MATH259 Mathematical Modelling
Yuri Litvinenko
June 24, 2011
Contents
1 First-order ordinary differential equations 2
1.1 Introduction . . . 2
1.2 Exponential growth and decay . . . 3
1.3 Linearisation . . . 8
1.4 Separable and linear equations . . . 9
1.5 Equilibrium and stability . . . 12
2 Second-order ordinary differential equations 19 2.1 Linear second-order equations . . . 19
2.2 The homogeneous equation with constant coefficients . . . 21
2.3 Linear oscillations . . . 25
2.4 Equilibrium and stability . . . 27
2.5 Inhomogeneous equations . . . 29
3 Elements of classical mechanics 33 3.1 Kinematics . . . 33
3.2 Dynamics and gravity . . . 35
3.3 Work and energy . . . 38
3.4 Planetary orbits and Kepler’s laws . . . 40
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 57 5.1 Geometric growth . . . 57
5.2 The equation xk+1= axk+ b . . . 58
5.3 Fibonacci numbers . . . 59
Chapter 1
First-order ordinary differential
equations
1.1
Introduction
Mathematical modelling is a part of applied mathematics. We use mathematical mod-elling 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. Mathemat-ical 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
dt = f (t, x) contains its first derivative or the rate of change
dx
dt = lim∆t→0
x(t + ∆t) − x(t)
∆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: x(t) = Z t 0 dx dsds = Z t 0 v(s)ds,
where s is a dummy variable of integration. Sometimes the definite integral is also written less formally as Rt
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 ˙x2 + 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 = 4πr3/3 of a spherical mothball decreases so that dV /dt
is proportional to the surface area A = 4πr2 of the ball. Thus observations imply that
−dVdt = kA, where k > 0 is a proportionality constant. We have
−4πr2˙r = k4πr2 or, as long as r > 0,
˙r = −k, which we integrate to obtain
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 x = adt, 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),
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 a ≈ 0.7 a , x(t) = x(0)2 t/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 |a| ≈ 0.7 |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 (14C) 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 of14C balances the rate of disintegration
of 14C. When an organism dies (t = 0), its 14C concentration starts to decrease:
N(t) = N(0)e−λt, where λ is the decay constant of 14C, corresponding to t
1/2 ≈ 5700 years. Consequently
the rate of decay is given by
We can measure the present rate R(t) of disintegration of 14C in a sample, say charcoal
from an archaeological site. Because the initial decay rate R(0) is equal to the rate of decay of14C in the same amount of living tissue, we can determine the age t of the sample:
t = 1 λln " R(0) R(t) # = t1/2 ln[R(0)/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 x ˙x), 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, C = y(0) = x(0) + b a. Hence x(t) = " x(0) + b a # eat− b 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 capaci-tance is C (not to be confused with an integration constant). The governing equation for the electric current I(t) in the circuit is
R ˙I + I C = 0, I(0) = I0, so I(t) = I0exp −RCt .
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
This is a linear first-order equation with the general solution I(t) = E R + C exp −RtL . If I(0) = 0, we obtain I(t) = E R 1 − exp −RtL .
We see that I → E/R as t → ∞. Hence Ohm’s law E = RI is nearly true for large 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
(temper-ature 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 propor-tional 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,
com-monly 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
˙
and its general solution is
N(t) = Q ν + Ce
νt.
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(0) = N0 = Q ν + C, N(T ) = 0 = Q ν + Ce νT.
On solving these two equations, we find C = −(Q/ν) exp(−νT ) and Q = νN0 1 − e−νT. It follows that N(t) = Q ν h 1 − eν(t−T )i= N0 eνT − eνt eνT − 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,
QT N0
= νT
1 − e−νT ≈ 1.9.
We can also calculate the fraction of the first payment which is interest: νN0dt
Qdt = νN0
Q = 1 − e
−νT.
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 dr =
˙s ˙r = 2,
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 for both the craft and the sub. Now, if we substitute x = r cos θ, y = r sin θ into ds =q(dx)2+ (dy)2, we get
ds =q(dr)2+ r2(dθ)2.
