MATH0008 Applied Mathematics 1

(1)

Based on lectures by Dr. Angelika Manhart

Notes taken by Robert Moye

Term 1 of 2020 – 2021

These notes are not endorsed by the lecturer. I have taken significant liberty in modifying them at my own discretion. As such, they should not be used as a replacement for the course material. Whist I have made every effort to ensure these notes are as accurate and useful as possible, I cannot guarantee that they are free from error. If you find any mistakes or anything you think should be changed about these notes, do not hesitate to email me at [email protected]. All my notes can currently be found at robmoye.neocities.org. These notes were last updated on January 9, 2021.

(2)

1 Forces and motion

1.1 Introduction to mechanics

1.1.1 Vectors and scalars

For a formal definition of vectors, including vector spaces, see MATH0005 – Algebra 1, and for scalars see MATH0003 – Analysis 1. Vectors are also covered in MATH0010 – Mathematical Methods 1. For this module, scalars can be described by a single number, and vectors can be described with a magnitude and direction. Scalars Vectors distance displacement speed velocity mass force height acceleration time

Some of these quantities are related to each other, such as distance being the magnitude of displacement and speed being the magnitude of velocity.

1.1.2 Forces as vectors

Force is measured in Newtons. One Newton (N) is the magnitude of force required to accelerate a one kilogram mass at 1 metre per second2_{. Forces are}

vectors. In 2D, F Fx Fy θ F= Fx Fy , Fx=Fcosθ, Fy =Fsinθ Magnitude of forceF =|F|=qF2

x+Fy2, Direction of force tanθ=

Fy

Fx

If multiple forces act on a single point particle, the total, resultant force is the vector sum of these forces. Vector sum is commutative, so it does not matter in which order this is done.

Fres=F1+F2+F3+. . .

F1

F2

F3

(5)

1.1.3 Newton’s laws

Newton’s laws apply to idealised point particles.

Law (Newton’s first law). If the resultant force on an object is zero, then its acceleration is zero. So it moves at a constant velocity.

Law(Newton’s second law). An object subject to a force moves with an accel-eration proportional to that force. SoF=ma.

Law(Newton’s third law). If objectAexerts a force on objectB, thenBexerts a force on Aequal in magnitude and opposite in direction.

1.1.4 Gravity

Gravity is a force that pulls two masses towards each other along the line con-necting their two centres of mass, with magnitude

F =G_·m1·m2 r2

where m1, m2 are the masses of the masses,r is the distance between the two

centres of mass, andGis the gravitational constant 6.674_×10−11_m3_kg−1_s−2_.

The gravitational forces experienced by an object near the earth’s sur-face, usually called its weight, can be calculated using the mass of the Earth

mEarth = 5.972×1024kg and the radius of the earth rEarth = 6.371×106m.

The direction is towards the centre of the Earth, and the magnitude is

FEarth≈G· mEarth·mobj

r2 Earth

≈g_·mobj where g≈9.8 m s−2

The actual value ofgvaries depending on where on Earth and how far above the centre of the Earth it is measured. The gravitational force the object exerts on the Earth is usually treated as negligible, since the mass of the earth and the mass of the object are usually so different.

1.1.5 Tension

Tension is the force transmitted by a rope, cable, string or other similar object when pulled by forces acting from opposite ends. These cables are usually assumed to be light, meaning that their mass is negligible.

Example. Consider suspending a weight of massm with three light cables as shown in the diagram.

T1 T2

T3

θ1 _θ₂

(6)

The tensionT3 has to be equal in magnitude to the weight of the mass so

that it remains in equilibrium. So T3=mg.

The point of intersection of the three cables must also be in equilibrium. The horizontal component and vertical components give

T1cosθ1=T2cosθ2

T1sinθ1+T2sinθ2=T3=mg

Solving forT1andT2 gives

T1= mgcosθ2 sin(θ1+θ2) and T2= mgcosθ1 sin(θ1+θ2)

Example. Consider two masses on frictionless inclined planes held up by a light string going over a frictionless pulley, as in the diagram.

m2

m1

θ1 _θ 2

The tension holding up each mass must be equal for both sides, and for each mass, for the masses to be in equilibrium, the tension must be equal and in an opposite direction to the component of weight acting down the slope. Hence

m1gsinθ1=m2gsinθ2

m1sinθ1=m2sinθ2

1.1.6 Springs

Law (Hooke’s law). The force needed to extend or compress a spring by some distance ∆l is proportional to that distance

F =k∆l

kis called the spring constant, whose units, from the previous equation, can be worked out to be

[k] = N m−1 The spring constantkcan also be expressed as

k= EA

l

Eis a property of a material called the Young’s modulus,Ais the cross-sectional area over which the force is applied andl is the natural, un-stretched length of the string.

Strings can be treated the same as springs under tension, but when com-pressed, they go slack and do not exert a restoring force.

(7)

Example. Consider three light springs, each of natural length l, and spring constant k, arranged vertically between two points 3l apart, and with weights of massmbetween them.

m m

3l

Let the upper weight be a distance l+x1 below the ceiling, and the second

weight be a distance 2l+x2 below the ceiling. Then equating forces for each

weight gives mg+k(x2−x1) =kx1 =⇒ 2x1−x2= mg k mg=k(x1−x1) =kx2 =⇒ 2x2−x1= mg k Hence x1=x2= mg k 1.1.7 Units

SI base units are the standard units of measurement used to measure seven base quantities. The SI units important for this course are the second (s) for time, the metre (m) for length and the kilogram (kg) for mass. Other units for this course are derived, such as N = kg m s−2.

The units of a term are denoted by square brackets around the term. So [t] = s

When differentiating, the units of the term are divided by what the units of what the term is being differentiated with respect to. So

_d_x₍_t₎

dt

= m s−1

When integrating, the units of the term are multiplied by what the term is being integrated with respect to. So

Z

x(t) dt

= m s These can be seen from their definitions.

Only quantities that have the same units can be added together. This means that both sides of an equation must to have the same units.

Many special functions only take dimensionless quantities as arguments, such as cos, exp, log.

(8)

1.2 Solving the equation of motion

1.2.1 The equation of motion

Consider a single particle of mass m moving in a straight line, such as the x -axis, under the action of a force F. This force may depend on the timet, the particles positionx=x(t) and the velocity of the particlev=v(t) =dx

dt = ˙x.

Newton’s second law relates the accelerationa= ¨xto this force.

mx¨=F(t, x,x˙)

This is called the equation of motion for the particle. It is a second order ordinary differential equation forx(t), which is not, in general, possible to solve. However, it is possible to solve it in several important cases.

The acceleration can be represented in several useful ways.

a=a(t) = ¨x= d 2_x dt2 = ˙v= dv dt = dx dt dv dx=v dv dx

Which form should be chosen depends on what techniques are used to solve the equation of motion.

Often initial conditions will be given in terms of initial times, displacements, velocities and forces, which may be written

t0, x(t0) =x0, v(t0) =v0, F(t0, x0, v0) =F0

1.2.2 Constant acceleration

Where acceleration is constant, force is constant, so it does not depend on any of

t, x,x˙. In this case, the equation of motion can be solved directly by integration. This can be done by putting in and working out integration constants.

dv dt =a Z _d_v dt dt= Z adt v=at+c1 v(t0) =v0=at0+c1 c1=v0−at0 v= dx dt =at+v0−at0 Z _d_x dt dt= Z at+v0−at0dt x=at 2 2 + (v0−at0)t+c2 x(t0) =x0= at2 0 2 + (v0−at0)t0+c2 c2=x0−at 2 0 2 −(v0−at0)t0 x=at 2 2 + (v0−at0)t+x0− at2 0 2 −(v0−at0)t0 x(t) =x0+v0(t−t0) +a 2(t−t0) 2

(9)

It can also be done by integrating with limits. When this is done, the integrand must be integrated with respect to a dummy variable ˜t, since this variable cannot be the same as the limit.

