Linear second-order differential equations with constantcoefficients and nonzero right-hand side

• We return to the damped, driven simple harmonic oscillator

d2y

dt2+ 2b

dy

dt+ ω2

0y = F sinωt

• We note that this differential equation is linear• We call yc the general solution which solves the homogeneousequation

d2yc

dt2+ 2b

dyc

dt+ ω2

0yc = 0

• We call yp the particular solution which solves theinhomogeneous equation

d2yp

dt2+ 2b

dyp

dt+ ω2

0yp = F sinωt

Linear equations with nonzero right-hand side continued

• Because the differential equation is linear, in generaly(t) = yc(t) + yp(t)• For the damped, driven simple-harmonic oscillator, withF (t) = F sinωt driving force,

y(t) = yc(t) +F√

(ω2 − ω20)2 + 4b2ω2

sin (ωt − φ)

• The yc depends on whether the system is underdamped,overdamped, or critically damped• To find a solution to a given situation, we need to know theposition and velocity of the oscillator at some time (eg t = 0), oralternately the position at two different times

Example: Damped, driven simple harmonic oscillator

• Example: At t = 0 we have y(t = 0) = 0 and dydt |t=0 = 0

• The system is underdamped, so that b < ω0

• The equation of motion is given by

d2yp

dt2+ 2b

dyp

dt+ ω2

0yp = F sinωt

• The solution is y(t) = yc(t) + yp(t)

yc(t) = ce−bt sin (βt + γ)

• Here β =√ω2

0 − b2

yp(t) =F√

(ω2 − ω20)2 + 4b2ω2

sin (ωt − φ)

tanφ =2bω

ω20 − ω2

Example continued

• So we have expressions for y(t) and dydt ,

y(t) = ce−bt sin (βt + γ) +F√

(ω2 − ω20)2 + 4b2ω2

sin (ωt − φ)

dy

dt= −ce−bt [b sin (βt + γ)− β cos (βt + γ)]+

Fω√(ω2 − ω2

0)2 + 4b2ω2cos (ωt − φ)

• We evaluate these at t = 0 to get expressions for c and γ

0 = c sin γ − F√(ω2 − ω2

0)2 + 4b2ω2sinφ

0 = −c [b sin γ − β cos γ] +Fω√

(ω2 − ω20)2 + 4b2ω2

cosφ

Example:continued

• Solving first for γ, we get

tan γ =2bβ

ω2 − ω20 + 2b2

• Then the value of c can be easily determined

c =F√

(ω2 − ω20)2 + 4b2ω2

sinφ

sin γ

Several terms on the right-hand side

• As long as the differential equation is linear, it is easy to treatmultiple terms on the right-hand side (principle of superposition• As an example, consider the damped, driven simple-harmonicoscillator, and for the moment ignore the complementary solutionyc(t)

d2y

dt2+ 2b

dy

dt+ ω2

0y = F1 sinω1t + F2 cosω2t

• We can solve separately for the particular solutions to theequations

d2yp1

dt2+ 2b

dyp1

dt+ ω2

0yp1 = F1 sinω1t

d2yp2

dt2+ 2b

dyp2

dt+ ω2

0yp2 = F2 cosω2t

Several terms on the right-hand side continued

• The particular solutions are then,

yp1(t) =F1√

(ω21 − ω2

0)2 + 4b2ω21

sin (ω1t − φ1)

yp1(t) =F2√

(ω22 − ω2

0)2 + 4b2ω22

cos (ω2t − φ2)

tanφ1 =2bω

ω20 − ω2

1

tanφ2 =2bω

ω20 − ω2

2

• And then finally, using the principle of superposition,yp(t) = yp1(t) + yp2(t)

Successive Integration of two first-order equations

• For example consider

(D − 1)(D + 2)y = ex

• Take u = (D + 2)y , then

(D − 1)u = ex

• We know how to solve this one, noting that P = −1 and Q = ex

• Then I =∫

Pdx is just I =∫

Pdx = −x• Then u = e−I

∫Qe I + c1e

−I becomesu = ex

∫e−xexdx + c1e

x = xex + c1ex

Successive integration continued

• Finally we have,

(D + 2)y = xex + c1ex

• Here we see P = 2 so I =∫

Pdx becomes I = 2x

Exponential right-hand side

• Consider the equation, with D = ddx , and assume c 6= a and

c 6= b

(D − a)(D − b)y = kecx

• First we have the complementary solution, yc(x) = c1eax + c2e

bx

• Next we set the particular solution yp(x) = Aecx , then we find Afrom

(c − a)(c − b)A = k

• So we get A = k(c−a)(c−b) and finally the solution (particular and

complementary)