Eliminating ds, we obtain
or dr r = dθ √ 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 = r0exp[(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
′(x
0), b = f (x0).
Assuming that a 6= 0, we integrate the linear equation for ξ(t) to get ξ(t) = b
ae
at
− ab, and so, approximately,
x(t) = x0+ ξ(t) ≈ " x(0) − b a # + b ae at.
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 = a 1 − N K N,
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
We can use the change of variables x = 1/N to integrate the left-hand side: Z KdN N(K − N) = − Z Kdx Kx − 1 = − ln(Kx − 1) + ln C. It follows that K N − 1 = Ce −at or N(t) = K 1 + C exp(−at).
The initial condition at t = 0 specifies the integration constant: K
N(0) − 1 = C, which yields
N(t) = KN(0)
N(0) + (K − N(0)) exp(−at). We see that, for any N(0), the exponential growth is limited:
N(t → ∞) → KN(0)N(0) = K.
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 Z t a(s)ds + y exp Z t a(s)ds d dt Z t a(s)ds = ˙y exp Z t a(s)ds + a(t)x,
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
exter-nal 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)ektdt = 1 k(kne+ αt)e kt − αk Z ektdt = ne+ α kt − α k2 ekt+ C. As usual, the initial condition specifies C, and we finally arrive at
n(t) = ne− α k2 + α kt + n0− ne+ α k2 e−kt.
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
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)fx1−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 = a " 1 − N 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 + CKsexp(−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 − 1 # e−sat )−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
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 perturba-tions 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′(x
0) > 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′(x
0) = x−10 = 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 − aV2/3,
where a is a constant. We solve f (V0) = 0 to obtain an equilibrium solution:
V0 =
r
a
3/2
. To determine its stability, we calculate f′(V ) = −2
3aV−1/3. Now f′(V0) = − 2 3aV −1/3 0 = − 2 3 a3/2 r1/2,
and we conclude that the equilibrium is stable because f′(V
Example. The logistic equation ˙ N = f (N) = a 1 − NK N
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∗
dt∗ = N∗(1 − N∗),
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 = β N − µ β − I ! I,
which we recognise as the logistic equation in different notation. If N > µ/β, the only stable equilibrium solution
I0 = N −
µ β > 0,
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.
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 N0 = 1 2 1 ±√1 − 4H,
hence H ≤ 1/4. Now f′(N) = 1 − 2N, and we see that N
0 = (1 − √ 1 − 4H)/2 is unstable because f′(N 0) = √1 − 4H > 0, whereas N0 = (1 + √1 − 4H)/2 is stable because f′(N
0) = −√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 ˙ N = (1 − N)N − 1 4 = − N −1 2 2 , and on separating variables and integrating, we get
1 N − 12
− 1
N(0) − 12
= t. For the initial condition
N(0) = 1 2− ǫ, ǫ > 0 we have N(t) = 1 2 − ǫ 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.
Example. We can prevent the collapse described in the previous example by introducing a feedback into the model:
˙
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 N
0 = 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
dκ = 1 − 2κ = 0.
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?
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 k2x.
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 m ˙v = −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 ˙ N = N0 2 sin πN 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 = kN2,
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
a(t) = a0+ a1t,
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 = k1exp(k2t) − 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 = bN2 − 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 r1N. Bugs on the perimeter, however, suffer from cold
and they die at a rate r2N1/2. Hence the differential equation satisfied by N is
˙
N = r1N − r2N1/2.
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 mathemat-ical model for dieting in which his weight W (t) changes due to food intake F (t), exercise E(t), and body metabolism M(t):
˙
W = F − E − M, M = bW3/4.
Assume that F − E and b are positive constants. Derive an equilibrium solution W0 and
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 = d2x/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 + p ˙x + 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 accom-plished 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.
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
C1dt = C1t + 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+ v0t.
