Applied Mathematics

MATH259 Mathematical Modelling Yuri Litvinenko June 24, 2011

Contents 1 First-order ordinary differential equations 1.1 Introduction . . . . . . . . . . . . . . . . . 1.2 Exponential growth and decay . . . . . . . 1.3 Linearisation . . . . . . . . . . . . . . . . 1.4 Separable and linear equations . . . . . . . 1.5 Equilibrium and stability . . . . . . . . . .

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2 Second-order ordinary differential equations 2.1 Linear second-order equations . . . . . . . . . . . . . 2.2 The homogeneous equation with constant coefficients 2.3 Linear oscillations . . . . . . . . . . . . . . . . . . . . 2.4 Equilibrium and stability . . . . . . . . . . . . . . . . 2.5 Inhomogeneous equations . . . . . . . . . . . . . . . .

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3 Elements of classical mechanics 3.1 Kinematics . . . . . . . . . . . . . 3.2 Dynamics and gravity . . . . . . . 3.3 Work and energy . . . . . . . . . . 3.4 Planetary orbits and Kepler’s laws

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4 Systems of first-order differential equations 47 4.1 Homogeneous linear systems with constant coefficients . . . . . . . . . . . 47 4.2 Equilibrium and stability: linear systems . . . . . . . . . . . . . . . . . . . 49 4.3 Equilibrium and stability: nonlinear systems . . . . . . . . . . . . . . . . . 50 5 Difference equations 5.1 Geometric growth . . . . . . . 5.2 The equation xk+1 = axk + b . 5.3 Fibonacci numbers . . . . . . 5.4 Nonlinear difference equations

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Chapter 1 First-order ordinary differential equations 1.1

Introduction

Mathematical modelling is a part of applied mathematics. We use mathematical modelling to describe and predict phenomena in the world around us. We may be interested in the temporal evolution of a phenomenon, its steady state, or its stability. Mathematical modelling always involves both mathematics and some area of science. “To isolate mathematics from the practical demands of the sciences is to invite the sterility of a cow shut away from the bulls” (P. L. Chebyshev). A differential equation is an equation involving an unknown function and its derivatives with respect to one or more independent variables. An ordinary differential equation contains only one independent variable. For example, a first-order ordinary differential equation for a function x(t) dx = f (t, x) dt contains its first derivative or the rate of change dx x(t + ∆t) − x(t) = lim . ∆t→0 dt ∆t The notation x˙ = dx/dt is used when the independent variable t is time. Differential equations is the basic tool for the study of change in the physical world. In this course we focus on the use of ordinary differential equations in modelling. Example. The speed of an object moving along the x-axis is given by v = x. ˙ If we know v(t) and initially x(0) = 0, we can determine the position of the object at a later time t by integration: Z t Z t dx x(t) = ds = v(s)ds, 0 ds 0 where s is a dummy variable of integration. Sometimes the definite integral is also written R less formally as 0t v(t)dt. 2

A model consists of a differential equation and an initial condition, say x(0) = x0 . We expect that the general solution of a first-order equation contains an arbitrary constant of integration. There are exceptions: for instance x = 0 is the only real solution to x˙ 2 + x2 = 0. Such equations, however, are very unlikely to be encountered in practice. Therefore we typically use an initial condition to specify the integration constant and thus obtain a unique solution. Example. Paradichlorobenzene (moth repellent) sublimes from the solid to the gaseous state. We want to predict the radius r(t) of a mothball at any time if initially r(0) = r0 . Observationally, the volume V = 4πr 3 /3 of a spherical mothball decreases so that dV /dt is proportional to the surface area A = 4πr 2 of the ball. Thus observations imply that −

dV = kA, dt

where k > 0 is a proportionality constant. We have −4πr 2 r˙ = k4πr 2 or, as long as r > 0, which we integrate to obtain

r˙ = −k, r(t) = C − kt = r0 − kt,

where C is an integration constant that is specified by the initial condition r(0) = r0 = C. Finally, r(ts ) = 0 defines the sublimation time ts = r0 /k. Because the radius r(t) cannot be negative, the correct solution for t > ts is simply r(t) = 0.

1.2

Exponential growth and decay

Suppose the rate of change of x(t) is proportional to x itself: x˙ = ax,

a = const.

We obtain the general solution to this equation by rewriting it in the form of differentials and integrating: dx = adt, x ln x = ln C + at, C = x(0), or x(t) = x(0)eat , where the integration constant C is specified by an initial condition. More generally, x(t) = x(t0 )ea(t−t0 ) , 3

where x(t0 ) is the value of x at some time t0 . Solutions with a > 0 describe unlimited growth, x → ∞ as t → ∞. T. R. Malthus (1798) emphasised the role of exponentially growing solutions in models for population growth, so such solutions are sometime said to describe the Malthusian growth. In reality, exponential growth is limited by finite resources. Still, exponentially growing solutions help us understand how quickly “things get out of hand”. Solutions with a < 0, x(t) = x(0)e−|a|t , describe exponential decay, x → 0 as t → ∞. It is often useful to define a doubling time t2 such that x(t2 )/x(0) = 2. For an exponentially growing x(t), exp(at2 ) = 2 and we obtain t2 =

ln 2 0.7 ≈ , a a

x(t) = x(0)2t/t2 .

Example. Presently the population of New Zealand grows annually by about 1%. If the trend continues, the population will double in about 0.7/0.01 = 70 years. Example. To double an investment in 8 years, it is necessary to achieve an annual return exceeding 0.7/8 ≈ 0.09 = 9%. Similarly, we define a half-life time t1/2 for a decreasing function, say the amount of a radioactive material, x(t1/2 )/x(0) = 1/2. For an exponentially decaying x(t), t1/2 =

ln 2 0.7 ≈ , |a| |a|

x(t) = x(0)2−t/t1/2 .

Example. Radium decays with a half-life time of about 1500 years, so after 1000 years there remains 2−1000/1500 = 2−2/3 ≈ 0.63 = 63% of the initial amount of radium.

Example. Carbon-14 dating, discovered by W. Libby around 1949, is one of the most accurate ways of dating archaeological finds. Radioactive carbon-14 (14 C) is continuously produced in the atmosphere of the earth by cosmic rays. This radiocarbon is incorporated in carbon dioxide and absorbed by plants. Animals, in turn, build radiocarbon by eating the plants. In living tissue, the rate of ingestion of 14 C balances the rate of disintegration of 14 C. When an organism dies (t = 0), its 14 C concentration starts to decrease: N(t) = N(0)e−λt , where λ is the decay constant of the rate of decay is given by

14

C, corresponding to t1/2 ≈ 5700 years. Consequently

R(t) = −N˙ = λN(0)e−λt = R(0)e−λt .

4

We can measure the present rate R(t) of disintegration of 14 C in a sample, say charcoal from an archaeological site. Because the initial decay rate R(0) is equal to the rate of decay of 14 C in the same amount of living tissue, we can determine the age t of the sample: #

"

1 ln[R(0)/R(t)] R(0) t = ln = t1/2 . λ R(t) ln 2 Consider next a generalisation of the equation for exponential growth or decay: x˙ = ax + b,

a = const,

b = const.

This is an example of a linear (meaning that the equation does not contain terms nonlinear in x, like x2 , sin x or xx), ˙ first-order differential equation with constant coefficients. If b 6= 0, the equation is called inhomogeneous (or nonhomogeneous). If b = 0, it is homogeneous. We note that x = −b/a is a particular solution of this equation, and we seek its general solution in the form x(t) = y(t) − b/a. Putting this into the differential equation yields y˙ = ay with the general solution y(t) = Ceat , Hence

b C = y(0) = x(0) + . a

"

#

b at b x(t) = x(0) + e − . a a

Linear first-order ordinary differential equations with constant coefficients are often encountered in practice. A few examples follow. Example. Consider an RC electric circuit in which the resistance is R and the capacitance is C (not to be confused with an integration constant). The governing equation for the electric current I(t) in the circuit is RI˙ + so

I = 0, C

I(0) = I0 ,

t I(t) = I0 exp − . RC 



Example. Consider an electric circuit consisting of an inductor of inductance L, a resistor of resistance R, and a battery producing a constant electromotive force (voltage) E. The governing equation for the electric current I(t) in the circuit is LI˙ + RI = E. 5

This is a linear first-order equation with the general solution E Rt I(t) = + C exp − . R L 



If I(0) = 0, we obtain Rt E 1 − exp − . R L We see that I → E/R as t → ∞. Hence Ohm’s law E = RI is nearly true for large t. I(t) =







Example. The rate of cooling of some body in air is approximately proportional to the difference between the temperature of the body and the temperature of the air. Suppose we move a thermometer from a warm room (temperature T0 ) to a cooler room (temperature T1 < T0 ). The thermometer reading T (t) is governed by T˙ = −k(T − T1 ),

T (0) = T0 ,

where k is a proportionality constant. The general solution is T (t) = T1 + Ce−kt . We use the initial condition T0 = T1 + C to find the integration constant C. Hence T (t) = T1 + (T0 − T1 )e−kt . It makes sense that T approaches T1 (the temperature in the cooler room) for large t. Example. In some chemical reactions, the rate of conversion of a substance is proportional to the amount of available substance. This is the case, for instance, when calcium carbonate releases carbon dioxide on heating above 840◦ C, to form calcium oxide, commonly called quicklime: CaCO3 → CaO + CO2 . If x(t) is the amount of the substance that has been converted in the time interval t, and a is the initial quantity of the substance, then we have x˙ = k(a − x),

x(0) = 0,

where k is a constant. Integrating this equation, we obtain x(t) = a(1 − e−kt ). Consequently, x → a as t → ∞.

Example. Suppose you borrow N0 dollars at an interest rate ν to be paid off at a constant rate Q over time T . What is Q? For continuously compounded interest, the amount N(t) that you owe at time t is governed by the equation N˙ = νN − Q, 6

and its general solution is

Q + Ceνt . ν Now two conditions are needed to specify both Q and an integration constant C. We require N(0) = N0 and N(T ) = 0. Therefore, N(t) =

N(0) = N0 =

Q + C, ν

N(T ) = 0 =

Q + CeνT . ν

On solving these two equations, we find C = −(Q/ν) exp(−νT ) and Q=

νN0 . 1 − e−νT

It follows that

i eνT − eνt Qh 1 − eν(t−T ) = N0 νT . ν e −1 The total amount you will pay, QT , is typically significantly greater than the borrowed amount N0 . For instance, if ν = 6% per year and T = 25 years,

N(t) =

QT νT = ≈ 1.9. N0 1 − e−νT We can also calculate the fraction of the first payment which is interest: νN0 dt νN0 = = 1 − e−νT . Qdt Q Not surprisingly, the fraction is close to 1 for most loans. Example. A patrol craft sights a surfaced enemy submarine that immediately submerges and proceeds on an unknown straight course. The craft is twice as fast as the sub. What path should the patrol craft follow to assure that it passes directly over the sub? To answer this question, it is convenient to use polar coordinates r and θ. Choosing the origin at the initial location of the sub, we have r˙ for its speed. Because the craft is twice as fast as the sub, we have ds s˙ = = 2, dr r˙ where s is the arc length of the curve, followed by the craft, and s˙ is the speed of the craft. We choose the radial speed r˙ to be the same q for both the craft and the sub. Now, if we substitute x = r cos θ, y = r sin θ into ds = (dx)2 + (dy)2, we get ds =

q

(dr)2 + r 2 (dθ)2 .

Eliminating ds, we obtain 2dr =

q

(dr)2 + r 2 (dθ)2 7

or

dr dθ =√ , r 3

and so

√ r(θ) = C exp(θ/ 3).

This path is a logarithmic spiral. Thus the best strategy for the patrol craft consists of two steps: (1) proceed in a straight line to some point (r0 , θ0 ) such that the craft and√the sub are the same distance r0 from the origin; (2) follow the spiral r = r0 exp[(θ − θ0 )/ 3]. The craft will “catch” the sub at some point during the 360◦ (= 2π) traversal on the spiral.

1.3

Linearisation

Can we learn anything about the solution of the differential equation x˙ = f (x) if f (x) is a complicated nonlinear function of x? Suppose we are interested in the solution x(t) only for x near x(0) = x0 . Then we only need to know how f (x) behaves near x0 . Example. The nonlinear equation x˙ = ax + bx2 is not easy to integrate. However, if bx2 is small compared with ax, we have approximately x˙ = ax. On integrating this simpler linear equation, we obtain an approximate solution x(t) = x(0) exp(at). In general, to simplify a nonlinear equation x˙ = f (x), we can use Taylor’s theorem and approximate f (x) by the first two terms of its Taylor expansion about x0 : f (x) ≈ f (x0 ) + f ′ (x)|x=x0 (x − x0 ). Introducing ξ(t) = x(t) − x0 , we have approximately ξ˙ = aξ + b,

ξ(0) = 0,

where a = f ′ (x)|x=x0 = f ′ (x0 ),

b = f (x0 ).

Assuming that a 6= 0, we integrate the linear equation for ξ(t) to get b b ξ(t) = eat − , a a and so, approximately, "

#

b b + eat . x(t) = x0 + ξ(t) ≈ x(0) − a a 8

Example. Suppose we are interested in the solution of x˙ = 1 − sin[ln(1 + x)],

x(0) = 0

only for x near x(0) = 0. We replace the right-hand side of the equation by the linear Taylor approximation of f (x) = 1 − sin[ln(1 + x)] about x = 0. We have f ′ (x) = − cos[ln(1 + x)]/(1 + x) and f ′ (0) = −1, so that approximately x˙ = 1 − x with the general solution x(t) = 1 + C exp(−t). The solution that satisfies the initial condition is x(t) = 1 − exp(−t).

Thus we simplified a problem by approximating a complicated function f (x) with a simpler linear function, defined by the first two terms in the Taylor series of f (x). We say that we linearised the original nonlinear differential equation. The solution of the linearised equation can often provide a good approximation to the exact solution x(t), at least for sufficiently small times. We basically gained some knowledge at the expense of some accuracy. This idea is often used in applied mathematics. Later we shall repeatedly use linearisation to study stability of equilibrium solutions to differential equations.