y(x) = yc(x) + yp(x) = c1eax + c2e

bx +k

(c − a)(c − b)ecx

Example: Exponential right-hand side

• Example, solve (D − 1)(D + 5)y = 7e2x

• We find y = c1ex + c2e

−5x + 7(2−1)(2+5)e

2x

• We can simplify, y = c1ex + c2e

−5x + e2x

Other cases with exponential right-hand side

• When c = a or c = b, we get a particular solution of the formyp = Axecx

• If c = a = b, then we get yp = Ax2ecx

Complex exponentials on the right hand side

• If the right hand side is of the form Ae ikx , we can use the sameapproach as above• For the case where the right hand side has terms like cos kx orsin kx , then we can use complex exponentials• We use form e ikx , for the right hand side, for example, withD = d

dx , and if we want to solve

(D − a)(D − b)y = A sin kx

• We first solve for Y from

(D − a)(D − b)Y = Ae ikx

• Take Y = Ce ikx , then we find C = A(ik−a)(ik−b) , and we get the

solution we want from• Then yp(x) = Re[Y ]• This was the technique we used for the damped-drivensimple-harmonic oscillator (see section 6, chapter 8)

Dirac delta function

• The Dirac delta function we can model by an analytic function

• For example the Gaussian δσ(x) = 1√2πσ2

e−x2

2σ2

• We can make the function infinitely narrow and high by takingthe limit σ → 0, however we always have∫ ∞

−∞

1√2πσ2

e−x2

2σ2 dx = 1

• The Dirac delta function has the following properties:∫ ∞−∞

δ(x)dx = 1

∫ ∞−∞

f (x)δ(x − x0)dx = f (x0)

Fourier transform of the Dirac delta function

• We often need a Fourier transform of δ(x) (or δ(t))

δ(x − a) =

∫ ∞−∞

g(k)e ikxdx

• The Fourier transform g(k) is then just

g(k) =1

2π

∫ ∞−∞

δ(x − a)e−ikxdx =1

2πe−ika

Solving differential equations with Fourier transforms

• Consider a damped simple harmonic oscillator with damping γand natural frequency ω0 and driving force f (t)

d2y

dt2+ 2b

dy

dt+ ω2

0y = f (t)

• At t = 0 the system is at equilibrium y = 0 and at rest so dydt = 0

• We subject the system to an force acting at t = t ′,f (t) = δ(t − t ′), with t ′ > 0• We take y(t) =

∫∞−∞ g(ω)e iωtdω and f (t) =

∫∞−∞ f (ω)e iωtdω

Example continued

• Substitute into the differential equation and we find[ω2

0 − ω2 + 2ibω]g(ω) = f (ω)

• We find also f (ω) = 12π

∫∞−∞ δ(t − t ′)e−iωtdt = 1

2π e−iωt′

• We find a relationship between the g(ω) and f (ω), and then wecan write for the response g(ω)

g(ω) =1

2π

e−iωt′

ω20 − ω2 + 2ibω

• Then with y(t) = 0 for t < t ′, we get y(t) for t > t ′

y(t) =1

2π

∫ ∞−∞

e iω(t−t′)

ω20 − ω2 + 2ibω

dω

Example continued

• The integral is hard to do (we might get to later), but the pointis we have reduced the problem to doing an integral• Assume b < ω0, then we find for y(t) with t > t ′,

y(t) = e−b(t−t′) sin [ω′(t − t ′)]

ω′

where ω′ =√ω2

0 − b2 and y(t) = 0 for t < t ′

• You can convince yourself that this is consistent with the b = 0case described in the book (see Eq. 12.5 in chapter 8)

Green functions: An introduction

• We can use as an example the damped simple harmonicoscillator subject to a driving force f (t) (The book examplecorresponds to γ = 0)

d2y

dt2+ 2b

dy

dt+ ω2

0y = f (t)

• Now that we know the properties of the Dirac delta function, wenotice that f (t) =

∫∞−∞ f (t ′)δ(t − t ′)dt ′

• This gives a hint that we can treat f (t) as a sequence ofdelta-function impulses• Let’s say f (t) is zero for t < 0, and also y(t) = 0 for t < 0, andthen we turn on the driving force f (t)

Solving differential equations with Fourier transforms

• Consider a damped simple harmonic oscillator with damping γand natural frequency ω0 and driving force f (t)

d2y

dt2+ 2b

dy

dt+ ω2

0y = f (t)

• At t = 0 the system is at equilibrium y = 0 and at rest so dydt = 0

• We subject the system to an force acting at t = t ′,f (t) = δ(t − t ′), with t ′ > 0• We take y(t) =

∫∞−∞ g(ω)e iωtdω and f (t) =

∫∞−∞ f (ω)e iωtdω

Example continued

• Substitute into the differential equation and we find[ω2

0 − ω2 + 2ibω]g(ω) = f (ω)

• We find also f (ω) = 12π

∫∞−∞ δ(t − t ′)e−iωtdt = 1

2π e−iωt′

• We find a relationship between the g(ω) and f (ω), and then wecan write for the response g(ω)

g(ω) =1

2π

e−iωt′

ω20 − ω2 + 2ibω

• Then with y(t) = 0 for t < t ′, we get y(t) for t > t ′

y(t) =1

2π

∫ ∞−∞

e iω(t−t′)