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) = C
1x + 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 T (x) = Ti+ (To− Ti)
x
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 = −κAdT
dx = κA
Ti− To
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 and
c = (1 − s)y, s = const,
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).
Hence x = 1 λ˙y + sy − g, which yields ˙x = 1 λy + s ˙y − ˙g.¨ The final assumption of the model is that
˙x = k(v ˙y − x), k = const, v = const,
where v is termed the accelerator. Eliminating x and ˙x in this equation, we get ¨
y + λs ˙y − λ ˙g = λk(v ˙y − λ1˙y − sy + g), or
¨
y + p ˙y + 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
coeffi-cients
We can determine a complete solution of the homogeneous linear second-order equation ¨
x + p ˙x + 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 = λ¨ 2Aeλt.
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: λ1 = − p 2 + √ p2− 4q 2 , λ2 = − p 2 − √ p2− 4q 2 .
Thus x(t) = A exp(λ1t) and x(t) = B exp(λ2t) are both solutions, whatever the values of
the constants A and B, and the general solution is x(t) = Aeλ1t
+ Beλ2t
If p2 > 4q, λ
1 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 λ
1,2 = ±1 and therefore
x(t) = Aet+ Be−t.
If p2 < 4q, then λ
1 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/2Aeiαt+ Be−iαt.
To specify a real-valued solution, we need Euler’s formula (L. Euler, 1748). We derive it formally by calculating
d dt
h
e−iαt(cos αt + i sin αt)i
= −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 x(t) = e−pt/2 C cos √ 4q − p2 2 t + D sin √ 4q − p2 2 t ! .
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
A snag arises if p2 = 4q. Because λ
1 = 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
˙x = ˙y − p2y e−pt/2, x =¨ y − p ˙y +¨ p 2 4y ! e−pt/2, and substitution into ¨x + p ˙x + qx = 0 yields
¨ y + q − p 2 4 ! y = 0 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 + 4 ˙x + 4x = 0 is a linear homogeneous equation with constant coefficients: p = 4, q = 4. Because p2 − 4q = 0, we have λ
1,2 = −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√klt+ Be−√klt. It follows that y(t) = ˙x k = s l k Ae√klt− Be−√klt.
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.
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.
Similarly for the y-force,
˙y = −cx − dy + Q(t), y(0) = y0.
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) = x0cosh( √ bct) − s b cy0sinh( √ bct), and similarly y(t) = y0cosh( √ bct) − rc bx0sinh( √ bct).
We can eliminate the explicit t-dependence by dividing one differential equation by the other: dy dx = ˙y ˙x = cx by.
This is a separable first-order equation. We integrate it to obtain an interesting result: by2− cx2 = by20− 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 by2
0 − cx20 > 0, and the x-force wins (x > 0 when
y = 0) if by2
0− 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 of 40,000 and 30,000. The first battle results in a y-force victory with a margin of √50, 0002− 40, 0002 = 30, 000, which is just enough to force a
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 + ω2x = 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+ B2 and a phase φ
0 = arctan(A/B) of the
oscillations, we can write the integration constants as A = a0sin φ0 and B = a0cos φ0.
Hence we can express the general solution as follows: x(t) = a0sin(ωt + φ0).
This form of the solution makes it clear that the oscillations are indeed periodic, x(t+T ) = x(t), with the period
T = 2π ω = 2π
rm
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 V = − πR2x 2πR3/3 = − 3x 2R. Hence m¨x = ∆F = −3x 2RF0 = − 3x 2Rmg or ¨ x + 3g 2Rx = 0,
and so the period of the resulting oscillations is T = 2πq2R/(3g).
Consider now the possibility that there is a friction force Ff that will damp the
oscil-lations. 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 + ω20x = 0. From now on, ω2
0 = 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 ω2 0− β2.
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 = a0sin(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 ω = 2π q ω2 0− β2 = q 2π k/m − r2/(4m2).
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 exponen-tially, 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.