1.4

Separable and linear equations

Generally speaking, it is difficult to solve ordinary differential equations. We consider two types of first-order equations for which routine methods of solution are available: separable and linear first-order equations. A first-order differential equation is separable if it can be written in the form x˙ =

f (t) . g(x)

To solve this, we have only to write it in the form g(x)dx = f (t)dt and integrate: Z

g(x)dx =

Z

f (t)dt + C.

Example. As we noted earlier, exponential growth cannot continue indefinitely in the real world. P. F. Verhulst (1838) suggested an extremely important model—the logistic growth model—that describes the limiting effect of finite resources on the size of a population: N N˙ = a 1 − N, K 



where a and K are positive constants, and K is called the carrying capacity. If K is infinity large, we have N˙ = aN and the population size N(t) is increasing exponentially, N(t) = N(0) exp(at). What is the effect of a finite K? The logistic equation is separable: KdN = adt. N(K − N) 9

We can use the change of variables x = 1/N to integrate the left-hand side: Z

It follows that

KdN =− N(K − N)

Z

Kdx = − ln(Kx − 1) + ln C. Kx − 1

K − 1 = Ce−at N

or

K . 1 + C exp(−at) The initial condition at t = 0 specifies the integration constant: N(t) =

K − 1 = C, N(0) which yields KN(0) . N(0) + (K − N(0)) exp(−at) We see that, for any N(0), the exponential growth is limited: N(t) =

N(t → ∞) →

KN(0) = K. N(0)

Thus the population can only reach a finite size equal to the carrying capacity K, which explains the meaning of the term. The logistic growth model also describes the spread of technological innovations, say substitution of electric locomotives for steam locomotives or adoption of high-speed bottle fillers in the brewing industry. Another important type of differential equation is the linear equation. The general linear first-order equation is x˙ = a(t)x + b(t). If the equation is homogeneous (b = 0), it is separable, and its solution is x(t) = x(0) exp

Z

t 0



a(s)ds .

Now we try to generalise this solution to solve a harder inhomogeneous equation. If b 6= 0, we search for a solution in the form x(t) = y(t) exp

Z

t



a(s)ds ,

where y(t) is a new unknown function. The method works because the equation for y(t) turns out to be simpler than that for x(t). We have x˙ = y˙ exp = y˙ exp

Z

Z

t

t



a(s)ds + y exp 

Z

a(s)ds + a(t)x, 10

t

a(s)ds



d dt

Z

t

a(s)ds



and substitution back into the original differential equation gives y˙ exp

Z

t



a(s)ds = b(t).

This is a separable equation. Its solution is y(t) =



Z

b(t) exp −

Z

t



a(s)ds dt + C,

which yields the general solution x(t) =

Z



b(t) exp −

Z

t





a(s)ds dt + C exp

Z

t



a(s)ds .

Example. Oxygen concentration in a lake, n(t), may be described by n˙ = k(ne − n) + S,

n(0) = n0 ,

where k is a positive constant, ne is an equilibrium concentration in the absence of external sources and sinks of oxygen (say, due to photosynthesis or pollution), and S(t) is a function describing the sources and sinks. Suppose that S(t) = αt. To solve the resulting inhomogeneous linear equation, n˙ + kn = kne + αt, we seek a solution in the form n(t) = f (t)e−kt . Substituting this into the equation yields a separable equation for the new function f (t): f˙ = (kne + αt)ekt . Using integration by parts, we get f (t) =

Z

(kne + αt)ekt dt

α 1 ekt dt (kne + αt)ekt − k k   α α kt = ne + t − 2 e + C. k k Z

=

As usual, the initial condition specifies C, and we finally arrive at n(t) = ne −

α α α + t + n0 − ne + 2 e−kt . 2 k k k 



Although solutions of nonlinear differential equations are not easy to obtain, some nonlinear equations can be solved by reducing them to simpler equations. For example, Bernoulli’s equation x˙ = f (t)x + g(t)xα 11

is nonlinear unless α = 0 or α = 1. The logistic growth model corresponds to α = 2. For an arbitrary α, we can reduce Bernoulli’s equation to a linear equation by the change of variable y = x1−α . We have y˙ = (1 − α)x−α x˙ = (1 − α)f x1−α + (1 − α)g, which yields a linear equation for y(t): y˙ = ay + b,

a = (1 − α)f,

b = (1 − α)g.

Example. An equation that generalises the logistic growth model is "

N N˙ = a 1 − K 

s #

N,

where a, K, and s are constants. This nonlinear equation is Bernoulli’s equation with α = s + 1. On substituting y(t) = N −s , we obtain the linear equation y˙ = −say + saK −s with the general solution y(t) = N −s = K −s + Ce−sat or N(t) =

K [1 +

CK s

exp(−sat)]1/s

.

We see that, as long as sa > 0, N → K as t → ∞. The initial condition N(0) = N0 specifies the integration constant C = N0−s − K −s , and so (

N(t) = K 1 +

"

K N0

s

#

−sat

−1 e

)−1/s

.

This model was used to describe the growth of pine trees in New Zealand (O. Garcia, 1983).

1.5

Equilibrium and stability

Equilibrium solutions to differential equations are defined as solutions that do not depend on time: x˙ = 0. We are often interested in equilibrium solutions. For example, we may want to determine the conditions required for sustainable harvesting of a renewable resource (say, fishing) or an equilibrium of supply and demand in economics. To model the global climate and climate change, we first need to describe an equilibrium between the solar radiation the Earth absorbs and the infrared radiation it emits. For a first-order equation of the form x˙ = f (x), 12

we find all equilibrium solutions x = x0 by solving f (x0 ) = 0. An equilibrium x = x0 is called stable if every solution x(t) that starts near x0 remains near x0 and x(t) → x0 as t → ∞. Otherwise, an equilibrium is called unstable. We are usually interested in stable equilibrium solutions. Nature is full of perturbations that destroy unstable equilibria, so they are rarely seen in natural systems. Suppose we have x˙ = f (x) and an equilibrium solution x = x0 satisfies f (x0 ) = 0. To investigate its stability, we need to determine the behaviour of a solution x(t) that is initially very close to x0 . Hence we only need to know how f (x) behaves near x0 . Using the first two terms of the Taylor expansion about x0 , we get f (x) ≈ f (x0 ) + f ′ (x0 )(x − x0 ) = f ′ (x0 )(x − x0 ), and we can approximate the original differential equation by x˙ = f ′ (x0 )(x − x0 ), which yields x(t) = x0 + C exp(f ′ (x0 )t). Clearly x(t) → x0 for large t, provided f ′ (x0 ) < 0, Conversely, x(t) deviates strongly from x0 for large t, if f ′ (x0 ) > 0. Thus we have the following stability criterion: if x˙ = f (x) and f (x0 ) = 0, then x = x0 is a stable equilibrium if f ′ (x0 ) < 0 and an unstable equilibrium if f ′ (x0 ) > 0. Example. The equilibrium solution x0 = 1 to the equation x˙ = ln x is unstable because f ′ (x) = d ln x/dx = x−1 , and therefore f ′ (x0 ) = x−1 0 = 1 > 0. Example. Suppose water flows into a lake at a constant rate r and that water evaporates from the lake at a rate that is proportional to (volume)2/3 . We have the following equation for the volume V of the lake: V˙ = f (V ) = r − aV 2/3 , where a is a constant. We solve f (V0 ) = 0 to obtain an equilibrium solution: V0 =

 3/2

r a

.

To determine its stability, we calculate f ′ (V ) = − 23 aV −1/3 . Now 2 −1/3 2 a3/2 f ′ (V0 ) = − aV0 = − 1/2 , 3 3r and we conclude that the equilibrium is stable because f ′ (V0 ) < 0. 13

Example. The logistic equation N N˙ = f (N) = a 1 − N K 



has two equilibrium solutions: N0 = 0 and N0 = K. Note that we can assume a = 1 and K = 1 without loss of generality. Indeed, using the variables N ∗ = N/K and t∗ = at, we get dN ∗ = N ∗ (1 − N ∗ ), ∗ dt which is the logistic equation with a = K = 1 for the new variables. So, assuming that a = K = 1, we have the equilibrium solutions N0 = 0 and N0 = 1. Now, f (N) = N(1 − N),

f ′ (N) = 1 − 2N,

and we conclude that N0 = 0 is unstable since f ′ (0) = 1 > 0, whereas N0 = 1 is stable since f ′ (1) = −1 < 0. This is consistent, of course, with the exact solution to the logistic equation. Example. The same mathematical equation can arise in very different contexts. Consider a model for HIV spread among drug users in a large housing complex. The total number of users N consists of I infected and S susceptible (uninfected) individuals. We assume that the rate of HIV transmission is proportional to the number of encounters between the infected and susceptible users, βSI, and that the death rate is proportional to the number of infected, µI, where β and µ are proportionality constants. We also assume that the total population N = I + S does not change (as tenants die, new ones move in). Then we have I˙ = βSI − µI = β(N − I)I − µI ! µ = β N − − I I, β which we recognise as the logistic equation in different notation. If N > µ/β, the only stable equilibrium solution µ I0 = N − > 0, β which means that the disease is constantly present in the population. By contrast, if N < µ/β, the solution I0 < 0 can be discarded as meaningless, and the other equilibrium solution I = 0 is now stable, which means that the disease disappears from the population. Hence the model suggests that the disease will be eradicated when N < µ/β. This can be accomplished by either of the following three methods: (1) decrease the population size N, say by resettling the tenants in smaller housing units; (2) decrease the transmission rate β, say by establishing a needle exchange program; (3) increase the death rate µ, say by withholding treatment. Clearly not all mathematically correct solutions are socially acceptable. 14

Example. If a fish population is growing logistically and is being harvested at a constant rate, the population size N(t) is described by N˙ = f (N) = (1 − N)N − H,

H = const.

Here N and t are measured in such units that a = K = 1, and H is the fish harvesting rate (the total catch per unit time) that we want to maximise. We describe a sustainable fishery by an equilibrium solution N0 : (1 − N0 )N0 − H = 0 or

 √ 1 1 ± 1 − 4H , 2 √ ′ hence H ≤ 1/4. Now f (N) = 1 − 2N, and we see that N0 =√(1 − 1 − 4H)/2 is √ unstable because f ′√ (N0 ) = 1 − 4H > 0, whereas N0 = (1 + 1 − 4H)/2 is stable because f ′ (N0 ) = − 1 − 4H < 0. When H = 1/4, the two equilibrium solutions merge into N0 = 1/2. It is tempting to choose H = 1/4 as the maximum sustainable harvesting rate, but is the corresponding population N0 = 1/2 stable? We cannot use our stability criterion because f ′ (1/2) = 0. We can integrate the differential equation though. When H = 1/4, we have

N0 =

1 N˙ = (1 − N)N − 4 2  1 , = − N− 2 and on separating variables and integrating, we get 1 N−

1 2

1 N(0) −

−

For the initial condition N(0) =

1 − ǫ, 2

1 2

= t.

ǫ>0

we have

1 ǫ − . 2 1 − ǫt If ǫ is small enough, we have N(0) that is very close to N0 = 1/2. Yet N(t) does not remain near N0 . Therefore the equilibrium N0 = 1/2 is unstable. In fact N(t) = 0 when t = ǫ−1 − 2, which means that the maximum harvesting rate H = 1/4 will lead to the collapse of a fishery. N(t) =

Example. We can prevent the collapse described in the previous example by introducing a feedback into the model: N˙ = f (N) = (1 − N)N − H, 15

H = κN.

Now the harvesting rate will decrease or increase depending on the current population size, which will have a stabilising effect on the population. Indeed, equilibrium solutions follow from (1 − N0 )N0 − κN0 = 0, which yields N0 = 0 and N0 = 1 − κ. Clearly we obtain a positive N0 if κ < 1. Since f ′ (N) = 1 − κ − 2N, we see that N0 = 0 is unstable because f ′ (N0 ) = 1 − κ > 0, whereas N0 = 1 − κ is stable because f ′ (N0 ) = κ − 1 < 0. We can express the harvesting rate H as a function of κ: H = κN0 = κ(1 − κ). We know from calculus that we can determine the maximum of a function by calculating its derivative and equating the derivative to zero. We obtain dH = 1 − 2κ = 0. dκ Hence the sustainable harvesting rate H is maximised by choosing κ = 1/2. Surprisingly, although the equilibrium is stable, the maximum H = κ(1 − κ) = 1/4 is the same as that for the unstable equilibrium of the previous example.

Exercises 1. Half of a spherical snowball melts in an hour. How long will it take for the remainder to melt? Assume that the snowball remains a sphere at all times and that its volume decreases at a rate proportional to its surface area. 2. Charcoal from a cave at an archaeological site is characterised by an average of 0.97 disintegrations of radioactive carbon-14 per minute per gram. Living wood is characterized by an average of 6.68 disintegrations of carbon-14 per minute per gram. The half-life of carbon-14 is approximately 5700 years. Estimate the date of occupation of the cave. 3. In an economic model, the price p of a commodity satisfies the equation p˙ = γ(D − Q), where the demand and supply are given by D(p) = a + bp and Q(p) = c + dp, and γ, a, b, c, and d are constants. You may assume γ > 0, b < 0, d > 0. Determine p(t) if p0 is the commodity price at t = 0. Use the solution to determine the limit of p(t) as t → ∞.