˙ v=a Z t t0 ˙ v(˜t) d˜t= Z t t0 ad˜t v(t)−v(t0) =a(t−t0) v(t) = ˙x=v0+a(t−t0) Z t t0 ˙ x(˜t) d˜t= Z t t0 v0+a(˜t−t0) d˜t x(t)₋x(t0) =v0(t−t0) + a 2(t−t0) 2 x(t) =x0+v0(t−t0) +a 2(t−t0) 2

The most common example of 2D motion with forces is projectiles. In this case, the acceleration horizontally is zero and the acceleration vertically is con-stant. Usually it is easiest to just solve the motion in these two axes separately, which gives xandy as parametric equations int.

1.2.3 Force as a function of time

If force is a function of time, then the equation of motion should be integrated with respect to time.

Example. Suppose F(t) = mF0sin(ωt), x(0) = 0 and ˙x(0) =v0. Then

inte-grating twice with respect to time gives ¨ x(t) =F0sin(ωt) ˙ x(t)−v0= F0 ω (cos(ωt)−1) x(t) = v0+ F0 ω t₋F0 ω2sin(ωt)

For this to be bounded for all time, v0=−F_ω0.

1.2.4 Force as a function of velocity

Forces that oppose motion, such as air resistance, often fall into this category. Take care to ensure that all the forces are in the right direction, since if force opposes motion and the direction of motion changes, then the direction of the force needs to change too.

Method.

1. Rewrite in separable form. Usually this involves substituting acceleration for dv_dt orvdv_dx.

2. Integrate with respect to to the relevant variable. 3. Solve for the required form of the answer.

(10)

Forf(v(t))₆= 0, this first step may look like ˙ v=f(v) 1 = v˙(t) f(v(t)) Z t t0 1 dt= Z t t0 ˙ v(t) f(v(˜t)) d˜t Letting ˜v=v(˜t), then d˜v= d˜tv˙(˜t). So t₋t0= Z v(t) v0 1 f(˜v) d˜v

Whilst not formally correct, as a memory help, it may be useful to imagine dv_dt as a fraction, whose denominator, when imagined to be multiplied with the dt

from the integration, cancels out.

Z ₁ f(v) dv dt dt= Z ₁ f(v) dv

Example. Consider a particle subject to a constant force and to air resistance, which is proportional to velocity. Then for constantsλ >0 andP,

¨

x=P₋λx˙

If ever ˙x= P_λ, then acceleration will be zero, so the velocity will not ever change. Otherwise, ¨ x=P−λx˙ dv dt =P−λv Z t 0 d˜t= Z v(t) v0 d˜v P₋λv˜ t=₋1 λln P₋λv P₋λv0 e−λt(P−λv0) =P−λv v(t) = ˙x(t) =e−λt v0−P λ +P λ x(t) =x0+ P λt+ 1 λ 1−e λt v0− P λ

The equation for velocity shows that the terminal velocity is P

λ. This can also

be seen by drawing the equation of motion graphically, in phase space. This is called qualitative analysis, and is covered in more detail in 3.1.3.

(11)

> > > > _| < < < < P λ v ˙ v=f(v) P ˙ v=P₋λv

Where the graph is above the v axis, the acceleration is positive, so v will increase. This is denoted by the arrows pointing to the right, pointing towards the point P_λ. Where the graph is below thev axis, the acceleration is negative, so v will decrease. For any value of v, including the initial condition v0, the

velocity will go towards P

λ, showing that it is a terminal velocity.

Example. Consider a particle launched upwards with initial velocity v0, and

air resistancekv2_{per unit mass. What is the maximum height}_h_{above ground}

level the particle reaches?

Wherey is the height above ground level, the equation of motion is

md

2_y

dt2 =−mg−mkv 2

As was seen previously, dv dt = dv dy dy dt = dv dyv

To remember this, it may be helpful to imagine these derivatives as fractions. If this were the case, then the dy from one fraction would cancel with the other.

However, what is really happening is thatv(t) is being replaced with another function,v(y), which, to differentiate it from the first, should be written ˜v(y).

˜

v y(t)

=v(t)

Differentiating this with respect tot, using the chain rule for the left hand side, gives d˜v dy y(t) · d_dy_t =dv dt d˜v dy˜v= ˙v

So the equation of motion can be written ˜ vd˜v dy =−g−k˜v 2 − Z 0 v0 ˜ v g+kv˜2 d˜v= Z h 0 dy h= 1 2kln 1 +k gv 2 0

(12)

1.3 Force as a function of position

1.3.1 Energy equation

In this case, the equation of motion has the form

md

2_x

dt2 =F(x)

It is usually more productive to work with the integral of₋F(x)

F(x) =₋dV(x) dx

The function V is called a potential. Given a force, the potential is defined up to an additive constant. As such, the particular value of this constant does not have any physical meaning, so it can be chosen to be a convenient value.

To the equation of motion can be rearranged to give

mvdv

dx =−

dV(x) dx

Integrating both sides gives the energy equation.

E=1 2mv

2₊_V₍_x₎

E is a constant called the energy. So the additive constant chosen when obtaining

V serves to shift the values of the energy. E has units of Joules. [E] = kg m2_s−2_{= N m = J}

1 2mv

2 _{is called the kinetic energy and} _V₍_x_{) is called the potential energy.}

The energy equation says that the sum of the kinetic and potential energies has to be constant.

Example. A particle under the influence of gravity has equation of motion

my¨=−mg

So its energy equation is

E= 1 2mv

2₊_mgy

This can be plotted in phase space.

y v

Each curve represents a different energy. The particle starts at some point on the graph, and as time progresses, the particle moves downwards along a curve.

(13)

1.3.2 Simple harmonic motion

In simple harmonic motion, the acceleration is directly proportional in magni-tude but opposite in direction to a particle’s displacement from a fixed point. So mx¨=−mω2x E=1 2mv 2₊1 2mω 2_x2

Plotting this in phase space gives

x v

Each ellipse is a different energy. The particles move clockwise around the graph.

IfE= 0, then the particle remains at rest atx= 0. Otherwise, ifE >0,

v2= 2E m 1−mω 2 2E x 2 ± Z dt= Z pm 2E dx q 1−mω2 2E x2 Letu= r mω2 2E x. Then ±t= 1 ω Z _d_u √ 1₋u2 = 1 ωsin −1_u₊_c x=± r 2E mω2sin(ωt−c)

For the positive part, defineφ=c. For the negative part, defineφ=c+π. Let

A=qmω2

2E . Then

x=Asin(ωt−φ)

The amplitude of the oscillator isA. φ is called the phase. Since sin has period 2π, the period ofxis 2π

ω. ω is called the angular frequency, andf = ω 2π

is called the frequency.

This equation can also be derived using the method in MATH0010 – Math-ematical methods 1 using ansatz.

1.3.3 Periodic motion and potentials

Solving the energy equation for the velocity gives

v=±

r

2

m E−V(x)

(14)

So motion is only possible when E_≥V(x). Consider the graph below.

x y y=V(x)

E

x1 x2 x3 x4

Since E _≥ V(x), the motion of the particle must be confined entirely to the interval [x1, x2] or [x3, x4]. The lower down on the graph the particle is, the

greaterE₋V(x) is, so the particle moves faster. Acceleration is related to the gradient of the graph, as can be seen by the equation of motion.

mx¨=₋dV(x) dx

Consider a particle that is atx=x1. SinceV(x1) =E,v= 0, so the particle

is momentarily at rest. At this instant, acceleration is ¨ x= 1 mF(x1) =− 1 mV 0₍_x 1)>0

So the particle moves to the right. Before the particle can move left again, it must first become zero, which it must do atx=x2. At this point, acceleration

is ¨ x= 1 mF(x2) =− 1 mV 0₍_x 2)<0

So the particle moves back to the left, towardsx1. Therefore the particle

oscil-lates betweenx1 andx2.

The time taken for the particle to move fromx1tox2 and back again might

be wanted. This is called the period of the motion. LetT1 be the time to go

from x1 and x2, and T2 be the time to go from x2 to x1. Since xis always

increasing going fromx1to x2, the equation for velocity is

dx dt = + r 2 m E−V(x) T1= r m 2 Z x2 x1 dx p E₋V(x)

Going from x2 tox1, the velocity is negative, so the time taken is

T2=− r m 2 Z x1 x2 dx p E₋V(x) =T1

(15)

So the period is T =T1+T2= 2T1= √ 2m Z x2 x1 dx p E−V(x)

In general, this integral cannot be solved explicitly. For simple harmonic motion, it can be solved explicitly, and gives the same period as was seen previously.