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 (x, ˙x) = f (x0, 0) + ∂f ∂x x0,0 (x − x0) + ∂f ∂ ˙x x0,0 ˙x + . . . ≈ ∂f ∂x x0,0 (x − x0) + ∂f ∂ ˙x x0,0 ˙x,
where we kept only the linear terms in the expansion and used f (x0, 0) = 0 and ˙x0 = 0.
If ξ(t) = x − x0 denotes a deviation from the equilibrium solution, we have ¨ξ =
f (x0 + ξ, ˙ξ), or approximately ¨ ξ + p ˙ξ + qξ = 0, where p = − ∂f∂ ˙x x 0,0 , q = − ∂f ∂x x 0,0 .
This is a second-order homogeneous equation with constant coefficients. Its general solu-tion is ξ(t) = A exp(λ1t) + B exp(λ2t) where as before
λ1,2 = 1 2 −p ±qp2− 4q .
We require for asymptotic stability that the deviation ξ remain small and ξ → 0 as t → ∞. Hence we require that the real parts of both λ1 and λ2 be negative. If q < p2/4, both
roots are real, and we require that the larger root be negative: (−p +√p2− 4q)/2 < 0,
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 ∂x x 0,0 < 0, ∂f ∂ ˙x x 0,0 < 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 ∂x x 0,0 = −1, ∂f∂ ˙x x 0,0 = −µ(x20− 1) = µ.
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 + ω2x = 0
is x0 = 0. Now f (x, ˙x) = −ω2x, 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′(x
0) < 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.
2.5
Inhomogeneous equations
We now consider a second-order inhomogeneous equation ¨
x + p ˙x + qx = f, f 6= 0
and suppose xp(t) is a particular solution of this equation, so that
¨
xp+ p ˙xp+ qxp = f.
On subtracting one equation from the other, we obtain a homogeneous equation ¨
xh + p ˙xh+ qxh = 0
for the function xh(t) = x(t) − xp(t). We see that the general solution x(t) of an
inho-mogeneous 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) = Aeλ1t+ Beλ2t+ 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
un-determined 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 x
p(t) = C0 + C1t + . . . + Cntn. If
f (t) = a exp(αt), we seek xp(t) = C exp(αt). If f (t) = a1cos(αt) + a2sin(αt), we seek
xp(t) = C1cos(αt) + C2sin(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(λ1t). 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) = Ct2exp(λ1t).
Example. A linear frictionless oscillator, driven by a periodic external force, is described by
¨
x + ω02x = sin ωt.
Consider first the case ω 6= ω0. We try xp(t) = C sin ωt. On substituting this into the
equation, we get
−Cω2sin ωt + Cω02sin ωt = sin ωt, which can be satisfied only if C = 1/(ω2
0 − ω2). Thus we have the general solution
x(t) = A cos ω0t + B sin ω0t +
1 ω2
0 − ω2
If the initial conditions are given by x(0) = 0 and ˙x(0) = 0, for example, we have A = 0, Bω0+ ω ω2 0 − ω2 = 0, and the solution is
x(t) = 1 ω2− ω2 0 ω ω0 sin ω0t − sin ωt .
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:
x(t) = lim ω→ω0 (ω/ω0) sin ω0t − sin ωt ω2− ω2 0 . On using L’Hospital’s rule to evaluate the limit, we find
x(t) = d dω[(ω/ω0) sin ω0t − sin ωt] d dω(ω2− ω02) ω=ω0 = 1 2ω2 0 sin ω0t − t 2ω0 cos ω0t.
Alternatively, we could find this solution using the method of undetermined coefficients. When ω = ω0, seek a particular solution
xp(t) = C1t cos(ω0t) + C2t sin(ω0t)
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 ω0t + B sin ω0t −
t 2ω0
cos ω0t.
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 fre-quency 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
force (voltage) E(t), all connected in series. The governing equation for the electric current I(t) in the circuit is
L ˙I + RI + Q C = E,
where Q is the electric charge. On differentiating this equation and using I = ˙Q, we have L ¨I + R ˙I + I
C = ˙E.
