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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 ′)