4. A new lecturer begins checking books out of the university library at the rate of one per day. Every week, he returns to the library 1/10 of the total number of books N(t) checked out. Assuming that N(t) can be approximated by a continuous function of time t, write down a differential equation for N(t). Specify a reasonable initial condition N(0) and solve the differential equation. How many library books does the lecturer have at any time after he has been around for a few years? 16

5. Suppose a quantity a − x of a substance A is present and changes into a substance B at a rate k1 (a − x), while the substance B is changing into the substance A at a rate k2 x. Determine x(t), assuming x(0) = 0. 6. The number of births per unit time in a population x(t) of zebras in a game park is Bx. The number of zebras that die per unit time is L + Dx where L is the number of deaths due to lions and Dx those due to other causes. Assume that B, L, and D are known constants (B 6= D). If x0 is the population at t = 0, find x(t) for t > 0. 7. The motion of a body falling in a viscous medium may be described by the equation mv˙ = −bv − mg, where m, g, b are positive constants. Given the initial conditions v(0) = 0 and x(0) = 0, determine the velocity v(t) and displacement x(t) for t ≥ 0. 8. The size of a population is described by

N0 πN , N˙ = sin 2 2N0 



N(t0 ) = N0 .

Determine N(t). What is the limit of N(t) as t → ∞? 9. The population size N(t) satisfies the differential equation N˙ = kN 2 , where k is a constant. Determine N(t) for t ≥ 0 if N(0) = N0 . Sketch the solution if k > 0 (explosive growth) and if k < 0 (catastrophic decay). 10. The seasonal growth of a tree is modelled by y˙ = α(1 + sin ωt)y,

y(0) = y0 ,

where α and ω are positive constants. Determine y(t). 11. A certain toxin destroys a strain of bacteria at a rate proportional to the product of the number of bacteria, N(t), and the amount of toxin present, T (t). If there were no toxin present, the bacteria would grow at a rate proportional to N(t). Solve the resulting ordinary differential equation for the number of bacteria at time t: N˙ = (a − T )N, assuming that T (t) = bt and N(0) = N0 , where a and b are known constants. Use the solution to determine the limit of N(t) as t → ∞. 12. The height h(t) of a growing person is modelled by h˙ = a(t)(h1 − h), 17

h(0) = h0 ,

a(t) = a0 + a1 t, where a0 , a1 , h0 , h1 are positive constants, and h0 < h1 . Determine h(t). What is the limit of h(t) as t → ∞?

13. A global warming model leads to the following equation for the average temperature of the Earth’s surface: T˙ = k1 exp(k2 t) − k3 (T − T0 ), where k1 , k2 , k3 are positive constants. Determine T (t), assuming that T (−∞) = T0 . 14. The rate of a chemical reaction A + B → AB is proportional to the product of the concentrations of A and B. At the same time, the compound AB itself breaks apart into A and B at a rate proportional to its own concentration x(t). The resulting equation for x(t) < 1 in appropriate units is as follows: x˙ = (1 − x)2 − kx. Derive the equilibrium solution x = x0 and determine whether it is stable. 15. The size N(t) of an insect population satisfies the differential equation N˙ = bN 2 − aN. Here a and b are known positive constants. Derive an equilibrium solution N0 > 0 and determine whether it is stable. 16. Consider a circular colony of bugs on a plate. If N is the total number of bugs, the Malthusian growth rate is given by r1 N. Bugs on the perimeter, however, suffer from cold and they die at a rate r2 N 1/2 . Hence the differential equation satisfied by N is N˙ = r1 N − r2 N 1/2 . Is there an equilibrium solution N0 > 0? If so, is it stable? 17. Your lecturer would like to lose some weight. He begins by constructing a mathematical model for dieting in which his weight W (t) changes due to food intake F (t), exercise E(t), and body metabolism M(t): ˙ = F − E − M, W

M = bW 3/4 .

Assume that F − E and b are positive constants. Derive an equilibrium solution W0 and determine whether it is stable.

18

Chapter 2 Second-order ordinary differential equations 2.1

Linear second-order equations

The order of a differential equation is the order of the highest derivative in the equation. A second-order equation for a function x(t) is an equation of the form x¨ = f (t, x, x). ˙ The notation x¨ = d2 x/dt2 is used when the independent variable t is time. Second-order differential equations arise quite often in applications. The most famous example is Newton’s second law of motion m¨ x = F (t, x, x), ˙ which governs the motion of a particle of mass m along the x-axis under the influence of a force F . Second-order differential equations are much more difficult to solve than first-order equations. Fortunately, many second-order equations of practical interest are linear. We only consider linear second-order differential equations: x¨ + px˙ + qx = f, where p(t), q(t), and f (t) can only depend on the independent variable t. No loss of generality results from taking the coefficient of x¨ to be 1 since this can always be accomplished by division. If f 6= 0, the equation is called inhomogeneous (or nonhomogeneous). If f = 0, it is homogeneous. Some theoretical questions about the existence and nature of solutions of differential solutions are very difficult. In applications, however, we typically expect that the general solution to a second-order equation contains two arbitrary constants that can be specified by initial conditions. For example, we can use the values of the function itself and its first derivative at a given time. 19

Example. Motion with a constant speed v0 corresponds to zero acceleration: x¨ = 0. On integrating twice, we obtain the general solution to this simple second-order equation: x˙ = C1 ,

x(t) =

Z

C1 dt = C1 t + C2 ,

where C1 and C2 are constants of integration. We use the initial conditions x(0) = x0 and x(0) ˙ = v0 to obtain x(t) = x0 + v0 t. Example. Suppose the temperature T (x) varies along a straight line (the x-axis) and does not depend on time. It turns out that in this case the heat equation simplifies to T ′′ (x) = 0. The general solution is T (x) = C1 x + C2 . For instance, if T (0) = Ti is the temperature inside a house, T (a) = To is the temperature outside, and a is the thickness of a window pane, then the temperature profile inside the window pane is x T (x) = Ti + (To − Ti ) , a

0 ≤ x ≤ a.

We have specified a unique solution by using the known values of T at two different points (boundaries): x = 0 and x = a. Problems of this type are called boundary value problems. Now we can calculate the rate of heat loss through the window, defined as Q = −κA

dT Ti − To = κA . dx a

Here κ is the coefficient of thermal conductivity of glass, and A is the area of the window pane. We see that Q′ (x) = 0, and so the equation T ′′ (x) = 0 means that the heat transfer rate Q is the same through each cross-section x = const. The thermal conductivity of air is about 16 times less than the thermal conductivity of glass. This is why the air gap in a double glazed window considerably reduces the heat loss through the window. Example. Second-order equations often arise in economics. Suppose z is the annual demand for goods and services in a national economy: z(t) = c(t) + x(t) + g(t), where c is the annual amount of consumption, x is the annual amount of investment, and g is the annual government expenditure. The so-called multiplier-accelerator model for the annual supply y(t) of those goods and services assumes that y˙ = λ(z − y),

λ = const

c = (1 − s)y,

s = const,

and where 0 < s < 1 and 1/s is termed the multiplier. Combining these equations, we have y˙ = λ[(1 − s)y + x + g − y] = λ(x + g − sy). 20

Hence x=

1 y˙ + sy − g, λ

which yields

1 y¨ + sy˙ − g. ˙ λ The final assumption of the model is that x˙ =

x˙ = k(v y˙ − x),

k = const,

v = const,

where v is termed the accelerator. Eliminating x and x˙ in this equation, we get 1 y¨ + λsy˙ − λg˙ = λk(v y˙ − y˙ − sy + g), λ or y¨ + py˙ + qy = f, where p = λ(s − kv) + k, q = λks, and f (t) = λ(kg + g). ˙ The model can describe a rich variety of types of behaviour, depending on the parameters p and q and the government policy, specified by f (t).

2.2

The homogeneous equation with constant coefficients

We can determine a complete solution of the homogeneous linear second-order equation x¨ + px˙ + qx = 0 for the special case in which p and q are constants. We begin by noting that the derivatives of the exponential function x(t) = A exp(λt) are constant multiples of the function itself: x˙ = λAeλt ,

x¨ = λ2 Aeλt .

We try to find the constant λ that makes x(t) = A exp(λt) a solution of the differential equation. On substituting this x(t) into the equation, we get (λ2 + pλ + q)Aeλt = 0. We see that x(t) = A exp(λt) is a nonzero solution if λ2 + pλ + q = 0. This is a quadratic equation, so there are two solutions: √ 2 √ 2 p p − 4q p − 4q p λ1 = − + , λ2 = − − . 2 2 2 2 Thus x(t) = A exp(λ1 t) and x(t) = B exp(λ2 t) are both solutions, whatever the values of the constants A and B, and the general solution is x(t) = Aeλ1 t + Beλ2 t . 21

If p2 > 4q, λ1 and λ2 are distinct real numbers, and the general solution x(t) contains two exponential functions. Example. The equation x¨ − x = 0 is a linear homogeneous equation with constant coefficients: p = 0, q = −1. Because p2 − 4q = 4 > 0, we have λ1,2 = ±1 and therefore x(t) = Aet + Be−t . √ If p2 < 4q, then λ1 and λ2 are complex numbers. If we denote α = 4q − p2 /2, we have x(t) = Ae−pt/2+iαt + Be−pt/2−iαt 



= e−pt/2 Aeiαt + Be−iαt . To specify a real-valued solution, we need Euler’s formula (L. Euler, 1748). We derive it formally by calculating i d h −iαt e (cos αt + i sin αt) dt = −iαe−iαt (cos αt + i sin αt) + e−iαt (−α sin αt + iα cos αt) = αe−iαt (−i cos αt + sin αt − sin αt + i cos αt) = 0.

Hence exp(−iαt)(cos αt + i sin αt) is a constant. We find its value by substituting t = 0, which yields exp(−iαt)(cos αt + i sin αt) = 1. On multiplying either side by exp(iαt), we obtain Euler’s formula eiαt = cos αt + i sin αt, and similarly (replacing α by −α) e−iαt = cos αt − i sin αt. Using Euler’s formula in the general solution x(t), we see that x(t) = e−pt/2 [(A + B) cos αt + i(A − B) sin αt] . To ensure that x(t) is real, we choose A and B such that A + B = C is real and A − B = −iD is imaginary. Remembering the definition of α, we finally obtain a real-valued solution ! √ √ 4q − p2 4q − p2 −pt/2 C cos x(t) = e t + D sin t . 2 2 Example. The equation x¨ + x = 0 is a linear homogeneous equation with constant coefficients: p = 0, q = 1. Because p2 − 4q = −4 < 0, we have α = 1 and therefore x(t) = C cos t + D sin t.

22

A snag arises if p2 = 4q. Because λ1 = λ2 , x(t) = A exp(−pt/2) contains only one arbitrary constant, so it cannot be the general solution. We find the general solution by searching for it in the form x(t) = y(t) exp(−pt/2). We have p2 x¨ = y¨ − py˙ + y e−pt/2 , 4 !

p x˙ = y˙ − y e−pt/2 , 2 



and substitution into x¨ + px˙ + qx = 0 yields p2 y¨ + q − y=0 4 !

or y¨ = 0,

y(t) = A + Bt.

Therefore, x(t) = (A + Bt)e−pt/2 is the general solution in the case p2 = 4q. Example. The equation x¨ + 4x˙ + 4x = 0 is a linear homogeneous equation with constant coefficients: p = 4, q = 4. Because p2 − 4q = 0, we have λ1,2 = −2 and therefore x(t) = (A + Bt) exp(−2t). If initial conditions are given by x(0) = 1 and x(0) ˙ = 3, for example, then we have 1 = A and 3 = B − 2A, and finally x(t) = (1 + 5t) exp(−2t). Example. L. F. Richardson (1939) suggested a model that describes the relation between two nations, each determined to defend itself against a possible attack by the other. Suppose x(t) is the war potential (armaments) of nation 1 and y(t) is the war potential of nation 2. The rate of change of x(t) depends on the war readiness y(t). In the simplest model, we represent this rate by ky where k > 0 is a constant. Similarly, we model the rate of change of y(t) by lx where l > 0 is a constant. Neglecting all other factors such as the cost of armaments, we write x˙ = ky,

y˙ = lx.

Differentiating the first equation with respect to t and using the second equation to eliminate y, ˙ we have x¨ − klx = 0. We solve this linear homogeneous equation to get √

x(t) = Ae It follows that

klt

s

+ Be−

√

klt

.

√  l  √klt x˙ − Be− klt . Ae y(t) = = k k The integration constants A and B are determined by initial conditions. If A > 0, both x(t) and y(t) tend to infinity. This can be interpreted as war.

23

Example. Suppose now that two armies are engaged in combat, with x(t) and y(t) denoting the number of combatants of the x and y forces. Who wins the battle? F. W. Lanchester (1916) tried to answer this question using the following model. The rate of change of x(t) depends on the fighting effectiveness of y, which we assume to be given by −by, operational losses (accidents, desertions), which we assume to be given by −ax, and reinforcements P (t). These assumptions enable us to write x˙ = −ax − by + P (t),

x(0) = x0 .

y˙ = −cx − dy + Q(t),

y(0) = y0 .

Similarly for the y-force,

If the operational losses and reinforcements are negligible (a = d = P = Q = 0), we have x˙ = −by,

x(0) = x0 ,

y˙ = −cx,

y(0) = y0 .

Note that we must also have x(0) ˙ = −by0 . Now, on differentiating the equation for x˙ and using the second equation to eliminate y, ˙ we get x¨ − bcx = 0, and the general solution is

√

x(t) = Ae

bct

+ Be−

√

bct

.

