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Solving dierential equations with Fourier transforms Consider a damped simple harmonic oscillator with damping and natural frequency!0and driving forcef(t) d 2ydt

2+ 2bdydt

+!20y=f(t) Att= 0 the system is at equilibriumy= 0 and at rest sodydt = 0

We subject the system to an force acting att=t0,

f(t) =(tt0), witht0>0

We takey(t) =R1

1g(!)ei!td!andf(t) =R1

1f(!)ei!td!

Example continued

Substitute into the dierential equation and we nd

!20!2+ 2ib!g(!) =f(!)

We nd alsof(!) =12R

1

1(tt0)ei!tdt=12ei!t0

We nd a relationship between theg(!) andf(!), and then we can write for the responseg(!) g(!) =12e i!t0!

20!2+ 2ib!

Then withy(t) = 0 fortt0

y(t) =12Z 1 1e i!(tt0)!

20!2+ 2ib!d!

Example continued

The integral is hard to do (we might get to later), but the point is we have reduced the problem to doing an integral

Assumeb< !0, then we nd fory(t) witht>t0,

y(t) =eb(tt0)sin[!0(tt0)]! 0 where!0=q!

20b2andy(t) = 0 fort You can convince yourself that this is consistent with theb= 0 case described in the book (see Eq. 12.5 in chapter 8)

Green functions: An introduction

We can use as an example the damped simple harmonic oscillator subject to a driving forcef(t) (The book example corresponds to = 0) d 2ydt

2+ 2bdydt

+!20y=f(t) Now that we know the properties of the Dirac delta function, we notice thatf(t) =R1

1f(t0)(tt0)dt0

This gives a hint that we can treatf(t) as a sequence of delta-function impulses

Green functions: Damped harmonic oscillator

d 2ydt

2+ 2bdydt

+!20y=f(t) Let's sayf(t) is zero fort<0, and alsoy(t) = 0 fort<0, and then we turn on the driving forcef(t) Using our insight, and the principle of superposition, we assume that the response (y(t)) depends on the entire history of the force f(t0) from 0G(t;t0)f(t0)dt0

Green function for damped oscillator

Substitute this into the equation of motion

d 2ydt

2+ 2bdydt

+!20y=f(t)

Usey(t) =Rt

0G(t;t0)f(t0)dt0andf(t) =R1

0f(t0)(t0t)dt0

Z t 0 f(t0)d2dt

2+ 2bddt

+!20

G(t;t0)

dt 0=Z 1 0 f(t0)(t0t)dt0 continued We see that theGreen function G(t;t0) solves the dierential equation, d2dt

2+ 2bddt

+!20

G(t;t0) =(t0t)

Note also thatG(t;t0) = 0 fort We already solved that! It was just the responsey(t) due to a -function impulse, with!0=q! 20b2

G(t;t0) =eb(tt0)sin[!0(tt0)]!

0 Notice that the response only depends ontt0, as we expect This was for the underdamped case (b< !0), and would not work for critical or overdamped cases!

Last one! Green function for damped oscillator

Finally we can write the solutiony(t) foranydriving forcef(t) turned on att= 0, for the damped oscillator in the underdamped regime, y(t) =Z t 0

G(t;t0)f(t0)dt0=Z

t 0 eb(tt0)sin[!0(tt0)]!

0f(t0)dt0

Green functions continued

Quite powerful! As long as dierential equation is linear, we can nd the Green (response) function which completely solves any problem

Another example: Electrostatics

We know that the electrostatic potential(~r) due to a continuous charge distribution(~r0) is simply additive (~r) =140Z (~r0)j ~r~r0jd3~r0 Because of this, Gauss' Law is a linear dierential equation, r ~E= 0

Then, sinceE=~r, we have

r 2= 0

Green function for electrostatics

We will see thatG(~r;~r0) =1401j

~r~r0j

First, take note that(~r) =R(~r0)(~r~r0)d3~r0

It might be more clear if we note that~r=x^i+y^j+z^kand r0=x0^i+y0^j+z0^k, and then (~r) =Z Z Z (~r0)(xx0)(yy0)(zz0)dx0dy0dz0 Next we use that the potential(~r) is found just by adding up the contributions due to each part of(~r0), so (~r) =Z

G(~r;~r0)(~r0)d3~r0

Green function for electrostatics

Substitute into the Gauss Law expressionr2=

0Z r

2G(~r;~r0)(~r0)d3~r0=1

0Z (~r0)(~r~r0)d3~r0 Noting that ther2is with respect to~r(and not~r0, we get the equation for the Green function r

2G(~r;~r0) =1

0(~r~r0)

ThenG(~r;~r0) is just the potential at~rdue to a unit charge located at ~r0 Since we know Coulomb's Law, we can see right away that

G(~r;~r0) =1401j

~r~r0j Solving Gauss' Law equation in dierential form to nd the

Green function

r

2G(~r;~r0) =1

0(~r~r0)

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