 100H will give 300 fish to harvest. BIFURCATION FILMSTRIPS and DIAGRAMS EXAMPLE. Logistic equation x'=(a-bx)x-h with parameters a,b,h. A filmstrip of phase diagrams, each frame a different value of the harvesting constant h. What does this diagram tell us about the harvesting problem? DEF. PITCHFORK BIFURCATION. The filmstrip starts with a phase diagram having 2 equilibria, and later frames look the same, until frame h=h0, where the 2 equilbria are lost and there is only one equilibrium. After that, h>h0, the filmstrip continues, each frame a phase diagram with no equilibrium solution at all. DEF. A BIFURCATION POINT for the logistic equation x'=(a-bx)x-h is a value of h such that the phase diagram makes a dramatic change in stability or in phase diagram character. DEF. A BIFURCATION DIAGRAM is a visual tool to summarize the results seen in a bifurcation filmstrip. In particular, it displays the change in character of the solutions near the bifurcation point. Disturbing on the first encounter with such diagrams is the absence of solutions of the differential equation. 2.3 Newton models, Jules Verne problem ==== EXAMPLE. A crossbow bolt is shot straight upwards along the gravity vector, on the ground y=0 at t=0, with initial velocity y'=49 meters per second. MODEL: my'' = -9.8m, y(0)=0, y'(0)=49 [mass m cancels] SOLUTION: y(t)=-9.8t^2/2+49t MAX HEIGHT. It happens when y'=0, which is -9.8t+49=0. Then y(t)=max height=122.5 meters FLIGHT TIME. It happens when y=0, but there are two answers for t: t=0 and -9.8t/2+49=0. Flight time = 10 seconds. IMPACT SPEED. This is the velocity at the end of the flight time, y'(10)=-49. NEWTON'S AIR RESISTANCE MODEL Isaac Newton proposed a resistant force F=-kv|v|^alpha where k is a constant and alpha>=0 is a constant. The two cases alpha=0 and alpha=1 are treated. DEF. The DRAG COEFFICIENT is rho=k/m, where k appears in Newton's resistive force formula and m is the mass. LINEAR MODEL v' = -rho v - g Solution v(t)=(v0+g/rho)exp(-rho t) - g/rho DEF. TERMINAL SPEED is limit v(t) as t goes to infinity. Answer is speed=g/rho (negative sign removed). EXAMPLE. Study the linear model with v(0)=49 m/s and rho=0.04. ANSWER: v(t)=294 exp(-t/25) - 245, y(t)=7350-245t-7350exp(-t/25) MAX HEIGHT. It happens when y'=0, which is v=0 or t=25 ln(294/245). Then y(t)=max height=108.28 meters, which is less than predicted by the no air resistance model. FLIGHT TIME. It happens when y=0, but there are two answers for t: t=0 and t=9.411 seconds. The no air resistance model predicted 10 seconds. IMPACT VELOCITY. This is the speed at the end of the flight, v(9.411)=43.227, a bit less than predicted by the no air resistance model. NONLINEAR MODEL v' = -rho v|v| - g Case 1. v0 > 0, v' = - rho v^2 - g Solution v(t)=sqrt(g/rho) tan(C1-t sqrt(rho g)), where C1 is defined by tan(C1)=v0 sqrt(rho/g). Case 2. v0 < 0, v' = rho v^2 - g Solution v(t)=sqrt(g/rho) tanh(C2-t sqrt(rho g)), where C2 is defined by tanh(C2)=v0 sqrt(rho/g). TERMINAL VELOCITY: Always sqrt(g/rho), linear and nonlinear models. EXAMPLE. Study the upward and downward nonlinear models for rho=0.0011 and compare with the linear model and no air resistance model using v(0)=49. ANSWER: See the table in section 2.3. The result is that it takes more time to fall than to rise, except in the case of no air resistance (rise=fall=5 seconds). The no resistance case over-estimates the flight time and the impact velocity. POSITION MODELS. Each is obtained by integration of the velocity model. See textbook formulas (14) and (17) for the nonlinear cases. NEWTON'S UNIVERSAL GRAVITATION LAW F = GMm/r^2, particles of mass M and m at distance r apart G = 6.6726/10^(11) in MKS units EXAMPLE. Lunar Lander Problem. A lunar lander model is r''(t) = T - GM/r^2 where M=7.35(10^(22)) kg is the mass of the moon, and T=4 m/s is the retrorocket deceleration. The lander travels along the moon's gravity vector with velocity 1477 km/h. Find the retrorocket activation height above the moon for a soft touchdown. ANSWER: Soft touchdown is velocity zero at impact. We solve for the height r0 assuming r''=4-GM/r^2, r(0)=r0, r'(0)= -1477(1000)(1/3600) r'(t0)=0 [soft landing] where the unit system is MKS and t0>0. The calculation gives r0=41870 meters, which is 26.17 miles. The single idea is to multiply the DE by r'(t), making the left side a perfect derivative, then apply the method of quadrature to get v^2/2 = 4r + GM/r+C. Constant C is found by setting t=0, r=r0, v=r'(0). Then set t=t0, r=R=1740000=moon radius, v=0 to get an equation for r0. EXAMPLE. Escape Velocity of the earth, Jules Verne problem Assume the model r''(t)=-GM/r^2 for the distance r(t) from the center of the earth to the center of the moon. This model ignores the gravitation attraction of the moon. Find the velocity v0 for the rocket needed to escape the gravitational pull of the earth and make a soft landing on the surface of the moon. ANSWER: v0=sqrt(2GM/R)=11180 m/s, about 25000 miles per hour. METHOD: Multiply the DE by r'(t) and apply the method of quadrature. The trick is to use limit r(t) = infinity to evaluate the integration constant. Then minimize v0 from the resulting equation. HISTORY: See exercise 30, section 2.3. ==== end