The initial conditions x(0) = x0 , x(0) ˙ = −by0 yield √ x(t) = x0 cosh( bct) −

s

√ b y0 sinh( bct), c

and similarly

r √ √ c y(t) = y0 cosh( bct) − x0 sinh( bct). b We can eliminate the explicit t-dependence by dividing one differential equation by the other: dy y˙ cx = = . dx x˙ by This is a separable first-order equation. We integrate it to obtain an interesting result:

by 2 − cx2 = by02 − cx20 = const. Since this quantity never changes sign, only one of x and y can ever be zero. Therefore, the y-force wins (y > 0 when x = 0) if by02 − cx20 > 0, and the x-force wins (x > 0 when y = 0) if by02 − cx20 < 0. If a 50,000 y-force is fighting a 70,000 x-force with an equal troop effectiveness, b = c, the larger x-force obviously wins. Suppose, however, that the y-force meets sequentially x-armies and 30,000. The first battle results in a y-force √ of 40,000 2 victory with a margin of 50, 000 − 40, 0002 = 30, 000, which is just enough to force a draw with the second x-army. This illustrates the importance of concentration in tactics. 24

2.3

Linear oscillations

We often need to describe systems that exhibit periodic, oscillatory behaviour, for instance the oscillations of a quartz crystal in a watch, the beating of the heart, the periodic rise and fall of the ocean level due to tides, or business cycles in economics. The simplest oscillatory system consists of a mass m attached to a spring, which moves without friction in a straight line. If x denotes the displacement of the mass from its equilibrium position and the spring exerts the restoring force F = −kx (k > 0) on m, then the equation of motion m¨ x = F is x¨ + ω 2 x = 0,

ω2 =

k m

with the general solution x(t) = A cos ωt + B sin ωt. Here ω is called the circular frequency, and the integration constants A and B are specified by initial conditions. Example. If x(0) = 2 and x(0) ˙ = 0, then A = 2, B = 0, and so x(t) = 2 cos ωt. √ If we define an amplitude a0 = A2 + B 2 and a phase φ0 = arctan(A/B) of the oscillations, we can write the integration constants as A = a0 sin φ0 and B = a0 cos φ0 . Hence we can express the general solution as follows: x(t) = a0 sin(ωt + φ0 ). This form of the solution makes it clear that the oscillations are indeed periodic, x(t+T ) = x(t), with the period r 2π m = 2π . T = ω k Example. Suppose a spherical buoy of radius R is floating half submerged in water. We have mg = F0 in equilibrium, where m is the mass of the buoy, g is the acceleration due to gravity, and F0 is an expulsion (Archimedes’s) force equal to the weight of the displaced water. If the buoy is depressed slightly and then released, a restoring force pushes it upward, causing oscillations. If the friction of the water is neglected, the oscillations are described by m¨ x = ∆F , where x(t) is the vertical displacement of the buoy, and ∆F = F − F0 is the change in the expulsion force. If the displacement is so small that x ≪ R, we can calculate ∆F = F0 ∆V /V , where ∆V is the small change in the submerged volume: ∆V πR2 x 3x =− =− . 3 V 2πR /3 2R Hence 3x 3x m¨ x = ∆F = − F0 = − mg 2R 2R or 3g x¨ + x = 0, 2R 25

q

and so the period of the resulting oscillations is T = 2π 2R/(3g). Consider now the possibility that there is a friction force Ff that will damp the oscillations. We assume that the force is proportional to the speed of the oscillating mass: Ff = −rv = −r x, ˙

r > 0.

The equation of motion becomes m¨ x + r x˙ + kx = 0 or, introducing β = r/(2m) and ω02 = k/m, x¨ + 2β x˙ + ω02 x = 0. From now on, ω02 = k/m denotes the natural frequency, that is the frequency at which the system would oscillate if there were no friction force. As usual, we solve this linear equation by assuming that x(t) = A exp(λt), which leads to λ2 + 2βλ + ω02 = 0 and so λ1,2 = −β ± iω,

ω=

q

ω02 − β 2 .

We observe that three different types of solution are possible, depending on whether β < ω0 , β > ω0 or β = ω0 . If the damping is weak, β < ω0 , we say that the system is underdamped. In this case the general solution is x(t) = (A cos ωt + B sin ωt)e−βt = a0 sin(ωt + φ0 )e−βt . Thus the mass executes oscillations, but their amplitude decreases exponentially with time. We define the period T of the decaying oscillations as the time between successive maxima of the function x(t): T =

2π 2π 2π =q . =q ω k/m − r 2 /(4m2 ) ω02 − β 2

If the damping is strong, β > ω0 , we say that the system is overdamped. In this case λ1 and λ2 are real and both negative. Hence the general solution is x(t) = Ae−|λ1 |t + Be−|λ2 |t . The friction force is so large that there are no oscillations. Instead x(t) decays exponentially, so that the mass approaches the equilibrium position x = 0 as t → ∞. Depending on initial conditions, the mass can cross x = 0 at most once. In the mathematically interesting case of critical damping, β = ω0 , we have x(t) = (A + Bt)e−βt . The behaviour of the solution is similar to its behaviour in the overdamped case since the exponential decay overcomes an algebraic growth. From a physical point of view, however, this case is insignificant because in practice it is very unlikely that β = ω0 exactly. 26

2.4

Equilibrium and stability

We are often interested in equilibrium solutions of second-order differential equations, which are solutions that do not depend on time: x = x0 is an equilibrium solution if x¨ = x˙ = 0. In mechanical problems, x¨ = 0 means that all forces cancel, and x˙ = 0 means that there is no motion. For a second-order equation of the form x¨ = f (x, x), ˙ we find all equilibrium solutions x = x0 by solving f (x0 , 0) = 0. An equilibrium x = x0 is called asymptotically stable if every solution x(t) that starts near x0 remains near x0 and x(t) → x0 as t → ∞. As for first-order differential equations, we can determine which equilibrium solutions are asymptotically stable by using a linearised stability analysis. Suppose an equilibrium solution x = x0 satisfies f (x0 , 0) = 0. To investigate the stability of x = x0 , we need to determine the behaviour of a solution x(t) that is initially very close to x0 . Hence we only need to know how f (x, x) ˙ behaves near x0 . We expand f (x, x) ˙ in a Taylor series as a function of two variables:



∂f ∂f f (x, x) ˙ = f (x0 , 0) + (x − x x˙ + . . . 0) + ∂x x0 ,0 ∂ x˙ x0 ,0 ≈





∂f ∂f (x − x0 ) + x, ˙ ∂x x0 ,0 ∂ x˙ x0 ,0

where we kept only the linear terms in the expansion and used f (x0 , 0) = 0 and x˙ 0 = 0. If ξ(t) = x − x0 denotes a deviation from the equilibrium solution, we have ξ¨ = ˙ or approximately f (x0 + ξ, ξ), ξ¨ + pξ˙ + qξ = 0, where





∂f q=− . ∂x x0 ,0

∂f , p=− ∂ x˙ x0 ,0

This is a second-order homogeneous equation with constant coefficients. Its general solution is ξ(t) = A exp(λ1 t) + B exp(λ2 t) where as before λ1,2

q 1 = −p ± p2 − 4q . 2 



We require for asymptotic stability that the deviation ξ remain small and ξ → 0 as t → ∞. 2 Hence we require that the real parts of both λ1 and λ2 be negative. If √ q2 < p /4, both roots are real, and we require that the larger root be negative: (−p + p − 4q)/2 < 0, which is the case if q > 0. If q > p2 /4, both roots are complex, but their real parts are 27

the same, and so we require that −p/2 < 0, which is the case if p > 0. Summarising these results, we obtain p > 0, q > 0. Thus we have the following stability criterion. If x¨ = f (x, x) ˙ and f (x0 , 0) = 0, then x = x0 is a stable equilibrium solution if



∂f < 0, ∂x x0 ,0

∂f < 0. ∂ x˙ x0 ,0

The equilibrium solution is unstable if either of the two inequalities is not satisfied. Example. The van der Pol equation x¨ + µ(x2 − 1)x˙ + x = 0,

µ>0

was encountered in the study of electric circuits (B. van der Pol, 1920). It is clear that x0 = 0 is the only equilibrium solution. Since f (x, x) ˙ = −x − µ(x2 − 1)x, ˙ we have

∂f = −1, ∂x x0 ,0



∂f = −µ(x20 − 1) = µ. ∂ x˙ x0 ,0

Thus the equilibrium solution is unstable.

Sometimes other definitions of stability may be useful. For example, if ξ(t) = x − x0 remains small but does not tend to zero as t → ∞, the equilibrium solution x0 is said to be neutrally stable. Example. The only equilibrium solution of the linear oscillator equation x¨ + ω 2 x = 0 is x0 = 0. Now f (x, x) ˙ = −ω 2 x, and so (∂f /∂x)|x0 ,0 = −ω 2 < 0 but (∂f /∂ x)| ˙ x0 ,0 = 0. Thus the equilibrium is not asymptotically stable. It is neutrally stable, however, since the general solution x(t) describes periodic oscillations about x0 . Generally, x = x0 is a neutrally stable equilibrium solution of the equation x¨ = f (x) if f ′ (x0 ) < 0. We can use linearisation to study asymptotic stability of equilibrium solutions to x¨ = f (x, x) ˙ and neutral stability of equilibrium solutions to x¨ = f (x). When the context is clear, we simply say that an equilibrium is either stable or unstable. It turns out, however, that linearisation cannot be used to study neutral stability of equilibrium solutions to x¨ = f (x, x). ˙ A. M. Lyapunov (1892) discovered a powerful but more complicated method for investigating stability in this case. A solution is said to be stable in the sense of Lyapunov if it is either neutrally stable or asymptotically stable.

28

2.5

Inhomogeneous equations

We now consider a second-order inhomogeneous equation x¨ + px˙ + qx = f,

f 6= 0

and suppose xp (t) is a particular solution of this equation, so that x¨p + px˙ p + qxp = f. On subtracting one equation from the other, we obtain a homogeneous equation x¨h + px˙ h + qxh = 0 for the function xh (t) = x(t) − xp (t). We see that the general solution x(t) of an inhomogeneous differential equation is the sum of the general solution xh of the homogeneous equation and a particular solution xp of the inhomogeneous equation: x(t) = xh (t) + xp (t). If both p and q are constants, we can determine a complete solution of the homogeneous equation, and therefore we can write the general solution of the inhomogeneous equation as x(t) = xh (t) + xp (t) = Aeλ1 t + Beλ2 t + xp (t) (suitably modified if λ1 and λ2 are complex numbers or if λ1 = λ2 ). For example, if f = const, we can simply choose xp = f /q = const. More generally, if f (t) is a simple function, we can find xp by the method of undetermined coefficients. The idea is to search for xp (t) in a form suggested by the form of f (t). For instance, if f (t) = atn , we seek xp (t) = C0 + C1 t + . . . + Cn tn . If f (t) = a exp(αt), we seek xp (t) = C exp(αt). If f (t) = a1 cos(αt) + a2 sin(αt), we seek xp (t) = C1 cos(αt) + C2 sin(αt). We determine the coefficients in xp (t) by substituting xp into the inhomogeneous differential equation. We should be careful if f (t) = a exp(αt) and α coincides with λ1 or λ2 . If α = λ1 but λ1 6= λ2 , we seek xp (t) = Ct exp(λ1 t). We can justify this assumption by an argument similar to that used in the derivation of the solution of the homogeneous equation in the case λ1 = λ2 . Finally, if α = λ1 = λ2 , we seek xp (t) = Ct2 exp(λ1 t). Example. A linear frictionless oscillator, driven by a periodic external force, is described by x¨ + ω02 x = sin ωt. Consider first the case ω 6= ω0 . We try xp (t) = C sin ωt. On substituting this into the equation, we get −Cω 2 sin ωt + Cω02 sin ωt = sin ωt, which can be satisfied only if C = 1/(ω02 − ω 2 ). Thus we have the general solution x(t) = A cos ω0 t + B sin ω0 t + 29

ω02

1 sin ωt. − ω2

If the initial conditions are given by x(0) = 0 and x(0) ˙ = 0, for example, we have A = 0, and the solution is

Bω0 +

1 x(t) = 2 ω − ω02



ω = 0, ω02 − ω 2

ω sin ω0 t − sin ωt . ω0 

Example. Now suppose the frequency of the external force is equal to the natural frequency of the system. We cannot just substitute ω = ω0 into the above solution for x(t) since that would lead to division by zero. We can, however, define the solution in this case as a limit: (ω/ω0 ) sin ω0 t − sin ωt . x(t) = lim ω→ω0 ω 2 − ω02

On using L’Hospital’s rule to evaluate the limit, we find x(t) =

d [(ω/ω0 ) sin ω0 t − dω d (ω 2 − ω02 ) dω



sin ωt]

1 t = sin ω0 t − cos ω0 t. 2 2ω0 2ω0



ω=ω0

Alternatively, we could find this solution using the method of undetermined coefficients. When ω = ω0 , seek a particular solution xp (t) = C1 t cos(ω0 t) + C2 t sin(ω0 t) and determine the constants C1 = −1/(2ω0 ) and C2 = 0 by substituting xp (t) into the differential equation. It follows that the general solution is x(t) = xh (t) + xp (t) = A cos ω0 t + B sin ω0 t −

t cos ω0 t. 2ω0

As before, the integration constants A and B are specified by the initial conditions. We see that even though the external force is periodic, the amplitude of the solution grows with time (strictly speaking, only until a friction force is no longer non-negligible). The phenomenon of a strong response of an oscillator when driven at the right frequency is called resonance. An impressive example of resonance is the vocal coach Jaime Vendera holding a note and shattering a wine glass when the frequency of the note is equal to the natural frequency of the glass. An everyday example of resonance is tuning. When we tune a radio, we change the parameters of an electric circuit to match the carrier frequency of a particular station. Example. Consider an electric circuit consisting of an inductor of inductance L, a resistor of resistance R, a capacitor of capacitance C, and a battery producing an electromotive

30

force (voltage) E(t), all connected in series. The governing equation for the electric current I(t) in the circuit is Q LI˙ + RI + = E, C ˙ we have where Q is the electric charge. On differentiating this equation and using I = Q, I ˙ LI¨ + RI˙ + = E. C For example, if E = E0 sin(ωt + β) where E0 and β are constants, the general solution of this inhomogeneous equation is I(t) = Ih (t) + Ip (t). Here the general solution of the homogeneous equation is √ ! 4CL − R2 C 2 −Rt/(2L) t+δ , Ih (t) = Ae sin 2CL where A and δ are integration constants, and we assume the underdamped case R < q 2 L/C. A particular solution of the inhomogeneous equation is Ip (t) =

ωCE0 [RCω sin(ωt + β) + (1 − CLω 2 ) cos(ωt + β)] . (RCω)2 + (1 − CLω 2 )2

The homogeneous solution Ih decays with time, so Ih is called the transient current, √ whereas Ip is called the steady state current. Resonance occurs if ω = 1/ CL, in which case the amplitude of Ip reaches the maximum value of E0 /R. Note that the solution remains bounded as long as the resistance R is nonzero, but its amplitude will be very large if R is very small.