1.3.4 Equilibrium points

Points at which particles can remain at rest for all time are called equilibrium points. The force acting on a particle must be zero at such a point. So for an equilibrium point x=cwith differentiableV,

V0(c) = 0

An equilibrium point is stable if any particle with small velocity starting sufficiently near the point remains near the point. Otherwise, it is unstable. Stable equilibrium points are minimums, whilst unstable equilibrium points are maximums (or points of inflection).

x y y=V(x)

stable

stable unstable

1.3.5 Small oscillations about stable equilibrium points

By Taylor expansion, for smallx₋c,

mx¨=F(x)_≈F(c) +F0(c)(x₋c)

Ifc is an equilibrium point, thenV0(c) = 0. LetX =x−c. Then for smallX,

mx¨≈ −V0(c)−V00(c)(x−c) ¨

X+V 00₍_c₎ m X ≈0

IfV00(c)>0, thenc is a stable equilibrium (this condition is sufficient, but not necessary). Letω= q V00_(c) m . Then ¨ X+ω2X_≈0

This is the equation for simple harmonic motion, which has period 2π

ω. So small

oscillations about the stable equilibrium pointchave approximate period 2π

r _m

(16)

Graphically, the curveV(x) is being approximated nearx=cby a positive quadratic, which is the shape of simple harmonic motion.

x y

y=V(x)

1.3.6 Degenerate energy values

SupposeV has a local maximum at c=x2 such thatV(x2) =E,V is smooth

andV00₍_c₎₆_{= 0.} x y y=V(x) E x1 _b x2

The time for the particle to move fromx1to x2 is

T1= r m 2 Z x2 x1 dx p E−V(x) Letα2_:= −V00(c)

2 . Then by Taylor expansion, forxnear c,

V(x)_≈V(c) +V0(c)(x₋c) +1 2V

00₍_c₎₍_x₋_c₎2

=E₋α2(x₋c)2 So for a pointb nearc,

Z c b dx p E−V(x) ≈ Z c b dx p α2₍_x₋_c₎2 = 1 α Z c b dx c−x = 1 αεlim→0+ −ln(ε) + ln(c−b) =∞

(17)

1.4 Collisions

1.4.1 Conservation of momentum

This topic considers idealised collisions between spheres. Suppose there are two spheres with masses m1 andm2 with position vectorsr1(t) and r2(t). If there

are no forces acting on the spheres before the collision, then

¨r1=¨r2=0

During the brief time the collision occurs, the first sphere exerts a forceF(t) on the second, and by Newton’s third law, the second sphere exerts a force −F(t) on the first. So

Before collision m1¨r1=0 m2¨r2=0

During collision m1¨r1=F(t) m2¨r2=−F(t)

After collision m1¨r1=0 m2¨r2=0

So at every moment in time

m1¨r1+m2¨r2=0

Integrating this gives

m1˙r1+m2˙r2= constant

The terms mv are called the momentums of the spheres, and this equation shows that momentum is conserved.

Law(Conservation of momentum). For spheres of massm1andm2with initial

velocities of U1 andU2 and final velocities after the collision ofV1andV2,

m1U1+m2U2=m1V1+m2V2

The left hand side of the equation is the initial momentum of the system, and the right hand side is the final momentum.

This equation does not provide enough information to determine the final velocities given the initial velocities.

1.4.2 Coefficient of restitution

Consider spheres colliding in one dimension. Then, the velocities,u1, u2, v1, v2

are one dimensional.

m1 u1 m2 u2 m1 u2 m2 u2 before collision after collision

The initial and final kinetic energies of the system are 1 2m1u 2 1+ 1 2m2u 2 2 and 1 2m1v 2 1+ 1 2m2v 2 2

(18)

If kinetic energy is conserved, then a collision is called perfectly elastic. Then 1 2m1u 2 1+ 1 2m2u 2 2= 1 2m1v 2 1+ 1 2m2v 2 2 m1(v12−u21) =−m2(v22−u22)

Conservation of momentum can be written in a similar form

m1(v1−u1) =−m2(v2−u2)

Clearly this has a solution v1 =u1, v2 =u2, but this is not interesting as it

corresponds to the case in which no collision occurs. Otherwise, the kinetic energy equation can be divided by the conservation of momentum equation to give

v2−v1=−(u2−u1)

This can be used with the conservation of momentum equation to findv1 and

v2 in terms ofu1and u2.

Suppose instead a proportion of energy is conserved. In this case, the colli-sion is called inelastic. The constant eis called the coefficient of restitution. e

only makes sense for 0_≤e_≤1. Ife= 0, then the collision is called perfectly inelastic. The difference between the final speeds of the spheres is 0, so the spheres are said to coalesce. Similarly to before,

v2−v1=−e(u2−u1)

This equation with the conservation of momentum equation form a system of linear equations m1 m2 −1 1 v1 v2 = m1 m2 e ₋e u1 u2

This has solution (so long asm1+m26= 0)

v1=

(m1−em2)u1+m2(1 +e)u2

m1+m2 and v2=

(m2−em1)u2+m1(1 +e)u1

m2+m1

1.4.3 Collisions in two dimensions

Consider two spheres moving in two dimensions. In the instant the spheres collide, chose a pair of of orthonormal (unit length and perpendicular) vectors

i,j such that i is in the direction of the line connecting the centre of the two spheres. Then jmust be tangent to the spheres at the point where they touch.

j i U2 V2 U1 V1

(19)

i,jform a basis over the 2D column vectors, so let

U1=Ui1i+Uj1j

U2=Ui2i+Uj2j

V1=Vi1i+Vj1j

V2=Vi2i+Vj2j

Note that the componentsU1

i . . . can be written as inner products, such as

Ui1=U1·i

After the collision, there is no change in the components of the velocities in thejdirection.

Vj1=Uj1 and Vj2=Uj2

The components of the velocities in the idirection act analogously to the one dimensional case. Hence

V1= (m1−em2)Ui1+m2(1 +e)Ui2 m1+m2 i+Uj1j V1= (m2−em1)Ui2+m1(1 +e)Ui1 m2+m1 i+Uj2j

(20)

2 Waves

2.1 Differential equations

2.1.1 Introduction to waves

Waves can be thought of as oscillations that transport energy. They can broadly be categorised as follows:

1. Mechanical waves. These are waves that require a medium to propagate through. Examples include sound waves, water waves and elastic waves. 2. Electromagnetic waves, such as light, radio waves and microwaves. 3. Gravitational waves.

Electromagnetic and gravitational waves can travel through a vacuum, unlike mechanical waves.

Ordinary differential equations, also called ODEs, are covered in MATH0010 – Mathematical Methods 1, so the theory on constant coefficient second order linear ODEs has been omitted here.

2.1.2 Damped oscillators

Hooke’s law and Newton’s law give the equation of motion of a weight of mass

mon a spring with spring constant kas

mx¨=₋kx

If a damping term with coefficientcis introduced, then the equation of motion becomes

mx¨+cx˙ +kx= 0

Since the equation can be divided through by m, m is taken to be one. This gives the equation of motion for a damped oscillator

¨

x+cx˙+kx= 0

This is a constant coefficient second order homogeneous linear ordinary dif-ferential equation. The rootsλ1, λ2to the characteristic equation are

λ=₋c 2 ± √ ∆ 2 where ∆ =c 2 −4k

Depending on the sign of the discriminant ∆, there are three cases.

1. Over-damping: ∆>0 whenc2_>₄_k_{. The two roots are real and distinct.}

Since c2 _> _∆ _> _{0, the roots are negative. So let} _µ

i = −λi. Then the

general solution is

x=c1e−µ1t+c2e−µ2t

This decays rapidly to zero.

2. Critical damping: ∆ = 0 when c2_{= 4}_k_{. There is one repeated real root}

−c

2. So the general solution is

x= (c1+c2t)e−

c

2t

(21)

3. Under-damping: ∆ < 0 when c2 _< ₄_k_{. There are two complex roots}

−c

2±νi, where ν= −

√

∆

2 . So the general solution is

x=e−c2(c₁cosνt+c₂sinνt)

With arbitrary constantsAandδ, the solution can also be written

x=Ae−c2sin(νt−δ)

This solution decays to zero, but it oscillates around aboutx= 0 whilst getting there.