ω20 − ω2 + 2ibω

dω

Example continued

• The integral is hard to do (we might get to later), but the pointis we have reduced the problem to doing an integral• Assume b < ω0, then we find for y(t) with t > t ′,

y(t) = e−b(t−t′) sin [ω′(t − t ′)]

ω′

where ω′ =√ω2

0 − b2 and y(t) = 0 for t < t ′

• You can convince yourself that this is consistent with the b = 0case described in the book (see Eq. 12.5 in chapter 8)

Green functions: An introduction

• We can use as an example the damped simple harmonicoscillator subject to a driving force f (t) (The book examplecorresponds to γ = 0)

d2y

dt2+ 2b

dy

dt+ ω2

0y = f (t)

• Now that we know the properties of the Dirac delta function, wenotice that f (t) =

∫∞−∞ f (t ′)δ(t − t ′)dt ′

• This gives a hint that we can treat f (t) as a sequence ofdelta-function impulses

Green functions: Damped harmonic oscillator

d2y

dt2+ 2b

dy

dt+ ω2

0y = f (t)

• Let’s say f (t) is zero for t < 0, and also y(t) = 0 for t < 0, andthen we turn on the driving force f (t)• Using our insight, and the principle of superposition, we assumethat the response (y(t)) depends on the entire history of the forcef (t ′) from 0 < t ′ < t,

y(t) =

∫ t

0G (t, t ′)f (t ′)dt ′

Green function for damped oscillator

• Substitute this into the equation of motion

d2y

dt2+ 2b

dy

dt+ ω2

0y = f (t)

• Use y(t) =∫ t0 G (t, t ′)f (t ′)dt ′ and f (t) =

∫∞0 f (t ′)δ(t ′ − t)dt ′

∫ t

0f (t ′)

[(d2

dt2+ 2b

d

dt+ ω2

0

)G (t, t ′)

]dt ′ =

∫ ∞0

f (t ′)δ(t ′−t)dt ′

continued

• We see that the Green function G (t, t ′) solves the differentialequation, (

d2

dt2+ 2b

d

dt+ ω2

0

)G (t, t ′) = δ(t ′ − t)

• Note also that G (t, t ′) = 0 for t < t ′

• We already solved that! It was just the response y(t) due to a

δ-function impulse, with ω′ =√ω2

0 − b2

G (t, t ′) = e−b(t−t′) sin [ω′(t − t ′)]

ω′

• Notice that the response only depends on t − t ′, as we expect• This was for the underdamped case (b < ω0), and would notwork for critical or overdamped cases!

Last one! Green function for damped oscillator

• Finally we can write the solution y(t) for any driving force f (t)turned on at t = 0, for the damped oscillator in the underdampedregime,

y(t) =

∫ t

0G (t, t ′)f (t ′)dt ′ =

∫ t

0e−b(t−t′) sin [ω′(t − t ′)]

ω′f (t ′)dt ′

Green functions continued

• Quite powerful! As long as differential equation is linear, we canfind the Green (response) function which completely solves anyproblem• Another example: Electrostatics• We know that the electrostatic potential φ(~r) due to acontinuous charge distribution ρ(~r ′) is simply additive

φ(~r) =1

4πε0

∫ρ(~r ′)

|~r −~r ′|d3~r ′

• Because of this, Gauss’ Law is a linear differential equation,

~∇ · ~E =ρ

ε0

• Then, since E = −~∇φ, we have

∇2φ = − ρε0

Green function for electrostatics

• We will see that G (~r ,~r ′) = 14πε0

1|~r−~r ′|

• First, take note that ρ(~r) =∫ρ(~r ′)δ(~r −~r ′)d3~r ′

• It might be more clear if we note that ~r = x i + y j + zk and~r ′ = x ′i + y ′ j + z ′k , and then

ρ(~r) =

∫ ∫ ∫ρ(~r ′)δ(x − x ′)δ(y − y ′)δ(z − z ′)dx ′dy ′dz ′

• Next we use that the potential φ(~r) is found just by adding upthe contributions due to each part of ρ(~r ′), so

φ(~r) =

∫G (~r ,~r ′)ρ(~r ′)d3~r ′

Green function for electrostatics

• Substitute into the Gauss Law expression ∇2φ = − ρε0∫

∇2G (~r ,~r ′)ρ(~r ′)d3~r ′ = − 1

ε0

∫ρ(~r ′)δ(~r −~r ′)d3~r ′

• Noting that the ∇2 is with respect to ~r (and not ~r ′, we get theequation for the Green function

∇2G (~r ,~r ′) = − 1

ε0δ(~r −~r ′)

• Then G (~r ,~r ′) is just the potential at ~r due to a unit chargelocated at ~r ′

• Since we know Coulomb’s Law, we can see right away thatG (~r ,~r ′) = 1

4πε01

|~r−~r ′|

Solving Gauss’ Law equation in differential form to find theGreen function

∇2G (~r ,~r ′) = − 1

ε0δ(~r −~r ′)