Exercises 1. Solve the following ordinary differential equation: u′′ + u′ + u = 1. Comment: the problem appears in “Heard on the Street: Quantitative Questions from Wall Street Job Interviews” by T. F. Crack. 2. Radium decays to radon which decays to polonium. This two-step process of radioactive decay is described by the following equations: N˙ 1 = −λ1 N1 ,

N1 (0) = N0 ,

N˙ 2 = λ1 N1 − λ2 N2 , 31

N2 (0) = 0.

Derive a second-order differential equation with constant coefficients for N2 (t). Determine N1 (t) and N2 (t). Investigate separately the cases λ1 6= λ2 and λ1 = λ2 .

3. A particle slides without friction in a long slender tube that rotates in a vertical plane about its centre with constant angular frequency ω. If the tube is horizontal at time t = 0 and g > 0 is a constant gravitational acceleration, then the particle displacement u(t) from the centre is governed by u¨ − ω 2 u = −g sin ωt. Find the general solution u(t) and solve the initial value problem u(0) = 0, 2ω u(0) ˙ = g. 4. Imagine a tunnel connecting two points on the surface of the Earth, which passes through the centre of the Earth. Neglecting air resistance, the only force acting on a body of mass m in the tunnel is the gravitational force: F = −mgr/R. Here g ≈ 9.8 m s−2 is the gravitational acceleration at the surface, r is the distance from the Earth’s centre, and R ≈ 6400 km is the radius of the Earth. Write down a second-order differential equation that describes the motion of a body in the tunnel. Determine the general solution r(t) to the equation and solve the initial value problem r(0) = R, r(0) ˙ = 0. For these initial conditions, calculate the time it takes for the body to travel between the two ends of the tunnel and the maximum speed reached in the course of the travel. 5. Determine whether x0 = 0 is a stable solution of the following equation: x¨ + 6x˙ − 4x = 0. 6. Suppose nonlinear oscillations are described by x¨ + x˙ = x − x2 . Determine all equilibrium solutions and investigate their stability. 7. A mathematically-minded vandal wishes to break a hot-water radiator of mass m away from its foundation, but finds that he cannot achieve this by applying steadily the greatest force of which he is capable. He finds, however, that he can apply a periodic force f (t) = f0 sin(ωt). The foundation of the radiator resists its movement by a force proportional to its displacement. The resulting differential equation for the displacement x(t) is as follows: k f (t) x¨ + x = . m m Assuming that the frequency ω is chosen so that k/m = ω 2 , determine x(t) for the initial conditions x(0) = 0, x(0) ˙ = 0. Will the vandal achieve his goal?

32

Chapter 3 Elements of classical mechanics 3.1

Kinematics

Classical mechanics involves solving Newton’s equations of motion to explain and predict how various objects move—how projectiles and aircraft move through the air, how spacecraft move around the Earth, how planets move around the Sun. Kinematics describes the motion of objects, without reference to the forces causing the motion. Dynamics deals with the forces that produce motion. Numerous applications of calculus arise in classical mechanics. When the size of a moving object is small enough, we can treat the object as a point mass, or particle. A particle is a convenient idealisation of a body with mass but no size. The laws of motion are simpler for particles than for extended bodies. If a particle moves along the x-axis, its displacement x(t) from the origin is a function of time t. We define the instantaneous velocity v = x˙ and acceleration a = v˙ = x¨. If a(t) is known, we can find v(t) and x(t) by integration: v(t) =

Z

a(t)dt,

x(t) =

Z

v(t)dt.

Example. If acceleration a = const, we have v(t) = v0 + at,

1 x(t) = x0 + v0 t + at2 , 2

where the initial conditions are v(0) = v0 and x(0) = x0 . Example. The acceleration of an aircraft during take-off is a = 3 m/s2 . If the initial speed v0 = 0 and the take-off speed is v1 = 60 m/s, then the take-off time is t1 = v1 /a = 20 s, and the required runway length is x(t1 ) − x(0) = at21 /2 = v12 /(2a) = 600 m. The gravitational attraction of the Earth accelerates all falling bodies downwards. The gravitational acceleration g does not depend on the mass of the body. Near the surface of the Earth, z¨ = −g, g ≈ 9.8 m/s2 , 33

where we use z to denote the height above ground and neglect air resistance. Example. Assuming z(0) = h and v(0) = 0, we integrate z¨ = −g and get z˙ = −gt and z(t) = h − gt2 /2. The time of the fall tf is defined by z(tf ) = 0. Hence we find q tf = 2h/g if air resistance is neglected. For instance, the height of the observation deck of Auckland Sky Tower is h = 192 m. Consequently the SkyJump duration should be q tf = 2h/g ≈ 6.3 s. The actual duration would be about 11 seconds because you would be attached to a wire that slows you down. Example. You drop a stone down a well of unknown depth h and hear the splash T = 4 s later. What is h, if the speed of sound isqcs ≈ 340 m/s and air resistance can be neglected? The time for the stone to fall down is 2h/g, and the time for the sound to travel up is h/cs . It follows that q T = 2h/g + h/cs . √ This is a quadratic equation for h, which we solve to obtain √ Remembering that

√

h=

q

− 2/g ±

q

2/g + 4T /cs

2/cs

.

h > 0, we choose the positive root and find 

h=

q

− 2/g +

q

2/g + 4T /cs



2/cs

When gT /cs ≪ 1 we have approximately s

2 4T = + g cs

s s

2gT 2 1+ ≈ g cs

2

s

≈ 72 m.

gT 2 1+ g cs 



√ (remember that 1 + x ≈ 1 + x/2 for small x). Therefore, in the limit cs → ∞, the formula for h reduces to h = gT 2 /2, as it should. In three dimensions, the position r of a particle is a vector. We specify a vector by its components. In Cartesian coordinates we write r = xi + yj + zk, where i, j, k are unit vectors, pointing along the three perpendicular axes. We can also write the same vector simply as r = (x, y, z). The velocity v(t) of a particle is the time derivative of its position r(t), v = dr/dt. Similarly the acceleration a(t) is the time derivative of the velocity, a = dv/dt. To differentiate a vector, we differentiate its components. The unit vectors i, j, k do not depend on time, so their time derivatives vanish. This enables us to write v = r˙ = (x, ˙ y, ˙ z), ˙ 34

a = v˙ = r¨ = (¨ x, y¨, z¨). Example. If r = (1 + t, t2 , 1 − t3 ), we calculate v = r˙ = (1, 2t, −3t2 ) and a = v˙ = (0, 2, −6t). Example. The cycloid is the curve traced out by a point on the rim of a wheel when it rolls along a straight line. If R is the radius of the wheel, and θ is the angle through which its radius vector rotates, then the position of a point on the circumference is r = R(θ − sin θ, 1 − cos θ, 0). This is a parametric representation of the cycloid. If the rolling speed Rθ˙ = const, we can introduce ω = θ˙ = const and calculate v = r˙ = ωR(1 − cos ωt, sin ωt, 0), a = v˙ = ω 2 R(sin ωt, − cos ωt, 0). We observe that v = 0 when ωt = 2πk, k = 0, ±1, ±2, ...: the point where the wheel touches the ground has a zero instantaneous velocity.

3.2

Dynamics and gravity

Isaac Newton’s second law of motion (I. Newton, 1687) states that the acceleration a of a body of mass m is equal to F /m, where F is the total force acting on the body. We can write ma = F . For motion in a straight line, we have a = x¨, and so m¨ x = F , where in general the force F depends on x, x, ˙ and t. We can integrate the resulting second-order differential equation without much trouble only in a few simple cases. Example. The slowing-down of a bus is described by m¨ x = −F0 where F0 = const. If v(0) = v0 and x(0) = x0 , then v(t) = x˙ = v0 − F0 t/m. Hence the stopping time, defined by v(ts ) = 0, is given by ts = mv0 /F0 . To determine the stopping distance x(ts ) − x(0), we integrate one more time: F0 2 mv0 F0 mv0 x(ts ) − x0 = v0 ts − ts = v0 − 2m F0 2m F0 

2

=

mv02 . 2F0

In three dimensions, a = r¨ and Newton’s law of motion for a particle, m¨ r = F , is equivalent to a system of three second-order differential equations. To describe the motion of N particles, we would need to solve a system of 3N equations. Numerical methods are typically necessary for solving such systems, although sometimes we are able to find an exact solution. 35

Example. The motion of a charged particle, say a proton of mass m and electric charge e > 0, in an electric field E is described by m¨ r = F , where F = eE. If the electric field E = α(y, x, 0) where α is a positive constant, then we have to solve the system m¨ x = αey,

m¨ y = αex,

m¨ z = 0.

˙ Suppose initially r(0) = (x0 , 0, z0 ) and r(0) = (0, z = 0 is q 0, 0). Then the equation m¨ easily integrated to give z = z0 . We define ω = αe/m and rewrite the remaining two equations as x¨ = ω 2y, y¨ = ω 2 x. The symmetry of the system suggests that we introduce new variables u1 = x + y and u2 = x − y. On taking the sum and the difference of the two equations, we get u¨1 = ω 2 u1 ,

u¨2 = −ω 2 u2.

The initial conditions are u1 (0) = u2 (0) = x0 and u˙ 1 (0) = u˙ 2 (0) = 0. On solving the resulting uncoupled equations and using the initial conditions, we find u1 (t) = x0 cosh(ωt), Finally,

u2 (t) = x0 cos(ωt).

1 x(t) = [u1 (t) + u2 (t)] = 2 1 y(t) = [u1 (t) − u2 (t)] = 2

x0 [cosh(ωt) + cos(ωt)], 2 x0 [cosh(ωt) − cos(ωt)]. 2

A body of mass m near the surface of the Earth experiences the gravitational force F = (0, 0, −mg), where the gravitational acceleration g = 9.8 m/s2 can be assumed to be constant in most applications. If no other forces are present, a = F /m reduces to the simple kinematic equations x¨ = 0, y¨ = 0, z¨ = −g. In practice, neglecting air resistance can lead to huge errors (unless you live on the Moon). Example. Taking air drag into account, the equation of motion for a falling body is m¨ z = −mg + Fr , where Fr is the resistive force of the air. It often turns out that Fr = bv 2 is a good approximation to the air drag, where the speed v = z˙ and b = const. Consequently, we have a separable differential equation: v˙ = or dt =

b 2 v −g m g dv, − g2

r2v2

36

where r 2 = bg/m is introduced for convenience. On integrating using partial fractions, we get t=

Z

1 g dv = 2 2 2 r v −g 2

Z

!





1 rv − g 1 1 dv = − ln + C. rv − g rv + g 2r rv + g

Suppose the initial condition is v(0) = 0. Then the integration constant C = 0 and g v(t) = − tanh(rt). r q

We see that v → −g/r = − mg/b as t → ∞. This limiting value of v is called the terminal speed. For instance, m ≈ 70 kg and b ≈ 700 kg/m for a falling parachutist, so the terminal speed is ≈ 1 m/s. Finally, replacing v by dz/dt and performing another integration, we find the displacement z − z0 =

Z

0

t

gZt g v(t)dt = − tanh(rt)dt = − 2 ln [cosh(rt)] . r 0 r

Although the gravitational acceleration can be considered constant near the surface of the Earth, this is not the case when the height above the surface is large enough. According to Newton’s law of gravitation, the force of gravity is inversely proportional to the square of the distance to the Earth’s centre: F = −G

mME . r2

Here the minus sign indicates attraction, G = 6.67 × 10−11 N m2 kg−2 is a proportionality constant, ME is the mass of the Earth, m is the mass of a falling body, and r is its distance from the centre of the Earth. The dimensional units of G mean that when we measure m and ME in kilograms and r in metres, the formula gives us F measured in the units of force called newtons. Now, as long as the height z above the surface of the Earth is small compared with the radius of the Earth RE , we can approximate r = RE + z ≈ RE , which leads to F ME ME . z¨ = = −G ≈ −G 2 m (RE + z)2 RE Numerically, we have RE = 6.38 × 106 m and ME = 5.98 × 1024 kg. Consequently, g=G

24 ME m m −11 5.98 × 10 = 6.67 × 10 ≈ 9.8 2 . 2 6 2 2 RE (6.38 × 10 ) s s

We see that g is independent of the mass m of the body, and the gravitational acceleration can indeed be assumed constant as long as z ≪ RE . So far we only discussed situations in which a force acts on a body of constant mass. If a force F is acting on a body of variable mass m moving with velocity v, Newton’s law of motion asserts that the force F is equal to the rate of change of momentum mv: F =

d(mv) . dt 37

This equation simplifies to m¨ r = F when m = const. For motion in one dimension, we have F = d(mv)/dt. Example. We have F = 0 for a rocket travelling in interplanetary space. If m(t) and v(t) are the mass and speed of the rocket, dm < 0 and dv > 0 are their differentials, and u = const is the speed of the exhaust gas, expelled by the rocket, then the differential of the rocket momentum mdv + vdm must be equal to the differential of the exhaust gas momentum (−dm)(u − v). Here we have to use (u − v) rather than u because u is defined as the speed relative to the moving rocket. Thus we have mdv + vdm = −udm + vdm or mdv + udm = 0. This is a separable first-order equation for the speed v as a function of the rocket mass m. We have Z m(t) Z v(t) dm dv = −u , v(0) m(0) m and so

"

#

m(0) v(t) − v(0) = u ln . m(t)

This is the so-called rocket equation (K. Tsiolkovsky, 1903). It shows that the speed of the rocket can be very high if m(0)/m(t) is made large enough.