Under-damped or critically damped systems may look like one of the curves below.

t x

Over-damping may look similar to one of these curves:

t x

2.1.3 Forced oscillators

An oscillator subject to an external time-dependent force F(t) has equation of motion

mx¨+cx˙+kx=F(t)

Consider the case in which an undamped oscillator (c= 0) is driven by the force

(22)

whereω >0 andF0 are constants. Then lettingω0= q k m andf = F0 m, ¨ x+ω2 0x=fcosωt

This has general solution, with constantsc1 andc2,

x(t) =c1cosω0t+c2sinω0t+

f ω2

0−ω2

cosωt

The initial conditions x(0) = 0 and ˙x(0) = 0 give the particular solution

x(t) = f

ω2 0−ω2

(cosωt−cosω0t)

Using the compound angle formula for cos gives

x(t) = f ω2 0−ω2 cos ω+ω0 2 t+ ω−ω0 2 t −cos ω+ω0 2 t− ω−ω0 2 t x(t) = 2f ω2 0−ω2 sin _ω₊_ω 0 2 t sin _ω −ω0 2 t

Supposeω _≈ω0 but ω 6= ω0. Then|ω−ω0| is much smaller thatω+ω0,

so sin ω−ω0

2 t

oscillates much slower than sin ω+ω0

2 t

. Sox(t) is a graph with fast oscillations and a slowly varying maximum amplitude. This phenomenon is called beats.

t x

Supposeω=ω0, with the same initial conditions as before. The solution is

x(t) = f 2ω0

tsinω0t

The graph ofx(t) oscillates with increasing maximum amplitudes. The motion is unbounded, which means the mathematical model will usually break down. This often corresponds to the physical structure also breaking down, such as a spring breaking. This phenomenon is called resonance.

(23)

t x

2.2 Wave equation

2.2.1 Derivation

Again, theory for partial derivatives is covered in MATH0010 – Mathematical Methods 1.

Imagine an elastic string stretched between two fixed end points at x= 0 andx=L, with mass per unit length ofµ. Denote the tension in the string by

T(x, t), the vertical displacement byu(x, t) and the angle above the horizontal byθ(x, t).

Consider a small section of the string on [x, x+∆x], where ∆xis small. This has a difference in vertical displacement of

∆u=u(x+ ∆x, t)₋u(x, t)

In this model, points on the string will only move vertically, not horizontally. This means that the horizontal components of the tension forces must cancel. So

T(x+ ∆x, t) cos θ(x+ ∆x, t)

=T(x, t) cos θ(x, t)

Newton’s second law gives an equation involving the vertical forces.

µp(∆u)2_{+ (∆}_x₎2 ∂ 2_u

∂t2 =T(x+ ∆x, t) sin θ(x+ ∆x, t)

−T(x, t) sin θ(x, t)

sinθ= tanθcosθ, and for small θ, tanθ_≈ ∆u

∆x. This is the slope of the string,

which can also be expressed, for small ∆u, as∂xu. So

µp(∆u)2_{+ (∆}_x₎2 ∂ 2_u ∂t2 = T(x+ ∆x, t) cos θ(x+ ∆x, t) ∂xu(x+ ∆x, t) −T(x, t) cos θ(x, t) ∂xu(x, t)

Substituting in the equation for horizontal forces gives

µp(∆u)2_{+ (∆}_x₎2 ∂ 2_u ∂t2 =T(x, t) cos θ(x, t) ∂xu(x+ ∆x, t)−∂xu(x, t) Now divide by ∆x. µ s ∆u ∆x 2 + 1 ∂ 2_u ∂t2 =T(x, t) cos θ(x, t) ∂xu(x+ ∆x, t)−∂xu(x, t) ∆x

(24)

Take the limit as ∆x_→0. Then by the definitions of partial derivatives, µp(∂xu)2+ 1 ∂2_u ∂t2 =T(x, t) cos θ(x, t) ∂ 2_u ∂x2

Similarly, dividing the equation for horizontal forces by ∆xand taking the limit as ∆x_→0 gives

∂x

T(x, t) cos θ(x, t)

= 0

Again, assume θ is very small. Then tanθ is also small, so (∂xu)2 ≈0. Also

cosθ≈1, so the previous equation shows that tension is constant in space. If tension is additionally assumed to be constant in time, thenT(x, t)≡T. So

µ∂ 2_u ∂t2 =T ∂2_u ∂x2 Letc2₌ T

µ. This gives the wave equation.

∂2_u

∂t2 =c 2∂2u

∂x2

c is called the wave speed, since it has units [c] = m s−1_.

2.2.2 Important waves

Definition (Travelling wave). A travelling or progressive wave is a wave that preserves its shape and travels at a constant speed. For some function f, any travelling wave can be written

y(x, t) =f(x₋vt)

f is called the profile or shape, and_|v_| is the speed of the wave. Ifv > 0, the wave moves to the right. Ifv <0, it moves to the left.

x y

t= 0 t >0

It can be seen thaty(x, t) solves the wave equation, assumingf is smooth enough, provided thatv=±c.

A simple harmonic wave has the form

Asin k(x−vt)−δ

or Acos k(x−vt)−δ

Ais the amplitude of the wave,v is the wave speed,kis the wave number and

δis the phase. The angular frequency isω=kv.

Definition (Standing waves). Standing waves are waves in which every par-ticle oscillates up and down in time, each with an amplitude dependent on its position. For some oscillating functionT, any standing wave can be written

(25)

x y _t_{= 0}

t >0

Example. Consider the wave

y(x, t) =acos(αx) sin(βt)

By calculating its partial derivatives, it solves the wave equation provided that

β

α =±c. Also, using double angle formulae,

acos(αx) sin(βt) =a 2 sin(αx+βt)−sin(αx−βt) =a 2sin α x+β α −a₂sin α x₋β α

So this standing wave can be expressed as the sum of two travelling waves, with wave speed β_α.

2.2.3 Boundary and initial conditions

Like any differential equation, in order to solve the wave equation, its it needs some conditions imposing on it.

When deriving the wave equation, it was assumed to have fixed endpoints atx= (0, L). Therefore the boundary conditions are

u(0, t) =u(L, t) = 0

Att= 0, the string must have some shape and velocity. These can be written as the initial conditions

u(x,0) =f(x)

∂tu(x,0) =g(x)

for some functionsf(x) andg(x).

2.2.4 Solving the wave equation

Suppose a solution has the form

u(x, t) =F(x)G(t) Then ∂2_u ∂f2 =F(x) ¨G(t) ∂2_u ∂x2 =F00(x)G(t) F(x) ¨G(t) =c2F00(x)G(t) ¨ G(t) c2_G₍_t₎ = F00₍_x₎ F(x)

(26)

The left hand side is a function of twhilst the right hand side is a function of

x. Therefore, they must be both equal to a constant, let this bek. So

F00(x)₋kF(x) = 0 ¨

G(t)₋c2_kG₍_t_{) = 0}

2.2.5 Applying boundary conditions to the solution

The boundary conditions give that for allt,

u(0, t) =G(t)F(0) = 0

u(L, t) =G(t)F(L) = 0 so

F(0) =F(L) = 0

The solutions toF00(x)−kF(x) = 0 can only satisfy these boundary conditions ifkis negative. So letk=−p2_{. So}

F00(x) +p2F(x) = 0

F(x) =c1cos(px) +c2sin(px)

The boundary conditions give c1 = 0 and sin(pL) = 0, so pL =nπ for some

integer n. So any constant multiple of

Fn(x) = sin

nπx

L

is a solution forn= 1,2,3, . . .. These spatial functions are called modes.

Fn(x) = 0 at x= 0, 1 nL, 2 nL, . . . , n₋1 n L, L

These points are called nodes. Notice they include the end points of the string.

x u F1(x) 0 L x u F2(x) 0 L L 2 x u F3(x) 0 L L 3 2L 3

The second equation is now ¨

G(t) +c2nπ

L

2

G(t) = 0 Forn= 1,2,3, . . ., this has solution

Gn(t) =Cncos nπc L t +Dnsin nπc L t

(27)

So for anyn= 0,1,2, . . . and any coefficientsCn andDn,

Fn(x)Gn(t)

solves the wave equation.

