Fourier transforms

• We can imagine our periodic function having periodicity taken tothe limits ±∞• In this case, the function f (x) is not necessarily periodic, but wecan still use Fourier transforms (related to Fourier series)• Consider the complex Fourier series, periodic with periodicity 2l

f (x) =∞∑

n=−∞cne

inπxl

• Write same thing in an equivalent form, using ∆n = 1,

f (x) =l

π

∞∑n=−∞

cneinπx

l

(∆nπ

l

)

Fourier transforms continued

• Next we take the limit l →∞, and the summation becomes anintegral

f (x) =

∫ ∞−∞

g(k)e ikxdx

• Here k = nπl , dk = ∆nπ/l , and g(k) = cnl/π• We say that f (x) is the inverse Fourier transform of g(k)• To get g(k) given f (x)(wesayg(k)istheFouriertransformof f(x)),we again start with the version for periodic functions,

cn =1

2l

∫ l

−lf (x)e−

inπxl dx

• Again use g(k) = cnl/π, k = nπ/l , and take the limits ofintegration from −∞ to ∞

g(k) =1

2π

∫ ∞−∞

f (x)e−ikxdx

Example: Problem 21

• Find the Fourier transform of f (x) = e−x2/2σ2

g(k) =1

2π

∫ ∞−∞

e−x2/2σ2e−ikxdx

• Notice that for the exponent, we can write

−x2/2σ2 − ikx = −(

x

21/2σ+

iσk

21/2

)2

− σ2k2

2

• Then the integral can be written

g(k) =1

2πe−

σ2k2

2

∫ ∞−∞

e−

h1

2σ2 (x+iσ2k)2idx

• Change variables to y = x + iσ2k, then dx = dy , and we get

g(k) =1

2πe−

σ2k2

2

∫ ∞−∞

e−y2

2σ2 dx =σ√2π

e−σ2k2

2

Fourier sine and cosine transforms

• Less commonly used are the Fourier sine/cosine transforms

fs(x) =

√2

π

∫ ∞0

gs(k) sin kxdk

gs(k) =

√2

π

∫ ∞0

fs(x) sin kxdx

fc(x) =

√2

π

∫ ∞0

gc(k) cos kxdk

gc(k) =

√2

π

∫ ∞0

fc(x) cos kxdx

• Here, it is assumed that fs(x) and gs(k) are odd functions of xand k respectively, and fc(x) and gc(k) are even functions of x andk respectively

Chapter 8: Ordinary differential equations: Introductionand goals

By the end of the chapter you should be able to

I Solve separable differential equations

I Solve linear first-order equations

I Solve linear second-order equations with constant coefficients

I Use Fourier transforms to solve differential equations

I Use the Dirac delta function

I Begin to apply Green functions for solving differentialequations

Note that we will skip Laplace transforms in this chapter, onlybecause of our time limit

Chapter 8: Ordinary differential equations: Introductionand goals

By the end of the chapter you should be able to

I Solve separable differential equations

I Solve linear first-order equations

I Solve linear second-order equations with constant coefficients

I Use Fourier transforms to solve differential equations

I Use the Dirac delta function

I Begin to apply Green functions for solving differentialequations

Note that we will skip Laplace transforms in this chapter, onlybecause of our time limit

Chapter 8: Ordinary differential equations: Introductionand goals

By the end of the chapter you should be able to

I Solve separable differential equations

I Solve linear first-order equations

I Solve linear second-order equations with constant coefficients

I Use Fourier transforms to solve differential equations

I Use the Dirac delta function

I Begin to apply Green functions for solving differentialequations

Note that we will skip Laplace transforms in this chapter, onlybecause of our time limit

Chapter 8: Ordinary differential equations: Introductionand goals

By the end of the chapter you should be able to

I Solve separable differential equations

I Solve linear first-order equations

I Solve linear second-order equations with constant coefficients

I Use Fourier transforms to solve differential equations

I Use the Dirac delta function

I Begin to apply Green functions for solving differentialequations

Note that we will skip Laplace transforms in this chapter, onlybecause of our time limit

Chapter 8: Ordinary differential equations: Introductionand goals

By the end of the chapter you should be able to

I Solve separable differential equations

I Solve linear first-order equations

I Solve linear second-order equations with constant coefficients

I Use Fourier transforms to solve differential equations

I Use the Dirac delta function

I Begin to apply Green functions for solving differentialequations

Note that we will skip Laplace transforms in this chapter, onlybecause of our time limit

Chapter 8: Ordinary differential equations: Introductionand goals

By the end of the chapter you should be able to

I Solve separable differential equations

I Solve linear first-order equations

I Solve linear second-order equations with constant coefficients

I Use Fourier transforms to solve differential equations

I Use the Dirac delta function

I Begin to apply Green functions for solving differentialequations

Note that we will skip Laplace transforms in this chapter, onlybecause of our time limit