3.3

Work and energy

For a freely falling body, we integrated z¨ = −g and obtained v(t) = z˙ = −gt and z(t) = h − gt2 /2. On eliminating t, we get z = h − v 2 /(2g), which yields 1 2 mv + mgz = mgh, 2 where the right-hand side does not depend on time. We see that the gain in the kinetic energy mv 2 /2 is equal to the work done by the gravitational force mg(h − z). This result illustrates the principle of conservation of energy. More generally, if we have m¨ x = F in one dimension and F = F (x), we can use the chain rule of differentiation to write d2 x dv dv dx dv d 1 2 = = = v= v . 2 dt dt dx dt dx dx 2 

The equation of motion becomes m

d 1 2 v = F (x). dx 2 



38



On integrating we see that the change in kinetic energy is equal to the work done by the position-dependent force F (x): m 2 m 2 v − v0 = 2 2

Z

x x0

F (s)ds = −U(x),

where the potential energy U(x) = − xx0 F (s)ds is introduced. We see that the total particle energy m E = v 2 + U(x) 2 is a constant of motion because R

dE d m 2 d m 2 = v + U(x) = v = 0. dt dt 2 dt 2 0 







The result is valid for any form of F (x). Example. The motion of a projectile moving radially away from the Earth is described by r˙ = v and mME mv˙ = −G 2 r or R2 v˙ = −g 2E , r 2 where r is the distance from the centre of the Earth and g = GME /RE . When r significantly exceeds the Earth’s radius RE , we can no longer assume that the gravitational acceleration is constant. To integrate the equation of motion, we rewrite v˙ as dv dr d dv = = dt dr dt dr On integrating d dr





1 2 v . 2 

1 2 R2 v = −g 2E 2 r 

we obtain for v = v(r):

 1 2 1 1 2 v − v02 = gRE , − 2 r RE where we used the initial condition v(RE ) = v0 . What is the minimal speed required to escape Earth’s gravity? If v → v1 in the limit r → ∞, we have v02 = v12 + 2gRE . We obtain the minimal v0 by setting v1 = 0. Hence the required escape speed is 

v0 =



q

2gRE ≈ 11.2 km/s,

where we used g = 9.8 m/s2 and RE = 6.38 × 106 m.

Example. More advanced methods are needed to derive constants of motion for motion in three dimensions. For instance, it turns out that the energy  m 2 E= x˙ + y˙ 2 + z˙ 2 − αexy 2 39

is a constant of motion for a charged particle of mass m and electric charge e moving in the electric field E = α(y, x, 0). We can use Newton’s equation m¨ r = F with F = eE to check that E˙ = m(x¨ ˙ x + y˙ y¨ + z¨ ˙ z ) − αe(xy ˙ + xy) ˙ = (m¨ x − αey)x˙ + (m¨ y − αex)y˙ + z¨z˙ = 0x˙ + 0y˙ + 0z˙ = 0.

3.4

Planetary orbits and Kepler’s laws

Although it is very natural to think that the Earth is fixed at the centre of the Universe and everything else moves around it, in reality the Earth and other planets move around the Sun. Johannes Kepler (1609, 1619) analysed a large collection of data on the position of the planets at various times and established three empirical laws of planetary motion. 1. The orbit of every planet is an ellipse with the Sun in one focus. 2. The line segment from the Sun to a planet sweeps out equal areas in equal times. 3. The square of the orbital period of a planet is proportional to the cube of the semimajor axis of the ellipse. Kepler did not have a theory for planetary motion. In a classical example of mathematical modelling, Isaac Newton (1687) created such a theory. He discovered how to derive the inverse square law of gravitational attraction from Kepler’s laws, and also how to derive Kepler’s laws from the inverse square law. We are going to derive Newton’s inverse square law of gravitational attraction from Kepler’s laws of planetary motion. The ratio of the mass of a planet to that of the Sun is very small—about 10−3 for Jupiter and even smaller for other planets. This is why we can neglect the motion of the Sun and assume that planets move around a fixed point coinciding with the centre of the Sun. We would need to relax this assumption for binary stellar systems, for instance, in which two stars of comparable masses orbit around their common centre of mass. We also assume that the Sun and planets are particles, that is, points at which mass is concentrated. Newton in fact showed that the Sun and planets behave like particles under the inverse square law of attraction. We begin by rewriting Kepler’s laws mathematically. It is convenient to describe the motion of a planet around the Sun using polar coordinates. We place the Sun at the origin of the coordinate system and express the position vector r of a moving particle (planet) in the form r = xi + yj = r cos θi + r sin θj. Kepler’s first law states that the orbit of a planet is an ellipse. The formula for an ellipse in polar coordinates is p r= , p > 0, 0 < e < 1. 1 + e cos θ 40

The parameter e is called the eccentricity (not to be confused with the base of natural logarithms). When e = 0, r = p describes a circle. To show that the formula describes an ellipse when e > 0, we write the formula as r = p − ex and square both sides. We have r 2 = x2 + y 2 = p2 − 2epx + e2 x2 or (1 − e2 )x2 + y 2 = p2 − 2epx, and so x2 +

y2 = pa − 2eax, (1 − e2 )

where we introduced the length a = p/(1 − e2 ) (not to be confused with the acceleration a). Now we have a x2 + 2eax + y 2 = pa, p or a (x + ea)2 + y 2 = pa + e2 a2 = a2 , p and finally

(x + ea)2 y2 + = 1. a2 ap

This is indeed the usual formula √ for an ellipse with the semimajor axis a = p/(1 − e2 ) and √ the semiminor axis b = ap = a 1 − e2 . If A = A(t) is the area swept out by the position vector r, so that the differential dA = 12 r 2 dθ, then Kepler’s second law states that the rate of change dA 1 dθ = r2 = const, dt 2 dt and so

r 2 θ˙ = h = const.

If T is the orbital period of a planet, Kepler’s third law states that T2 = const, a3 where the constant on the right-hand side is the same for any planet. Example. The orbital period of Pluto is TP = 248 years. The Earth’s orbital period, of course, is TE = 1 year. Kepler’s third law gives the ratio of Pluto’s semimajor axis to that of the Earth: aP /aE = (TP /TE )2/3 = 2482/3 ≈ 39.5. To determine the velocity and acceleration of a particle in polar coordinates, we write r = r cos θi + r sin θj = rˆ r, 41

where rˆ = cos θi + sin θj ˆ perpendicular to is a unit vector in the direction of r. The corresponding unit vector θ, rˆ in the direction of increasing θ, is given by ˆ = − sin θi + cos θj θ (we can check that the two unit vectors are perpendicular by calculating the scalar product ˆ = 0). When the angle θ depends on time t, we have rˆ · θ d dθ dˆ r = (cos θi + sin θj) = (− sin θi + cos θj) , dt dt dt and so

dˆ r ˆ = θ˙θ. dt

Similarly

ˆ dθ ˙r. = −θˆ dt We now use the product rule of differentiation to obtain the particle velocity v=

d(rˆ r) dr dˆ r dr ˆ = = rˆ + r = rˆ ˙ r + r θ˙θ dt dt dt dt

and acceleration  dv d  ˆ rˆ ˙ r + r θ˙θ = dt dt ˆ + r˙ θ˙θ ˆ + r θ¨θ ˆ − r θ˙2 rˆ = r¨rˆ + r˙ θ˙θ ˙ θ. ˆ = (¨ r − r θ˙2 )ˆ r + (r θ¨ + 2r˙ θ)

a =

ˆ where We observe that a = ar rˆ + aθ θ, ar = r¨ − r θ˙2 ,

aθ = r θ¨ + 2r˙ θ˙

are the radial and azimuthal components of acceleration. We now use Newton’s law of motion ˆ F = ma = m(ar rˆ + aθ θ) and determine the gravitational force F from the acceleration a of a planet. The azimuthal component of acceleration is 1 d  2 ˙ 1 dh aθ = r θ¨ + 2r˙ θ˙ = r θ = = 0, r dt r dt 42

since h = const by Kepler’s second law. Hence the force F has no component perpendicular to r: Fθ = maθ = 0, F = F rˆ = mar rˆ . Next we calculate the component of acceleration ar in the radial direction. We need expressions for r˙ and r¨. We use Kepler’s first and second laws to calculate r˙ =

d p dt 1 + e cos θ 



=

ep sin θ e ˙ sin θ = eh sin θ. θ˙ = (r 2 θ) 2 (1 + e cos θ) p p

On differentiating again with respect to time, we find eh eh eh2 sin θ = cos θθ˙ = 2 cos θ. p p pr !

d r¨ = dt Also,

r θ˙2 =

˙ 2 h2 (r 2 θ) = . r3 r3

Consequently, h2 eh2 cos θ − pr 2 r3 ! e cos θ 1 − p r ! e cos θ 1 + e cos θ − p p

ar = r¨ − r θ˙2 = h2 = 2 r h2 = 2 r = −

h2 , pr 2

and so F = mar = −

mh2 . pr 2

To complete the derivation, we need to show that h2 /p is the same for all planets. From Kepler’s second law, dt = r 2 dθ/h, and the orbital period is T =

Z

T 0

1 Z 2π 2 r dθ. dt = h 0

On using Kepler’s first law to express r in terms of θ, we get p2 T = h

Z

0

2π

dθ p2 2π = . 2 (1 + e cos θ) h (1 − e2 )3/2

(The trigonometric integral is difficult, but you may want to try and evaluate it using the substitution u = tan(θ/2).) Now Kepler’s third law gives T2 p4 4π 2 p 4π 2 = = = const, a3 h2 (1 − e2 )3 a3 h2 43

where we used a = p/(1 − e2 ). We see that h2 /p is indeed the same for all planets. Finally, the force of gravitational attraction should be proportional to the masses of both gravitating bodies, so we set h2 /p = GmM, where M is the mass of the Sun and G is a proportionality constant. We arrive at Newton’s inverse square law of gravitational attraction: GmM F = − 2 rˆ . r Newton postulated that the law is universal: it describes the gravitational force not only between a planet and the Sun, but also between the Earth and the Moon, for instance, and in fact between any two masses. In practice, we can often simplify a problem by assuming that an orbit is circular, which corresponds to e = 0. For example, the eccentricity of the Earth’s orbit is only 0.017. If e = 0, we have r = const and r˙ = r¨ = 0. Furthermore, θ˙ = h/r 2 = const and θ¨ = 0. The orbital speed is v = r θ˙ = const and the acceleration ar = −r θ˙2 = −v 2 /r. Newton’s equation of motion mar = −GmM/r 2 simplifies to v2 =

GM , r

v=

2πr . T

Example. A telecommunications satellite in a geosynchronous orbit appears to stay above the same spot. For an orbit in the plane of the equator, we have T = 24 hours, and q the distance between the satellite and the centre of the Earth is r = vT /(2π), where v = GME /r and ME is the mass of the Earth. On eliminating v, we find r=

GME T 2 4π 2

!1/3

≈ 42, 200 km,

where G = 6.67 × 10−11 N m2 kg−2 and ME = 5.98 × 1024 kg. The radius of the Earth is RE = 6, 371 km, so the satellite is located approximately 35, 800 km above the sea level. Example. For a low-orbit satellite, we may assume r = RE . We have v2 = and so

2 GME 4π 2 RE , = 2 T RE

s

3 RE ≈ 84 minutes. GME Alternatively, we can express the low-orbit period T in terms of the gravitational accel3 2 ): eration g = GME /RE or in terms of the average density ρE = ME /( 43 πRE

T = 2π

s

RE T = 2π = g

s

3π . GρE

Example. Periods of satellites in elliptical orbits are longer than 84 minutes. For instance, the distance r for the first Earth-orbiting artificial satellite (Sputnik 1, launched 44

in 1957) varied between 6,586 km and 7,310 km. Hence the semimajor axis was a = (6, 586 + 7, 310)/2 = 6, 948 km, and we obtain from Kepler’s third law that the period was T = 84 (6, 948/6, 371)3/2 ≈ 96 minutes. Newton’s inverse square law of gravitational attraction implies that when two bodies are close to each other, gravitational forces are greater in the nearer parts of the bodies. The resulting tidal forces can be large enough to tear the less massive body apart. The rings of Saturn, for instance, probably had been produced when the tidal forces ripped apart its ancient moon.

Exercises 1. The motion of a slowing train is described by mv˙ = −α − βv 2 ,

v(0) = v0 ,

where α and β are positive constants. Calculate the stopping time of the train ts . Determine the limit of ts as v0 → ∞.

2. At what altitude above the surface of a planet is the gravitational acceleration one half of that at the surface? The radius of the planet is R. 3. Calculate the work required to raise a body of mass m from the surface of the Earth to a height h above the surface. Assume that the radius of the Earth is R and the gravitational acceleration near the surface is g. Also calculate the required work in the limit h → ∞. 4. The Earth orbits around the Sun with a period of 1 year. The average Sun–Earth distance is about 1.5 × 108 km. Use these facts to determine the mass of the Sun.

Comment: how can we determine the Sun–Earth distance? If you are curious, read about it at http://curious.astro.cornell.edu

5. (a) The orbital period of a planet is 27 years. What is the ratio of its semimajor axis to that of the Earth? (b) Determine the orbital period (in years) of a planet whose mean distance from the Sun is 25 times that of the Earth. 6. The Vostok-1 spacecraft carried the first human into outer space in 1961. The height of the orbit was 181 km in its perigee and 327 km in its apogee. Determine the orbital period of the spacecraft. You may assume that the period of a low-orbit satellite is 84 minutes, and the average radius of the Earth is 6371 km. 7. If the Earth suddenly stopped orbiting the Sun, it would be pulled in by the gravitational force of the Sun. How long would it take for the Earth to hit the Sun? 8. Suppose you are travelling to the antipodes in a low-orbit satellite. What would be your travel time? Express your answer in terms of the gravitational acceleration at the surface, g, and the radius of the Earth, R. 45

9. A spaceship is in a low circular orbit around a newly discovered planet which is made of pure platinum. The density of platinum is four times greater than the average density of the Earth. The period of a low-orbit satellite orbiting around the Earth is about 90 minutes. What is the orbital period of the spaceship?