The wave equation is linear, meaning any solutions can be added or multi-plied by scalers to give a new solution. So

u(x, t) = ∞ X n=0 Fn(x)Gn(t) = ∞ X n=0 Cncos _nπc L t +Dnsin nπc L t sinnπx L

is a solution to the wave equation for any coefficientsCnandDn,n= 0,1,2, . . ..

2.2.6 Applying initial conditions to the solution

To find a specific solution, the initial conditions need to be used. The initial conditions are u(x,0) =f(x) and ∂tu(x,0) =g(x) So it is required that ∞ X n=0 Cnsin nπx L =f(x) and ∞ X n=0 Dn ncπ L sin nπx L =g(x) To find these coefficientsCn andDn requires Fourier series.

2.2.7 Using Fourier series to find a solution

Using Fourier sine series on [0, L] gives

Cn= 2 L Z L 0 f(x) sinnπx L dx and Dn= 2 ncπ Z L 0 g(x) sinnπx L dx

If the string is released from rest, then G(x) = 0 soDn= 0. Then

u(x, t) = ∞ X n=1 Cncos nπc L t sinnπx L = 1 2 ∞ X n=1 Cnsin nπ L (x−ct) +1 2 ∞ X n=1 Cnsin nπ L (x+ct) = 1 2 f(x−ct) +f(x+ct)

Therefore the solution to the wave equation for a string released from rest with formf(x) is the mean of two travelling waves with formf travelling to the left and right with speed c.

2.3 Fourier series

2.3.1 Periodic, even and odd functions

Definition(Periodic functions). A functionf :R→Ris periodic with period

pif

(28)

Definition(Even and odd functions). A functionf :R→Ris even if

∀x_∈R f(−x) =f(x)

A functionf :R→Ris odd if

∀x_∈R f(−x) =−f(x)

2.3.2 Deriving the Fourier series

Suppose constants an and bn are independent of x, and the following series

converges for allx:

a0 2 + ∞ X n=1 ancos nπx L +bnsin nπx L

Then this defines a 2L periodic functionf(x)

f(x) = a0 2 + ∞ X n=1 ancos nπx L +bnsin nπx L

To findan andbn, the following identities will be needed

(F1) Z L −L cosnπx L dx= 0 (F2) Z L −L sinnπx L dx= 0 (F3) Z L −L sinmπx L cosnπx L dx= 0 (F4) Z L −L sinmπx L sinnπx L dx=Lδmn (F5) Z L −L cosmπx L cosnπx L dx=Lδmn

δij is the Kronecker delta. δij = 1 if i = j, and δij = 0 otherwise. This is

covered in MATH0010.

Proof. (F1) and (F2) are obvious. (F3) is a consequence of sin being odd and cos being even. For (F4), the compound angle formula for cos can be used.

Z L −L sinmπx L sinnπx L dx =1 2 Z L −L cos ₍_m −n)πx L −cos ₍_m₊_n₎_πx L dx = ( 0 ifm₆=n 1 2 RL −L 1−cos 2nπx L dx ifm=n

(29)

To find the coefficients inf(x), start by integrating it. Then by (F1), Z L −L f(x) dx= Z L −L a0 2 + ∞ X n=1 ancos nπx L +bnsin nπx L ! dx =a0L

Now multiplyf(x) by sin mπx_L

for some positive integerm.

Z L −L f(x) sinmπx L dx = Z L −L a0 2 sin mπx L + ∞ X n=1 ancos nπx L sinmπx L +bnsin nπx L sinmπx L ! dx = Z L −L a0 2 sin mπx L dx+ ∞ X n=1 Z L −L ancos nπx L sinmπx L +bnsin nπx L sinmπx L ! dx =Lbm Similarly, Z L −L f(x) cosmπx L dx=Lam

2.3.3 Definition of the Fourier series

Definition (Fourier series). The Fourier series of f on (₋L, L), denoted by

s[f](x) is the series a0 2 + ∞ X n=1 ancos nπx L +bnsin nπx L where forn= 0,1,2, . . . an = 1 L Z L −L f(x) cosnπx L dx and forn= 1,2,3, . . . bn= 1 L Z L −L f(x) sinnπx L dx

Whethers[f](x) =f(x) on (₋L, L) is true or not is beyond the scope of this module, so it will be assumed to be true.

Definition (Fourier cosine series). If f is even, bn = 0 for all n = 1,2, . . ..

Then the Fourier series for f on (−L, L) becomes the Fourier cosine series

s[f](x) = a0 2 + ∞ X n=1 ancos nπx L

(30)

where forn= 0,1,2, . . . an= 2 L Z L 0 f(x) cosnπx L dx

This formula only uses values of f on the interval (0, L) since iff is even, the interval (₋L,0) is also known. The Fourier cosine series can be calculated even for functions that are not even.

Definition(Fourier sine series). Iff is odd,an= 0 for alln= 0,1,2, . . .. Then

the Fourier series for f on (₋L, L) becomes the Fourier sine series

s[f](x) = ∞ X n=1 bnsin nπx L where forn= 1,2,3, . . . bn= 2 L Z L 0 f(x) sinnπx L dx

This formula only uses values off on the interval (0, L) since iff is odd, the interval (−L,0) is also known. The Fourier sine series can be calculated even for functions that are not even.

2.3.4 Examples

Example. Find the Fourier series on [₋π, π] for the step function

f(x) =

(

1 if 0< x < π −1 if −π < x <0

This function is odd, so the Fourier sine series on [0, π] can be used. So

bn= 2 L Z L 0 f(x) sinnπx L dx = 2 π Z π 0 sinnxdx = −_nπ2 cosnx π 0 = 2 nπ(1−cosnπ) = 2 nπ(1−(−1) n₎ = ( 0 ifnis even 4 nπ ifnis odd

Hence the Fourier series off on [₋π, π] is

s[f](x) = ∞ X n=1 bnsinnx= 4 π ∞ X j=1 1 2j−1sin (2j−1)x

(31)

Example. Find the Fourier series on (₋π, π) for f(x) = x2_{. Assuming that}

this series converges to the 2π-periodic extension off, evaluate ∞

X

n=1

1

n2

Sincef is an even function,bn= 0 for alln.

a0= 2 L Z L 0 f(x) cos ₀_πx L dx= 2 π Z π 0 x2dx=2π 2 3 Using integration by parts, forn= 1,2, . . .

an = 2 L Z L 0 f(x) cosnπx L dx= 2 π Z π 0 x2_cos_nx_d_x₌ 4 cosnπ n2 = 4(₋1)n n2

Hence the Fourier series off on (−π, π) is

s[f](x) =π 2 3 + ∞ X n=1 4(₋1)n n2 cosnπ Evaluating atx=πgives π2₌π2 3 + ∞ X n=1 4(₋1)n n2 cosnπ ∞ X n=1 1 n2 = π2 6 2.3.5 Parseval’s identity

Consider the Fourier series of a functionf on [₋L, L]

s[f](x) = a0 2 + ∞ X n=1 ancos nπx L +bnsin nπx L

(32)

If it converges tof, then 1 L Z L −L f(x)2 dx =1 L Z L −L a2 0 4 + a0 2 ∞ X n=1 ancos nπx L +bnsin nπx L ! dx + 1 L Z L −L ∞ X n=1 ∞ X m=1 anamcos nπx L cosmπx L + ∞ X n=1 ∞ X m=1 bnbmsin nπx L sinmπx L ! dx + 2 L Z L −L ∞ X n=1 ∞ X m=1 anbmcos nπx L sinmπx L ! dx = a 2 2 + 0 + 0 + ∞ X n=1 ∞ X m=1 anamδmn+ ∞ X n=1 ∞ X m=1 bnbmδmn+ 0 = a 2 2 + ∞ X n=1 a2n+ ∞ X m=1 b2m

This is analogous to saying that the dot product of a vector with itself is the sum of its components squared.