For example, if E = E0sin(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 Ih(t) = Ae−Rt/(2L)sin √ 4CL − R2C2 2CL t + δ ! ,
where A and δ are integration constants, and we assume the underdamped case R < 2qL/C. A particular solution of the inhomogeneous equation is
Ip(t) =
ωCE0[RCω sin(ωt + β) + (1 − CLω2) cos(ωt + β)]
(RCω)2+ (1 − CLω2)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 radioac-tive decay is described by the following equations:
˙
N1 = −λ1N1, N1(0) = N0,
˙
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 − ω2u = −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 + 6 ˙x − 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) = f0sin(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: ¨ x + k mx = f (t) m .
Assuming that the frequency ω is chosen so that k/m = ω2, determine x(t) for the initial
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 space-craft 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, x(t) = x0+ v0t +
1 2at
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,
¨
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 t
f is defined by z(tf) = 0. Hence we find
tf =
q
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 tf =
q
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 is cs ≈ 340 m/s and air resistance can be neglected?
The time for the stone to fall down is q2h/g, and the time for the sound to travel up is h/cs. It follows that
T = q2h/g + h/cs.
This is a quadratic equation for√h, which we solve to obtain √ h = − q 2/g ±q2/g + 4T /cs 2/cs . Remembering that √h > 0, we choose the positive root and find
h = −q2/g +q2/g + 4T /cs 2/cs 2 ≈ 72 m. When gT /cs≪ 1 we have approximately
s 2 g + 4T cs = s 2 g s 1 + 2gT cs ≈ s 2 g 1 + gT cs
(remember that √1 + x ≈ 1 + x/2 for small x). Therefore, in the limit cs → ∞, the
formula for h reduces to h = gT2/2, 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
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 R ˙θ = const, we can introduce ω = ˙θ = const and calculate
v = ˙r = ωR(1 − cos ωt, sin ωt, 0), a= ˙v = ω2R(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− F0t/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: x(ts) − x0 = v0ts− F0 2mt 2 s = v0 mv0 F0 − F0 2m mv 0 F0 2 = mv 2 0 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.
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, 0, 0). Then the equation m¨z = 0 is
easily integrated to give z = z0. We define ω =
q
αe/m and rewrite the remaining two equations as
¨
x = ω2y, y = ω¨ 2x.
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
¨
u1 = ω2u1, u¨2 = −ω2u2.
The initial conditions are u1(0) = u2(0) = x0 and ˙u1(0) = ˙u2(0) = 0. On solving the
resulting uncoupled equations and using the initial conditions, we find u1(t) = x0cosh(ωt), u2(t) = x0cos(ωt). Finally, x(t) = 1 2[u1(t) + u2(t)] = x0 2 [cosh(ωt) + cos(ωt)], y(t) = 1 2[u1(t) − u2(t)] = x0 2 [cosh(ωt) − cos(ωt)].
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 = bv2 is a good
approximation to the air drag, where the speed v = ˙z and b = const. Consequently, we have a separable differential equation:
˙v = b mv 2 − g or dt = g r2v2− g2dv,
where r2 = bg/m is introduced for convenience. On integrating using partial fractions, we get t = Z g dv r2v2− g2 = 1 2 Z 1 rv − g − 1 rv + g ! dv = 1 2rln rv − g rv + g + C. Suppose the initial condition is v(0) = 0. Then the integration constant C = 0 and
v(t) = −grtanh(rt).
We see that v → −g/r = −qmg/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 t 0 v(t)dt = − g r Z t 0 tanh(rt)dt = − g r2 ln [cosh(rt)] .
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 = −GmME 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 ¨ z = F m = −G ME (RE + z)2 ≈ −G ME R2 E .
Numerically, we have RE = 6.38 × 106 m and ME = 5.98 × 1024 kg. Consequently,
g = GME R2 E = 6.67 × 10 −11 5.98 × 1024 (6.38 × 106)2 m s2 ≈ 9.8 m s2.
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 .