46

Chapter 4 Systems of first-order differential equations 4.1

Homogeneous linear systems with constant coefficients

In many applications, we need to determine several functions that are described by a system of differential equations. We have already encountered examples of Richardson’s theory of conflict and Lanchester’s model of combat, which lead to such systems. Another example is provided by two-species models in biology (deer and the vegetation they consume, foxes and rabbits, sharks and smaller fish). Thus we are often interested in solving a system of two first-order differential equations: x˙ = f (x, y), y˙ = g(x, y), where f and g are some known functions. We start with homogeneous linear systems with constant coefficients: x˙ = ax + by, y˙ = cx + dy, where a, b, c, d are constants. We can always solve this system by eliminating one of the unknown functions and reducing the system to a single second-order equation with constant coefficients. If c 6= 0, for example, we can eliminate x by rewriting the second equation as d 1 x = y˙ − y, c c which we differentiate to find 1 d x˙ = y¨ − y. ˙ c c 47

On substituting these x and x˙ into the first equation, we get a homogeneous second-order equation y¨ − (a + d)y˙ + (ad − bc)y = 0.

As usual, we seek solutions of the form y(t) ∼ exp(λt) and obtain the quadratic equation λ2 − (a + d)λ + (ad − bc) = 0

with the two roots λ1,2

q 1 = (a + d) ± (a + d)2 − 4(ad − bc) . 2 



Assuming that λ1 6= λ2 , the general solution for y(t) is y(t) = c1 eλ1 t + c2 eλ2 t , where the constants c1 and c2 are specified by initial conditions. It follows that the general solution for x(t) is 1 d y˙ − y c c λ2 − d λ2 t λ1 − d λ1 t e + c2 e . = c1 c c

x(t) =

Of course, we would obtain the same general solution of the system if we eliminated y and solved the resulting second-order equation for x. Example. In order to solve x˙ = 3x − y, y˙ = x + y,

x(0) = 3, y(0) = 0,

we eliminate y. From the first equation, we have y = −x˙ + 3x, and so y˙ = −¨ x + 3x. ˙ Substitution into the second equation yields x¨ − 4x˙ + 4x = 0, which leads to λ2 − 4λ + 4 = 0.

Because λ1 = λ2 = 2, the general solution is

x(t) = (c1 + c2 t)e2t and y(t) = −x˙ + 3x = (c1 − c2 + c2 t)e2t

On using the initial conditions x(0) = c1 = 3 and y(0) = c1 −c2 = 0, we obtain c1 = c2 = 3, and thus the unique solution is x(t) = 3(1 + t)e2t , 48

y(t) = 3te2t .

4.2

Equilibrium and stability: linear systems

For a general system x˙ = f (x, y), y˙ = g(x, y), we define an equilibrium solution, say x = x0 , y = y0 , as a solution that does not vary in time: x˙0 = y˙0 = 0. An equilibrium is called stable if an arbitrary small displacement x(t), y(t) from the equilibrium remains in the vicinity of the equilibrium. Otherwise, the equilibrium is called unstable. The equilibrium is called asymptotically stable if, in addition to stability, x(t) → x0 and y(t) → y0 as t → ∞. These definitions for a system are similar to those for a single differential equation. It will be seen later that the stability of equilibrium solutions of a nonlinear system can be determined by studying a related linear system. This is why it is instructive to begin by analysing linear systems which we can always solve exactly. We observe that x0 = y0 = 0 is an equilibrium solution of the linear system with constant coefficients x˙ = ax + by, y˙ = cx + dy. We can express both x(t) and y(t) as linear combinations of exponential solutions exp(λ1 t) and exp(λ2 t), where   q 1 2 λ1,2 = (a + d) ± (a + d) − 4(ad − bc) . 2 Because we can reduce the linear system to a second-order homogeneous equation with constant coefficients, we can obtain a stability criterion for the linear system using the argument that we used to obtain the stability criterion for a second-order equation. We require for stability that the real parts of both λ1 and λ2 be negative. This is equivalent to λ1 + λ2 < 0, λ1 λ2 > 0, or a + d < 0,

ad − bc > 0.

The equilibrium x0 = y0 = 0 is asymptotically stable if and only if both these conditions are satisfied. It is important that we can determine whether or not the equilibrium is stable without writing down the general solution of the system. Unless otherwise noted, we do not investigate the more difficult question of neutral stability. Qualitative theory of ordinary differential equations is used to describe the behaviour of solutions in the vicinity of equilibrium points. Example. For the system x˙ = 3x − y, y˙ = x + y, we have a + d = 3 + 1 = 4 > 0 and ad − bc = 3 × 1 − (−1) × 1 = 4 > 0. Because a + d > 0, the equilibrium x0 = y0 = 0 is unstable. 49

4.3

Equilibrium and stability: nonlinear systems

To determine an equilibrium solution x = x0 , y = y0 of the system x˙ = f (x, y), y˙ = g(x, y), we solve f (x0 , y0) = 0, g(x0 , y0 ) = 0. In applications, we are usually interested in stable equilibrium solutions that are not destroyed by small perturbations. To determine whether an equilibrium is stable, we need to know the behaviour of a solution x(t), y(t) that is initially very close to the equilibrium. Therefore, just as for firstorder and second-order differential equations, we can use a linearised stability analysis to investigate whether an equilibrium solution is asymptotically stable. We define a timedependent deviation from the equilibrium u(t) = x(t) − x0 ,

u˙ = x, ˙

v(t) = y(t) − y0 ,

v˙ = y, ˙

and we expand f (x, y) and g(x, y) in Taylor series as functions of two variables:







∂f ∂f (x − x0 ) + (y − y0 ) + . . . , f (x, y) = f (x0 , y0) + ∂x x0 ,y0 ∂y x0 ,y0 ∂g ∂g g(x, y) = g(x0 , y0) + (x − x0 ) + (y − y0 ) + . . . . ∂x x0 ,y0 ∂y x0 ,y0

Keeping only the linear terms in the expansions and using f (x0 , y0) = 0 and g(x0 , y0 ) = 0, we approximate the original nonlinear system by a linear system with constant coefficients: u˙ = au + bv, v˙ = cu + dv, where



∂f , a= ∂x x0 ,y0





∂f b= , ∂y x0 ,y0

∂g c= , ∂x x0 ,y0



∂g d= . ∂y x0 ,y0

We have already derived the asymptotic stability criterion for equilibrium solutions of linear systems: a + d < 0, ad − bc > 0. Now we can determine whether the solution of a nonlinear system is asymptotically stable by applying this criterion to a linearised system. 50

We can only use this approach to investigate asymptotic stability. As the following example shows, the linearised system cannot be used to establish whether an equilibrium solution is neutrally stable. Example. The solution x0 = 0, y0 = 0 is the only equilibrium of the system x˙ = −y + x(x2 + y 2), y˙ = x + y(x2 + y 2 ). The linearised system u˙ = −v,

v˙ = u

is equivalent to u¨ + u = 0,

v¨ + v = 0.

Because its solutions are oscillatory and hence remain small for all t, we might draw the wrong conclusion that x0 = 0, y0 = 0 is a neutrally stable equilibrium of the nonlinear system as well. However, if we consider the function s(t) = x2 + y 2 , we see that 1 ds = xx˙ + y y˙ = x[−y + x(x2 + y 2 )] + y[x + y(x2 + y 2 )] = (x2 + y 2)(x2 + y 2 ) 2 dt or s˙ = 2s2 . On integrating this separable first-order equation, we find s(t) =

s(0) . 1 − 2s(0)t

It is clear that s = x2 + y 2 → ∞ as t → [2s(0)]−1 , which can only happen if either x or y or both become infinite as t → [2s(0)]−1 . Therefore, x0 = 0, y0 = 0 is an unstable equilibrium. Example. Previously we used the logistic equation N˙ = a(1 − N/K)N, where a and K are positive constants, to describe the limiting effect of finite resources on the size of a population. Now consider two species that are competing for the same limited food supply. Introducing a parameter α > 0 that measures the degree of competition between the two species for the resources, we can write for the two populations N1 and N2 N1 − αN2 N1 , N˙ 1 = a1 1 − K1 



N2 N˙ 2 = a2 1 − − αN1 N2 . K2 For simplicity, we assume K1 = K2 = ∞. The resulting nonlinear system 



N˙ 1 = f (N1 , N2 ) = a1 (1 − αN2 ) N1 , 51

N˙ 2 = g(N1 , N2 ) = a2 (1 − αN1 ) N2

has two equilibrium solutions: N10 = N20 = 0 and N10 = N20 = α−1 . To investigate the stability of the equilibria, we introduce the perturbations u(t) = N1 (t) − N10 , v(t) = N2 (t) − N20 and linearise the system about an equilibrium. For the solution N10 = N20 = 0, the linearised system is u˙ = a1 u, v˙ = a2 v, and we conclude that the equilibrium is unstable. For the nonzero equilibrium solution N10 = N20 = α−1 , we calculate the partial derivatives ∂f a= = a1 (1 − αN20 ) = 0, ∂N1 eq

∂f b= = a1 N10 (−α) = −a1 , ∂N2 eq

∂g = a2 N20 (−α) = −a2 , c= ∂N1 eq

∂g d= = a2 (1 − αN10 ) = 0, ∂N2 eq

where the subscript ‘eq’ indicates equilibrium values. The linearised system is u˙ = 0u − a1 v, v˙ = −a2 u + 0v. To apply our stability criterion, we calculate a + d = 0 + 0 = 0 and ad − bc = 0 × 0 − (−a1 ) × (−a2 ) = −a1 a2 . We see that neither a + d < 0 nor ad − bc > 0 is satisfied. Hence the model predicts that the equilibrium is unstable: two similar species in the same habitat cannot coexist. The same result is obtained for finite K1 and K2 . Therefore we should expect that the struggle for existence between two similar species nearly always leads to the complete extinction of one of them. This result is known as the principle of competitive exclusion in population biology. Example. Suppose x(t) is the number of males infected with gonorrhoea, y(t) is the number of infected females, a1 and a2 are the rates with which the infectives are cured, c1 = const is the number of promiscuous males, and c2 = const is the number of promiscuous females. We assume that new male infectives are added at a rate b1 (c1 − x)y that is proportional to the number of the male susceptibles c1 − x and to the number of the female infectives y. Similarly, we assume that new female infectives are added at a rate b2 (c2 − y)x that is proportional to the number of the female susceptibles c2 − y and to 52

the number of the male infectives x. Then a model for the spread of gonorrhoea due to heterosexual contacts is x˙ = −a1 x + b1 (c1 − x)y, y˙ = −a2 y + b2 (c2 − y)x. For practical reasons, we are interested in the equilibrium solution x0 = 0, y0 = 0. Is it stable? The linearised system is x˙ = −a1 x + b1 c1 y, y˙ = b2 c2 x − a2 y. We observe that a + d = −(a1 + a2 ) < 0, and thus the first part of our stability criterion is always satisfied. Now, ad − bc = (−a1 )(−a2 ) − (b1 c1 )(b2 c2 ), and the equilibrium is stable if ad − bc > 0 or b1 b2 c1 c2 < a1 a2 . This is when the disease disappears from the population. Otherwise, if b1 b2 c1 c2 > a1 a2 , it can be checked by a tedious calculation that x0 =

b1 b2 c1 c2 − a1 a2 , a1 b2 + b1 b2 c2

y0 =

b1 b2 c1 c2 − a1 a2 a2 b1 + b1 b2 c1

is a stable equilibrium. In this case the total number of infective males and females ultimately levels off. It should not surprise us that public health data indicate that b1 b2 c1 c2 > a1 a2 . Example. Consider now two species, one of which feeds on the other. Suppose x(t) is the prey population, say small fish, and y(t) is the predator population, say sharks. If the species did not interact, we could write x˙ = rx, where r > 0 implies abundant food supply for the prey population, say algae for the small fish. By contrast, the predators would die off without food, and so y˙ = −ky. In reality the two species do interact—the small fish are eaten by the sharks—and the number of contacts between them is proportional to both x and y. Hence we arrive at the following system describing predator–prey interactions: x˙ = f (x, y) = rx − αxy, y˙ = g(x, y) = −ky + βxy. These equations are called the Lotka–Volterra equations (A. J. Lotka, 1925; V. Volterra, 1926). We find equilibrium solutions by solving f (x0 , y0) = x0 (r − αy0) = 0, g(x0 , y0) = (βx0 − k)y0 = 0. 53

The system has two equilibrium solution. The solution x0 = 0, y0 = 0 is unstable since the corresponding linearised system is x˙ = rx,

y˙ = −ky,

and so a small nonzero x will deviate strongly from the equilibrium. For the nonzero equilibrium solution x0 = k/β, y0 = r/α, we calculate the partial derivatives

∂f a= = r − αy0 = 0, ∂x eq

∂f αk b= = −αx0 = − , ∂y eq β

βr ∂g = βy0 = c= , ∂x eq α

∂g = βx0 − k = 0, d= ∂y eq

and we use them to obtain the linearised system

u˙ = 0u + (−αk/β)v, v˙ = (βr/α)u + 0v. To apply our stability criterion, we calculate a + d = 0 + 0 = 0 and ad − bc = 0 × 0 − (−αk/β)(βr/α) = rk > 0. We see that a + d < 0 is not satisfied, and thus the equilibrium is not asymptotically stable. More detailed analysis, however, shows that the equilibrium is neutrally stable. On differentiating the linearised equations, we obtain u¨ + kru = 0,

v¨ + krv = 0.