Example. Use Parseval’s Identity to find ∞

X

n=1

1

n2

From the previous example, the Fourier series forf(x) =x2 _{is known. So}

1 π Z π −π x4dx= 2 π π5 5 = 2 5π 4₌a20 2 + ∞ X n=1 a2n= 1 2 2π2 3 2 + ∞ X n=1 16 n4 ∞ X n=1 1 n4 = 1 16 2 5π 4 −2₉π4 = π 4 90

(33)

3 Population models

3.1 First order ODE models

3.1.1 Exponential growth

Let P0 ≥ 0 be the initial population size and let P(t) be the population size

at time t. After a time interval ∆t later, the population size has changed due to births and deaths. These are both modelled to be proportional toP(t) with ratesbanddrespectively. So

P(t+ ∆t) =P(t) +b∆tP(t)₋d∆tP(t)

P(t+ ∆t)−P(t)

∆t = (b−d)P(t)

Taking the limit as ∆t→0, and letting g be the overall growth rateg=b−d

gives

dP

dt =gP

This has solution

P(t) =P0egt

If P0 = 0 then P(t)≡ 0. If P0 >0 then for g < 0, the population decreases

exponentially, forg >0, the population increases exponentially.

This model has a few issues. Forg <0, the population will never die out completely, it will just get arbitrarily close to zero. For g >0, the population continues to grow exponentially without bound. These issues can be solved with more complex models, such as the next one, but in some situations, such as the start as of an epidemic, this model is still useful.

3.1.2 Logistic growth

Unlike the previous model, this one will assume that for a high population, density and competition for resources would limit population growth. This is done by writingg as a function ofP, such as forr, a >0,g(P) =r₋aP.

LettingK= r

a be the carrying capacity gives the logistic growth model.

dP

dt =rP

1−_KP

Let P0 =P(0) be the initial population. Then this ODE can be solved using

separation of variables to give

P(t) = KP0

P0+ (K−P0)e−rt

IfP0= 0 thenP(t)≡0. Otherwise, P(t)→Kas t→ ∞.

3.1.3 Qualitative analysis

The exponential growth model is linear, making it easy to solve. The logistic growth model is non-linear, making it harder to solve. Many non-linear models

(34)

cannot be solved explicitly. However, their qualitative behaviour (when the function increases or decreases) and its long term behaviour can still be analysed.

Reconsider the Logistic growth model: dP

dt =rP

1−_KP

The solutionP(t) will increase ifP < K and it will decrease ifP > K. It will stay constant if P = 0 or P =K. For 0< P0< K, the solution will increase

and converge toK from below, and forP0> K, the solution will decrease and

converge to K from above. The next section will show thatK is a stable fixed point, and 0 is an unstable fixed point.

This behaviour can be seen by plotting P(t) against t for different values. These graphs can be made even when the solution is not known.

P t

Similarly to the qualitative analysis in 1.2.4, the behaviour of the logistic growth model can also be seen by plotting ˙P = f(p) in phase space. Where

f(P) is above theP axis, ˙P is positive, soP increases. Similarly, where f(P) is below theP axis, ˙P is negative, so P decreases.

0 > > > > _| < < < <

K _P

˙

P

Qualitative analysis works for any scalar first order autonomous ODE. This means that the rate of change of population depends only the population, not time.

˙

P =f(P)

f andf0 _{are also assumed to be continuous.} 3.1.4 Linear stability

Steady states and fixed points were already seen in 1.3.4. They are similar for population models.

(35)

Definition(Fixed point). A real numberP∗ _{is an equilibrium or steady state} or fixed point of ˙P=f(P) if

f(P∗) = 0

Definition(Stable and unstable fixed points). A fixed point is stable if, given an initial population near the fixed point, the solution stays close to the fixed point for all time. A fixed point is unstable if it isn’t stable.

LetP∗ _{be a fixed point of ˙}_P ₌_f₍_P_{), and let} _P₍_t_{) =} _P∗₊_p₍_t_{) with} _p₍₀₎ close to zero. Then p(t) is initially small. p is called a perturbation. Using Taylor expansion, f(P) =f(P∗+p)≈f(P∗) +f0(P∗)p(t) =f0(P∗)p(t) ˙ p= ˙P, so nearP∗, dp dt =f 0₍_P∗₎_p₍_t₎

This is the model for exponential growth, so its solution will exponentially grow or decay depending on the sign off0₍_P∗_).

LetP∗ _{be a fixed point of ˙}_P ₌_f₍_P_{). Then} − Iff0₍_P∗₎_<_{0, then}_P∗ _{is a stable fixed point.} − Iff0₍_P∗₎_>_{0, then}_P∗ _{is an unstable fixed point.}

− Iff0₍_P∗_{) = 0, then more terms in the Taylor expansion are needed.} Example. Consider the logistic growth model.

dP dt =f(P) =rP 1₋P K

At 0, f0(0)>0, so 0 is unstable. AtK,f0(K)<0, soK is unstable. This can be seen in the graph of f.

0 P ˙ P _˙ P =f(P) K 3.1.5 Competition model

Suppose there are two populationsX(t) andY(t). They may be related by the equations

˙

X =f(X, Y) ˙

(36)

Suppose this is a linear system of ODEs for two unknown functions x1(t)

andx2(t). This has the general form, with constants a,b,c,d,

˙

x1=ax1+bx2

˙

x2=cx1+bx2

This can be written in matrix form

˙x=Ax where x= x1 x2 , A= a b c d

It is easy to check that ifxandyare solutions, then for any constantsα, β,

αx+βy is a solution. As a guess, suppose

x(t) =eλt_v _where _λ ∈R, v= v1 v2

is a solution, where not both v1, v2 = 0. It can be shown that such solutions

exist ifv andλfulfil

Av=λv (A₋λI)v=0 det a₋λ b c d₋λ = 0 (a₋λ)(d₋λ)₋bc= 0 λ2 −(a+d)λ+ (ad₋bc) = 0

The vector v is called an eigenvector ofA, andλis called an eigenvalue ofA. If there are two non-parallel eigenvectorsv andw with eigenvaluesλ1 andλ2,

then the general solution can be written

x(t) =c1eλ1tv+c2eλ2tw

Note that λ1 andλ2could be complex. The long term behaviour can be found

depending on the values ofλi.

Re(λ1)<0 and Re(λ2)<0 =⇒ x(t)→0ast→ ∞

Re(λ1)>0 or Re(λ2)>0 =⇒ |x(t)| → ∞as t→ ∞

3.1.6 Linear stability for systems of equations Definition (Fixed point). A point (X∗_{, Y}∗_{) in}

R2is an equilibrium or steady state or fixed point if

f(X∗, Y∗) = 0

g(X∗, Y∗) = 0

The definitions of stable and unstable are the same as previously. Suppose there is a fixed point at (X, Y) = (X∗_{, Y}∗_{). Then}

(37)

Near the fixed point let X(t) = X∗₊_x₍_t_{) and}_Y₍_t_{) =}_Y∗₊_y₍_t_{), where} _x₍₀₎ andy(0) are small. Again,x(t) andy(t) are called perturbations. Using Taylor expansion, ˙ x= ˙X =f(X, Y)_≈f(X∗, Y∗) + ∂f ∂X(X ∗_{, Y}∗₎_x₊ ∂f ∂Y (X ∗_{, Y}∗₎ ˙ y= ˙Y =g(X, Y)≈g(X∗, Y∗) + ∂g ∂X(X ∗_{, Y}∗₎_x₊ ∂g ∂Y (X ∗_{, Y}∗₎

The partial derivatives are evaluated at the fixed point, so are constants. In the order they are listed, let them beA, B, C, D. So

˙

x=Ax+By

˙

y =Cx+Dy

This is a linear system. So the behaviour at the fixed points depends on the signs of the real parts of its eigenvaluesλ1, λ2, which fulfil

λ2−(A+D)λ+ (AD−BC) = 0 Re(λ1)<0 and Re(λ2)<0 =⇒ (X∗, Y∗) is stable

Re(λ1)>0 or Re(λ2)>0 =⇒ (X∗, Y∗) is unstable

Example. LetX(t) andY(t) be competing species modelled by ˙

X =X(5−X−Y) =f(X, Y) ˙

Y =Y(8₋2Y ₋X) =g(X, Y)

with initial conditionsX(0) =X0≥0 andY(0) =Y0≥0. The−XY terms

rep-resent population decline proportional to the frequency of encounters of species

X and speciesY. This model has fixed points at

P0= (0,0), P1= (0,4), P2= (5,0), P3= (2,3)

The partial derivatives are

∂Xf(X, Y) = 5−2X−Y, ∂Yf(X, Y) =−X

∂Xg(X, Y) =−Y, ∂Yg(X, Y) = 8−X−4Y

The linearised system atP0 is

˙

x= 5x

˙

y= 8y

which has eigenvaluesλ1= 5 and λ2= 8, henceP0 is unstable.