Separable equations

• If we have a differential equation with two variables, we candirectly integrate if the equation is separable• Consider the decay/growth equation

dN

dt= ±λN

• We can rewrite and get only N on the left side, and t on theright side

dN

N= −λdt

• Integrate then, and we get up to a constant

ln N = ±λt + C

N = N0e±λt

Example: Problem 1

• Simple example, problem 1, solve xy ′ = y if y = 1 when x = 2• Separate the equation

dy

y=

dx

x

• Solution is y = Cx , so for the conditions given C = 1/2, andy = (1/2)x• Check it! dy

dx = 1/2, so xy ′ = (1/2)x = y

Linear first-order equations

• We consider equations here of the form

y ′ + Py = Q

• P and Q are functions of x (cases where coefficients areconstants are dealt with later)• First consider Q = 0, then equation is separable

dy

y= −Pdx

• We can solve as y = Ae−R

Pdx

• Write I =∫

Pdx , the ye I = A

d

dx(ye I ) = e I y ′ + e I yP = e I (y ′ + Py) = e IQ

Linear first-order equations continued

• We can integrate the equation on the previous slide

ye I =

∫e IQdx + c

• We can multiply by e−I to obtain

y = e−I

∫e IQdx + ce−I

Example: First order linear equation

• Example: Section 3, problem 11, solve y ′ + y cos x = sin 2x• Here we see P(x) = cos x and Q(x) = sin 2x• We see that I =

∫Pdx =

∫cos xdx = sin x

• The solution we can write as,

y = e− sin x

∫sin 2xesin xdx + ce− sin x

• In the integral, we can make the substitution u = sin x , anddu = cos xdx , and also sin 2x = 2 sin x cos x , so that we get∫

sin 2xesin xdx = 2

∫ueudx = 2(sin x − 1)esin x

• Finally we get the solution,

y = 2(sin x − 1) + ce− sin x

Example, section 3 problem 11

• Test the solution! (make sure it is correct)

y ′+ Py = 2 cos x − c cos xe− sin x + 2 cos x(sin x −1) + c cos xe− sin x

• Finally the is equal to 2 cos x sin x = sin 2x

Second-order linear equations with constant coefficients

• We will consider first homogeneous equations, with a2, a1, anda0 constants (independent of x and y)

a2d2y

dx2+ a1

dy

dx+ a0y = 0

• Example: Simple harmonic oscillator, a2 = 1, a1 = 0, a0 = ω20

d2y

dt2+ ω2

0y = 0

• We solved this before,

y = A cosω0t + B sinω0t = Ce iω0t

• Let’s try to systematically solve differential equations of this type

Characteristic equation

• Define the linear operator, D = ddx , and the D2 = d2

dx2 , then

(a2D2 + a1D + a0)y = 0

• We can in general factor this equation

(D − a)(D − b)y = 0

• We can solve then if we take that if y = c1eax + c2e

bx , since

D(c1eax + c2e

bx) = c1aeax + c2bebx

Characteristic equation continued

(a2D2 + a1D + a0)y = 0

• We get from this the characteristic equation, treating D now asa number

a2D2 + a1D + a0 = 0

• A quadratic equation!

D1,2 =−a1 ±

√a21 − 4a2a0

2a2

• Then the solution is y = c1eD1x + c2e

D2x (in the last slide,D1 = a, D2 = b)• Also note that if 4a2a0 > a1 then D1 and D2 are complex!

A special case: Equal roots for characteristic equation

• We might end up with equal roots for the characteristicequation, for example

(D − a)(D − a)y = 0

• Here the operator D = ddx

• Apparently only gives solution y = c1eax , but we will see there is

one more solution• Take u = (D − a)y , then the u satisfies

(D − a)u = 0

• Therefore we find u = Aeax = (D − a)y , which we then solve fory• This is a first-order linear equation and we get

y = (Ax + B)eax

Example: Simple harmonic oscillator

• As an example, consider the simple harmonic oscillator, withD = d

dt and D2 = d2

dt2

d2y

dt2+ ω2

0y = (D2 + ω20)y = (D + iω0)(D − iω0)y = 0

• The roots of the characteristic equation then are ±iω0, so finally

y = c1eiω0t + c2e

−iω0t

• As we saw before, we can also use the Euler relation and writethis in terms of sines and cosines,

y = A cosω0t + sinω0t

Example: Damped simple harmonic oscillator

• When there is damping, we need to consider the differentialequation

d2y

dt2+ 2b

dy

dt+ ω2

0y = 0

• We have from this, with D = ddt the characteristic equation

D2 + 2bD + ω20 = 0

• The roots of the characteristic equation we find to be

D1,2 = −b ±√

b2 − ω20

• There three distinct regimes: b < ω0 (underdamped), b > ω0

(overdamped), and b = ω0 (critically damped)

Damped simple-harmonic oscillator: Underdamped regime

• In the underdamped regime, b < ω0 and the system showsdamped oscillatory behavior

y(t) = e−bt(c1e

iβt + c2e−iβt

)= e−bt (A sinβt + B cosβt)

• Here β =√ω2

0 − b2 is the effective oscillation frequency

Damped simple-harmonic oscillator: Overdamped regime

• In the overdamped regime b > ω0 and the system shows nooscillatory behavior• The roots of the characteristic equation are real and equal to

D1,2 = −b ±√

b2 − ω20

• The equation of motion is given by

y = AeD1t + BeD2t = Ae

h−b+√

b2−ω20

it

+ Be

h−b−√

b2−ω20

it

• If we give the system a kick at t = 0, the displacement willexponentially decay to zero in a time faster than one period ofoscillation

Damped simple-harmonic oscillator: Critical regime

• In the critical regime b = ω0, the system is on the edge betweenthe over and under-damped cases• We get equal roots for the characteristic equation D1,2 = −b• From what we saw in the case of equal roots, we find

y = (At + B)e−bt