Thus the key prediction of the Lotka–Volterra √ model is that the predator and prey populations should oscillate with period T = 2π/ kr. In the real world, of course, interaction between species are a lot more complicated than in this one-predator-one-prey model. Example. The Italian biologist Umberto D’Ancona noticed that the total catch in the Mediterranean had a larger fraction of sharks during World War I when the level of fishing was greatly reduced. To explain this phenomenon, the mathematician Vito Volterra incorporated the effect of fishing into the Lotka–Volterra model. If h is the fraction of sharks and small fish being caught, we can write x˙ = rx − αxy − hx, y˙ = −ky + βxy − hy. 54

It follows that the equilibrium solution is x0 =

k+h , β

y0 =

r−h . α

Therefore, as long as h < r, harvesting increases the equilibrium prey population and decreases the equilibrium predator population. (There is no biologically meaningful equilibrium in the case h > r that corresponds to overfishing.) This is why reduced fishing during the war (smaller h) had increased the shark population (larger y0 ) and decreased the average fish population (smaller x0 ). More recently, an unanticipated effect of DDT spraying was to destroy not the harmful insects but rather their natural predators.

Exercises 1. Suppose R(t) measures Romeo’s love for Juliet and J(t) measures Juliet’s love for Romeo. Romeo’s love grows in response to Juliet’s love for him and vice versa. At the same time, both lovers try to rein in their feelings for each other. The resulting system of equations is as follows: R˙ = −R + J, J˙ = −J + R. Find the general solution of the system. Solve the initial value problem R(0) = 1, J(0) = 0. 2. We wish to model the outbreak of an infectious disease. The number of individuals susceptible to infection at time t is S(t), and the number of infected individuals is I(t). We postulate that the population increases at a constant rate in the absence of an epidemic. We also postulate that the infection rate is proportional to the product of S and I and that the death rate is proportional to I. Consider the resulting system of coupled differential equations for S(t) and I(t): S˙ = −rSI + µ, I˙ = rSI − γI. Here r, µ and γ are positive constants. Derive an equilibrium solution of the system, S = S0 and I = I0 , and determine its stability. 3. We wish to model the size of an insect population. The population at time t is P (t), and the total food supply is F (t). We postulate that the birth rate is proportional to the product of the population size and the food supply and that the death rate is proportional to the population size. We also postulate that F (t) is described by a linear first-order differential equation in the absence of the insects and that the food consumption rate is proportional to the population size. Consider the resulting system of coupled differential equations for P (t) and F (t) in appropriate units: P˙ = F P − kP, 55

F˙ = c − F − P. Here k and c are positive constants, and we assume c > k. Derive an equilibrium solution of the system, P = P0 > 0 and F = F0 , and determine its stability. 4. We wish to model the size of a fish population in a lake. The fish population at time t is F (t), and the number of fishermen is M(t). We postulate that fish grow logistically in the absence of fishing, and that the presence of fishermen decreases the growth rate by an amount proportional to the product of the fish and fishermen numbers. We also postulate that fishermen are attracted to the lake at a rate proportional to the amount of fish in the lake, and that fishermen are discouraged from the lake at a rate proportional to the number of fishermen already there. Consider the resulting system of coupled differential equations for F (t) and M(t) in appropriate units: F˙ = aF (1 − F ) − F M, M˙ = bF − M. Here a and b are positive constants. Calculate the equilibrium solutions of the system and determine their stability.

56

Chapter 5 Difference equations 5.1

Geometric growth

In many applied problems, we are interested in variables that are only known at certain time intervals. For example, economic quantities (income, savings, gross domestic product, unemployment) can be reported weekly, monthly or annually. Learning results are evaluated by tests and exams. Population sizes of annual plants vary from year to year. These problems lead to mathematical models that involve difference equations. The difference equation for geometric growth is xk+1 = axk ,

a = const,

where xk is some variable at time step k. Given an initial value x0 , we can determine x1 = ax0 , x2 = ax1 = a2 x0 , and in general xk = x0 ak . Example. If a sum of money accumulates at compound interest, S0 is the initial deposit, and Sk is the sum on deposit after k compounding intervals, then Sk+1 = Sk + iSk = (1 + i)Sk , where i is the interest rate per interval. Hence Sk = S0 (1 + i)k . Example. In a model for economic growth, we consider annual national income Yk for year k. The total income is divided into consumer expenditures Ck and private investments Ik , Yk = Ck + Ik . We assume that consumption is proportional to the available income, Ck = mYk ,

0 < m < 1,

and also that income growth is proportional to the invested amount, Yk+1 − Yk = rIk , 57

r > 0.

We have Yk+1 = = = =

Yk + rIk Yk + r(Yk − Ck ) Yk + r(Yk − mYk ) (1 + r − rm)Yk

or Yk = Y0 (1 + r − rm)k . The model neglects many important factors, such as unemployment, so its prediction of a robust economic growth is not very realistic.

5.2

The equation xk+1 = axk + b

The difference equation xk+1 = axk + b,

a = const,

b = const

is a linear first-order equation with constant coefficients. The equation is first-order because it only involves xk and xk+1 . If a = 1, its solution is xk = x0 + kb. Example. If a sum of money earns simple interest at the rate r, S0 is the initial deposit, and Sk is the sum on deposit after k years, then Sk+1 = Sk + rS0 and Sk = S0 (1 + kr). If a 6= 1, we seek a solution of the form xk = C1 + C2 λk , where C1 , C2 , and λ are constants to be determined. On substituting into the equation, we get C1 + C2 λk+1 = aC1 + aC2 λk + b or (C1 − aC1 − b) + C2 λk (λ − a) = 0, which we satisfy by choosing

b , λ = a. 1−a We use the initial condition x0 = C1 + C2 to specify the constant C2 = x0 − C1 = x0 − b/(1 − a). The resulting solution is C1 =

!

b b ak + x0 − xk = 1−a 1−a or

1 − ak . 1−a If b = 0, we recover the formula for geometric growth. xk = x0 ak + b

58

Behaviour of solutions to first-order difference equations is very different for different numerical values of a and b. A solution xk can monotonically decrease or increase (with or without a finite limit as k → ∞), oscillate with a constant amplitude, or oscillate with increasing or decreasing amplitude. Example. The solution of the difference equation xk+1 = 2xk + 1,

x0 = 5

is xk = 6 × 2k − 1. We observe that xk → ∞ as k → ∞. Example. The solution of the equation

1 xk+1 = xk + 2, 2

x0 = 3

is xk = 4 − (1/2)k . We observe that xk → 4 as k → ∞. Example. The solution of the equation

xk+1 = −xk + 1,

x0 = 1

is xk = [1 + (−1)k ]/2. We observe that xk oscillates with a constant amplitude. Example. The quantity supplied Sk+1 of some commodity (say, crop harvested and brought to market) is an increasing function of its price Pk in the previous year. The quantity demanded Dk+1 of the same commodity is a decreasing function of the current price. Suppose that Sk+1 = aPk , Dk+1 = b − cPk+1, where a, b, c are positive constants. Market equilibrium is reached when supply equals demand, that is Sk+1 = Dk+1 for any k. Then we have aPk = b − cPk+1 or Pk+1 = and so

b a − Pk , c c

"

b b + P0 − Pk = a+c a+c

#

a − c

k

.

We note that price equilibrium can be achieved only if a/c < 1, in which case Pk → b/(a + c) as k → ∞. Otherwise, the model predicts that the commodity price will oscillate with an increasing amplitude.

5.3

Fibonacci numbers

The Fibonacci numbers Fk are defined by Fk+2 = Fk+1 + Fk , 59

F0 = F1 = 1,

so that Fk = 1, 1, 2, 3, 5, 8, 13, . . .. This equation is an example of a linear second-order difference equation with constant coefficients. In 1202, the Italian merchant and mathematician Leonardo Fibonacci introduced the sequence Fk to describe the growth of a (biologically unrealistic) rabbit population: Fk is the population after k months if rabbits start to breed after two months and each month each pair begets a new pair. We may not care about the peculiar habits of incestuous rabbits, but the Fibonacci numbers appear in more plausible contexts. For instance, Fk is the number of branches on a tree if a new shoot has to grow two months before it branches. Just as we solve homogeneous linear differential equations with constant coefficients by seeking solutions ∼ exp(λt), we can solve linear difference equations by seeking solutions ∼ λk . On substituting Fk = Cλk into the Fibonacci equation, we obtain the quadratic equation λ2 − λ − 1 = 0 with the roots

√ √ 1 1 λ1 = (1 + 5), λ2 = (1 − 5). 2 2 Therefore the general solution of the Fibonacci equation is Fk = C1 λk1 + C2 λk2 .

We use the initial conditions F0 = C1 + C2 = 1 and F1 = C1 λ1 + C2 λ2 = 1 to specify the constants √ 1 1+ 5 1 − λ2 = √ = √ λ1 , C1 = λ1 − λ2 5 2 5 √ 1 5−1 C 2 = 1 − C 1 = √ = − √ λ2 . 5 2 5 Therefore a formula for the Fibonacci sequence is √ !k+1 √ !k+1 1 1+ 5 1 1− 5 Fk = √ −√ . 2 2 5 5 It is perhaps not obvious that this formula gives a natural number for any k.

5.4

Nonlinear difference equations

There is no general method for solving nonlinear difference equations, but sometimes we are able to find a solution by guessing a change of variables that reduces a nonlinear equation to a simpler equation. Example. R. J. H. Beverton and S. J. Holt (1957) suggested a nonlinear model for the size of a fish population nk in year k: nk+1 =

Rnk , 1 + nk /M 60

where R > 1 and M > 0 are constants, and (1 + nk /M)−1 is the probability of survival till next year. The Beverton–Holt equation is a discrete analogue of the logistic growth equation. Substitution xk = 1/nk leads to a linear first-order equation xk+1 =

1 1 xk + R RM

with the solution xk = x0 R−k + or nk =

1 − R−k (R − 1)M

Qn0 1 = , xk n0 + (Q − n0 )R−k

where Q = (R − 1)M. Because nk → Q as k → ∞, we see that Q is the carrying capacity in the Beverton–Holt model. Example. The nonlinear equation xk+1 = 2xk (1 − xk ) can be rewritten as 1 − 2xk+1 = (1 − 2xk )2 . It follows that 1 − 2xk = (1 − 2x0 )2k or

1 − (1 − 2x0 )2k . 2 as k → ∞, if |1 − 2x0 | < 1 or 0 < x0 < 1. xk =

We observe that xk →

1 2

Example. Consider the nonlinear equation xk+1 = 4xk (1 − xk ). Using the substitution xk = sin2 θk , we get sin2 θk+1 = 4 sin2 θk cos2 θk or sin θk+1 = sin(2θk ), and so, as long as −π/2 < θk < π/2, we have θk = θ0 2k and xk = sin2 (θ0 2k ). We observe that a very small change in θ0 would lead to a very different solution after only a few steps because of the rapidly increasing factor 2k . 61

When we cannot solve a nonlinear difference equation analytically, we can try to learn about the properties of its solutions by studying equilibrium points and their stability. For a difference equation xk+1 = f (xk ), where f is a known function, we define an equilibrium point as a constant xe such that xe = f (xe ). A small deviation from the equilibrium is described by xk+1 − xe = f (xk ) − f (xe ) ≈ f ′ (xe )(xk − xe ). Here f ′ (xe ) is the derivative of f (xk ) with respect to xk , evaluated at xk = xe . We see that xk → xe as k → ∞, if |f ′ (xe )| < 1. Hence we can use the criterion |f ′(xe )| < 1 to determine whether an equilibrium is stable. Example. The nonlinear difference equation xk+1 = f (xk ) = rxk (1 − xk ),

0 3.569 . . .. “Chaotic” means that the solution xk never repeats itself and is extremely sensitive to the initial condition x0 . Thus even simple nonlinear equations can have solutions with very complicated behaviour.

Exercises 1. Suppose that a species of animal only breeds during the spring and all adults die before the next breeding season. Every female produces on average R female offspring that survive to breed in the next year. Find the female population as a function of time. 62

2. A model for the price of bonds leads to the difference equation   E E pt = 1 − pt−1 + P, t = 1, 2, 3, . . . , D D where pt denotes the bond price at time t, p0 is known, and E, D, and P are constants. Solve this equation and thus show that pt converges to P if 0 < E/D < 2. 3. A sum P is deposited in a bank at equal intervals. The interest rate per interval is i. If Sk is the sum on deposit after k intervals, show that Sk+1 = (1 + i)Sk + P. Find the solution of this difference equation if S0 = 0. 4. Suppose that B dollars are borrowed from a bank at a fixed annual interest rate R%. The loan is to be repaid in n years in equal monthly instalments of P dollars. Find P . Fk+1 5. The Fibonacci numbers Fk are defined by Fk+2 = Fk+1 + Fk . Calculate lim . k→∞ Fk 6. Consider the difference equation xk+2 − 3xk+1 + 2xk = 0.

Show that the general solution to this equation is xk = C1 + C2 2k . Find the particular solution such that x0 = 1, x1 = 2. 7. A model for the number of red blood cells circulating the blood is given by Rk+1 = (1 − f )Rk + Mk , Mk+1 = f Rk , where Rk is the number of cells in the circulatory system on day k, Mk is the number of cells produced by marrow on day k, and f is a fraction of cells removed by spleen daily. Derive a single second-order equation for Rk and find its general solution. 8. Solve the nonlinear difference equation xk+1 = xk /(1 + xk ). 9. Find the nonzero equilibrium point of the difference equation xk+1 = (xk )2 and determine whether it is stable. 10. A model for the size of fish populations leads to the nonlinear difference equation Nk+1 = αNk exp(−βNk ), where α and β are positive constants. Find the nonzero equilibrium solution of this equation and determine when it is stable. 11. The nonlinear difference equation xk+1 = f (xk ) = rxk (1 − xk ) is known to have stable period-2 solutions, xk+2 = xk , when the parameter r lies in the range r1 < r < r2 . Calculate the values of r1 and r2 by considering the equation xk+2 = f2 (xk ) = f (f (xk )). 63