˙

x=x

˙

(38)

which has eigenvaluesλ1=−8 andλ2= 1, henceP1 is unstable.

˙

x=₋5x₋5y

˙

y= 3y

which has eigenvaluesλ1=−5 andλ2= 3, henceP2 is unstable.

˙

x=₋2x₋2y

˙

y=−3x−6y

which has eigenvaluesλ1=−4−

√

10 andλ2=−4 +

√

10, henceP4 is stable.

3.1.7 Phase Portraits

A phase portrait shows the dynamics in phase space, which has axesX andY. Solutions can be represented as curves t _7→ (X(t), Y(t)). Fixed points should be marked on.

Nullclines are curves on which either ˙X or ˙Y are zero. So X-nullclines are where the direction field is vertical andY-nullclines are where the direction field is horizontal.

The direction field is a field of vectors

_˙

X

˙

Y

. Representative vectors should be calculated or worked out using the continuity of the system of equations to give an idea of how the entire vector field looks.

Example. Continuing the previous example, the fixed points P0, P1, P2, P3

should be marked on. TheX-nullclines are

X= 0

X+Y = 5 and the Y-nullclines are

Y = 0

X+ 2Y = 8

Notice this includes both axes.

At (X, Y) = (1,1), ˙X > 0 and ˙Y > 0. So the entire bottom left sector bounded by the nullclines must point up and right. Similarly for large X and

Y, ˙X < 0 and ˙Y <0. So the entire top right bounded by the nullclines must point down and left. Looking at the area between the nullclines, the vectors must point towardsP3.

(39)

0 2 4 6 8 0 1 2 3 4 5 X Y

From this diagram, the behaviour can be understood globally.

3.1.8 Lotka-Volterra system

The Lotka-Volterra system is a predator-prey model, and unlike the previous models studied, it may have non-constant long term behaviour. Letx(t) be the population size of predators andy(t) be the population size of prey.

This model says that in the absences of predators, the prey population creases exponentially, and in the absence of prey, the predator population in-creases exponentially. In addition, encounters of prey and predators increase the predator population and decrease the prey population. This gives the following equations, wherea, b, c, d are positive constants:

˙

x= (₋a+by)x=f(x, y) ˙

y= (c−dx)y =g(x, y)

Clearly there are two fixed points P0 = (0,0) and P1 = c_d,a_b. The partial

derivatives are ∂Xf(x, y) =−a+by, ∂Yf(x, y) =bx ∂Xg(x, y) =−d, ∂Yg(x, y) =c−dx The eigenvalues of −a 0 0 c

areλ1=−a <0 andλ2=c >0, soP0is unstable.

The eigenvalues of

0 bc_d

−da_b 0

areλ=_±i√ac, soP1stability is undetermined.

The x-nullclines are x= 0 and y = a

b. The y-nullclines arey = 0 andx= c d.

From this information, and plotting a few vectors on the direction field, the phase portrait can be plotted.

Clearly the solutions on their way to their long term behaviour will rotate aroundP1, however, the long term behaviour is still unclear. Several possibilities

could be that solutions spiral inwards towardsP1, or spiral outwards away from

(40)

c d a b x y 0

Therefore, to find the long term behaviour, the solutions need to be found. Define a new function ˜y(x) by ˜y x(t)

=y(t). Taking the derivative with respect to time with the chain rule gives

d˜y dx dx dt = dy dt

So from the system of equations defining the model, d˜y dx = (c₋dx)˜y (₋a+by˜)x= c x−d −a ˜ y +b Z −a_y_˜+bd˜y= Z _c x−ddx K1−aln ˜y+by˜=clnx−dx alny(t)−y(t)b+clnx(t)−dx(t) =K1

So solution curves are level sets onH(x, y), where

H(x, y) :=alny₋yb+clnx₋dx

These solution curves can be examined graphically to understand solution tra-jectories, or analysed analytically.

Taking the exponential of the previous equation, whereKis a new constant, gives

xce−xd

yae−yb

=K

It can be shown that all solutions withx(0)>0, y(0)>0 and (x(0), y(0))₆= (c_d,a_b) are periodic in time, and their trajectories lie on the closed curve above.

(41)

c d a b x y 0

LetT be the period of one cycle. Then using ˙y= (c−dx)yandy(T) =y(0), the average predator population is

¯ x= 1 T Z T 0 x(t) dt= 1 dT Z T 0 c−y_y˙ dt= c d− 1 dT lny(T)−lny(0) = c d Similarly ¯y=a_b.

Harvest can be incorporated into this model by adding harvest terms

h1, h2≥0.

˙

x= (₋a+by₋h1)x=f(x, y)

˙

y= (c−dx−h2)y =g(x, y)

This is the same as the previous model except replacingaandc bya+h1 and

c−h2. So the effect of harvesting is to decrease the average predator population

from _dc to c−h2

d and to increase the prey population from a b to

a+h1

b .

3.1.9 Epidemics

Initially, an epidemic can be modelled by exponential growth, however, this quickly becomes inaccurate, since in most epidemics, infected, dead and recov-ered people cannot catch the epidemic again. For this, extra terms need to be added.

The most common model is called the SIR model. The population is split into three categories: Susceptibles (those who have not yet caught the disease, so can catch the disease), infectious (those with the disease, so can spread it), and removals (those who have recovered, died, or isolated, so can no longer transmit the disease). LetS(t),I(t) andR(t) be the number of people in each category at timet, andN be the total number of people. This must fulfilN=S+I+R. The progress of the individuals is represented by

S_7→I_7→R

Letαbe the contagion parameter. Infection is proportional toS andI, and inversely proportional toN. This term removes people fromSand adds them to

(42)

I. Letβ be the recovery parameter. Recovery is proportional toIand removes people fromI and adds them to R. Hence the model has system of ODEs

˙ S=₋αSI N ˙ I=αSI N −βI ˙ R=βI

Adding these equations together reveals that ˙

N= ˙S+ ˙I+ ˙R= 0

Hence the total population is constant. Initially R = 0, S(0) = S0 > 0 and

I(0) =I0 >0, soN =S0+I0. R =N−S−I, so only the following system

needs to be considered: ˙ S=−αSI N ˙ I= αS N −β I ˙ I >0 when α β S N >1. The numberR0= α

β is called the basis reproduction

number. This is the number of secondary infections one infected person causes, assuming the population is fully susceptible (S = N). The higher R0 is, the

faster the disease spreads.

If the fraction of susceptible people _NS decreases below _R1

0, then the number

of infected people decreases. This situation is called herd immunity. It can be used to determine what fraction of the population has to be vaccinated to stop the disease outbreak.

Like the previous examples, the dynamics can be shown in phase space, with axes S and I. The solutions are curves t→(S(t), I(t)). To find these curves, define a function ˜I(S) such that

˜

I S(t)

=I(t) Then by the chain rule

d ˜I dS · dS dt = dI dt d ˜I dS = αSI_N −βI −αSI N d ˜I dS =−1 + βN α 1 S ˜ I=I=−S+βN α lnS+c

The integration constantccan be found by substituting in the initial conditions

c=I0+S0+βN

(43)

Hence the solution fulfils

I(t) =N₋S(t) +βN

α ln S(t)

S0

From the equation ˙I = α_NS ₋β

I, the maximum of I occurs when S =

βN α =:ρ. So Imax=N−ρ+ρln ρ S0 I S S=ρ N N

Epidemiologists are often more concerned with R than I, since these indi-viduals are more likely to be recorded. Define a function ˜S such that

˜ S R(t) =S(t) So therefore d ˜S dR · dR dt = dS dt d ˜S dR = −βSI_N αI =− S ρ ˜ S(R) =S(t) =S0e− R ρ

From this, a single autonomous equation for R(t) can be found. ˙

R=βN−R−S0e−

R ρ

This equation is hard to solve.

The SIR model has some limitations and shortcomings. It requiresαandβ

to be estimated for the entire population, and does not take into account how it may vary across the population. It does not take into account people who may be asymptomatic or infected but not yet infectious, the possibility of recovered people being infected once again, or different age groups.

(44)

3.2 Discrete dynamical systems

3.2.1 Exponential growth

Discrete dynamical systems, as the name may suggest, unlike first order ODE models, consider discrete time steps, which are described by finite difference equations. They may be used, for example, to model populations in which reproduction occurs at a particular time of year.

Consider a population of animals whose size at generationt= 0,1,2, . . . is given byNt. Each animal lives for one year and has on averageroffspring. The

number of animals in the next generation is given by

Nt+1=rNt

By induction, whereN0=N(0)>0

Nt=rtN0

So as t → ∞, if r > 1, then the population tends to infinity, if r < 1, the population converges to 0, and ifr= 1, then the population is constant.

If r is not an integer, then Nt will typically not be an integer. Note the

number rbehaves differently to the growth rategseen in 3.1.1.

3.2.2 Logistic map

Similarly to logistic growth, for large population, reproduction may be lower due to factors such as competition for resources. For this the logistic map is used, where Kis a parameter:

Nt+1=rNt

1−N_Kt

Khas a related but not equivalent meaning as for the ODE based logistic growth model.

It is convinient to letxt= N_Kt for allt. Then

xt+1=rxt(1−xt)

It is always assumed that r _∈ (0,4]. Then x0 ∈ [0,1] =⇒ xt ∈ [0,1] for

t= 0,1,2, . . ..

3.2.3 Difference equations

A one-dimensional difference equation has the form

xt+1=f(xt)

for some functionf. The initial conditionx0also needs to be specified.

The cobweb method is a graphical method for constructing solutions of dif-ference equations. It works by plotting a graph of f on the same axes as the graph ofy=x. First construct a vertical line fromx0 to f. Then a horizontal

(45)

y=f(x) y=x y x x1 f(x2) x2 f(x3) x3 f(x4) x4 f(x5)

Often the long term behaviour of the system is of interest. The simplest behaviour is when allxi are equal.

Definition (Fixed point). A real number x∗ _{is an equilibrium or steady state} or fixed point of f if

f(x∗) =x∗

These are solutions toxt+1=xt.

3.2.4 Stability of fixed points

The definition of stability is the same as in 3.1.4. To find stability, consider a small perturbation ytfrom a steady state and see whether it grows wheny0 is

small:

yt+1=f(x∗+yt)−x∗

By Taylor expansion off aroundx∗_,

yt+1≈f(x∗) +ytf0(x∗)−x∗=ytf0(x∗)

ytf0(x∗) is a constant, so this equation is just exponential growth. So

yt= f0(x∗) t

y0

So if_|f0₍_x∗₎_|_<_{1, then the steady state is stable, and if} _|_f0₍_x∗₎_|_>_{1, then the} steady state is unstable. Iff0₍_x∗_{) = 1 then the behaviour is unknown. The sign} off0₍_x∗_{) indicates if the solution oscillates.}

Definition (Bifurcation). Iff(x) depends on a parameterr, and the system’s long term behaviour changes significantly for small changes inr, then there is a bifurcation.

For discrete systems, if the stability of the fixed pointx∗ _{changes at some} parameter value, such as _|f0₍_x∗₎_| _{= 1 at} _r ₌ _r∗_{, then the system exhibits a} bifurcation atr=r∗

Generally, a bifurcation means that a small change in a parameter leads to qualitatively different long term behaviour of the system.

(46)

y=x y x x0 x0 0< f0(x∗)<1 y=x y x x0 x0 1< f0(x∗) y=x y x x0 x0 −1< f0₍_x∗₎_<₀ y=x y x x0 x0 f0₍_x∗₎_<₋₁

Example. Reconsider the logistic map.

f(x) =rx(1₋x) It has fixed points atx∗0= 0 and ifr >1,x∗1= 1−1r.

f0(x) =r(1−2x) so f0(0) =randf0 1−1

r

= 2−r. Hence ifr <1,x∗0 is the only fixed point,

and it is stable. For 1< r <3,x∗

1 is a stable fixed point andx∗0 is an unstable

fixed point.

So there is a bifurcation atr= 1, when x∗

0 changes stability, and atr= 3,

whenx∗

(47)

r <1 y=x y x x0 r >1 y=x y x x0 3.2.5 Period doubling

Another possibility for the long-term dynamics of a finite difference equation are periodic orbits. This means that a sequence of values of xtrepeat themselves.

The simplest example is a system which oscillates between two values. This is called a period-2 cycle.

A period-2 cycle satisfies for allt,

xt+2=xt

Suppose there is a period-2 cycle_{x∗

2, x∗3}. Thenx∗2 =f(x∗3) andx∗3 =f(x∗2).

So every point satisfies

x∗i =f f(x∗i)

So these points are steady states of the map x_→f f(x)

. Reconsider the logistic map.

g(x) :=f f(x)

=rf(x) 1₋f(x)

=r2x(1₋x) 1₋rx(1₋x)

g has rootsx= 0,1₋1

r and roots that satisfy

x2−x 1 +1 r +1 r+ 1 r2 = 0 x=r+ 1± √ r2₋₂_r₋₃ 2r

The first two roots are also roots of the logistic map, so the logistic map has a period-2 cycle whenx >3. This period-2 cycle must also be in [0,1].

The stability of a period-2 cycle of f is equivalently the stability of fixed points of the mapg. By the chain rule

g0(x) =f0 f(x)

f0(x) For the logistic map,

g0(x∗2) =f0 f(x∗2) f0(x∗2)f0(x∗3) =g0(x∗3) f0₍_x_{) =}_r₍₁₋₂_x_{), and}_x∗ 2, x∗3 are roots ofx2−x 1 + 1r +1 r+ 1 r2 = 0, so their

sum and product must be 1 +1 r and

1 r +

1 r2. So

(48)

g0(x∗2) =r2(1−2x∗2)(1−2x∗3) =r2 1₋2(x∗2+x∗3) + 4x∗2x∗3 =r2 1₋2 1 +1 r + 4 ₁ r + 1 r2 = 5₋(r₋1)2

ghas bifurcations wheng0₍_x∗

2) = 1, which is whenr= 3,1+

√

6. So the period-2 cycle is stable if

3< r <1 +√6

Asrincreases pastr= 3, the steady state offloses stability and the period-2 gains stability. This bifurcation is called a period doubling bifurcation. This in fact happens again atr= 1 +√6, where a period-4 cycle off gains stability. The logistic map undergoes infinitely many period-doubling bifurcations be-tweenr= 3 andr≈3.57. Asr increases,f goes from having a stable period-2 cycles, to stable period-4 cycles, to period-8,16, . . ..

Forr >3.57, there are no stable periodic orbits for many values of r. The dynamics are aperiodic and bounded, and have sensitive dependence on initial conditions (Butterfly effect). These three properties in a deterministic system define chaos. The bifurcations betweenr= 3 andr_≈3.57 are therefore called the route to chaos.

There are some values ofr >3.57 that have dynamics that are not aperiodic. These can be seen as white bands in the diagrams below.

The diagrams below are called bifurcation diagrams, here drawn for the logistic map, where it is called the Feigenbaum Cascade. The x-axis recordsr, whilst they axis records the possible limiting values (long term behaviour).

0 1 2 3 4 0.0 0.2 0.4 0.6 0.8 1.0 r Limit

(49)

3.0 3.2 3.4 3.6 3.8 4.0 0.0 0.2 0.4 0.6 0.8 1.0 r Limit 3.4 3.5 3.6 3.7 3.8 3.9 4.0 0.0 0.2 0.4 0.6 0.8 1.0 r Limit

MATH0008 Applied Mathematics 1

Based on lectures by Dr. Angelika Manhart

Term 1 of 2020 – 2021

Contents

1

Forces and motion

1.1

Introduction to mechanics

1.2

Solving the equation of motion

1.3

Force as a function of position

1.4

Collisions

2

Waves

2.1

Differential equations

2.2

Wave equation

2.3

Fourier series

3

Population models

3.1

First order ODE models

3.2

Discrete dynamical systems