Forced Wave Equations, Resonant Response, and Modal Green Functions

(Topic)

Forced Wave Equations, Resonant Response, and Modal Green Functions

M06-19 converted the homogeneous wave equation into a spatial eigenvalue problem.

We now add a source.

Consider the forced scalar wave equation

|----------------------| -ψtt −-c2∇2-ψ-=-S(r,t).|

(1)

The function

S(r,t)

(2)

represents an external drive.

It may describe:

a force density on a string or membrane,
an acoustic source,
a distributed actuator,
an idealized impulse,
a localized point drive.

The crucial simplification is to expand the field in the eigenfunctions of the corresponding homogeneous boundary value problem.

− ∇2 Φn = λnΦn,

(3)

with

∘ --- ωn = c λn,

(4)

write

|---------∑--------------| |ψ(r,t) = qn(t)Φn (r).| | n | -------------------------

(5)

Projection onto one eigenfunction gives

|------------------| |¨qn + ω2nqn = Fn(t), --------------------

(6)

where

|--------∫-------------------| | | | S (r,t)Φn(r)dV | |Fn (t) = --V∫-------------- .| | Φ2 (r)dV | --------------V--n-----------|

(7)

Thus a forced wave equation becomes a collection of driven harmonic oscillators.

Figure 1. The forcing is projected onto the spatial eigenfunctions. Each modal projection drives one independent oscillator, and the total field is reconstructed by summing the modal responses.

1 The homogeneous eigenfunctions

Let the boundary conditions define a self-adjoint spatial problem:

|----------------| |− ∇2 Φn = λnΦn. | ------------------

(8)

Assume the eigenfunctions are orthogonal:

∫----------------------| | | | Φm Φn dV = Nn δmn.| --V---------------------

(9)

Here

|------∫---------| |Nn = Φ2 dV | | V n | -----------------

(10)

is the mode norm.

The corresponding natural angular frequency is

|------------| | ∘ --- | -ωn-=-c--λn.--

(11)

2 Modal expansion of the forced field

Write

∑ ψ(r,t) = q (t)Φ (r). n n n

(12)

Then

∑ ψtt = ¨qnΦn. n

(13)

Also,

− c²∇²ψ	= −c² ∑ _nq_n∇²Φ_n	(14)
	= c² ∑ _nλ_nq_nΦ_n	(15)
	= ∑ _nω_n²q_nΦ_n.	(16)

Therefore the PDE becomes

∑ ( ) q¨n + ω2nqn Φn = S(r,t). n

(17)

3 Projection onto one mode

Multiply by

Φ m

(18)

and integrate over the domain:

	∑ _n∫ _VΦ_mΦ_n dV	(19)
	= ∫ _VS(r,t)Φ_m(r) dV.	(20)

Using orthogonality,

( ) ∫ Nm ¨qm + ω2 qm = SΦm dV. m V

(21)

Thus

|--------------------| |¨qm + ω2mqm = Fm (t),| ----------------------

(22)

where

|-------------∫------------------| | -1-- | |Fm (t) = N S(r,t)Φm (r)dV. | -----------m---V-----------------

(23)

4 Spatial overlap

The integral

|∫-----------------| | S(r,t)Φn(r) dV | --V-----------------

(24)

measures the spatial overlap between the drive and the mode.

A source can strongly excite one mode and fail to excite another even when the forcing frequency matches both closely.

The temporal frequency controls resonance.

The spatial overlap controls whether the source actually couples to that mode.

These are separate conditions.

5 Localized point forcing

Suppose

|----------------------| -S(r,t) =-s(t)δ(r-−-r0).-

(25)

Then

F_n(t)	= ∫ _Vδ(r − r₀)Φ_n(r) dV	(26)
	= Φ_n(r₀).	(27)

Therefore a point force applied at a node of mode

n

(28)

does not excite that ideal mode:

|------------------------------| -Φn(r0)-=-0---=⇒-----Fn(t) =-0.-

(29)

6 Example 1: center forcing on a fixed string

Consider a string on

0 < x < L

(30)

with fixed ends.

The eigenfunctions are

|-----------(-----)--| | nπx- | |Φn(x) = sin L .| ----------------------

(31)

Apply a point force at

x0 = L ∕2.

(32)

The modal coupling contains

( ) Φn (L∕2) = sin nπ- . 2

(33)

For even

n,

(34)

|--------------| |Φn (L∕2) = 0. | ---------------

(35)

Thus a centered point drive excites only the odd string modes.

Figure 2. A point drive at the center of a fixed string has zero overlap with every even mode because those modes have a node at the forcing location.

7 Harmonic modal forcing

Suppose one modal force is harmonic:

|--------------------| |F (t) = F cos(Ωt ).| --n-------n0---------

(36)

The modal equation is

|------2-----------------| -¨qn +-ω-nqn-=-Fn0-cos(Ωt).

(37)

For

Ω ⁄= ωn,

(38)

try

qn,p = An cos(Ωt ).

(39)

Substitution gives

(ω2 − Ω2 )A = F . n n n0

(40)

Therefore

|--------------------------| | --Fn0--- | |qn,p(t) = ω2 − Ω2 cos(Ωt ).| -----------n---------------

(41)

The modal response becomes large when

Ω

(42)

approaches

ωn.

(43)

8 Undamped resonance

At exact resonance,

Ω = ωn,

(44)

the ordinary harmonic trial solution fails because it duplicates the homogeneous solution.

A resonant particular solution is

|---------F------------| qn,p(t) = --n0tsin(ωnt).| ---------2-ωn-----------

(45)

The oscillation amplitude grows linearly with time.

This does not mean a real physical system develops infinite amplitude.

It means the ideal undamped linear model continuously accumulates energy from a perfectly phase-matched drive.

Figure 3. At exact resonance in the ideal undamped model, the envelope grows linearly with time because the drive adds energy coherently on every cycle.

9 Energy growth at ideal resonance

For

q ∼ t sin(ωt),

(46)

both displacement amplitude and velocity amplitude grow approximately linearly with

t.

(47)

Oscillator energy is quadratic in displacement and velocity.

Therefore the modal energy grows approximately as

|------2-| -En-∝--t-|

(48)

during ideal exact resonance.

Real damping, nonlinearities, finite source power, or detuning eventually alter this behavior.

10 Adding modal damping

A simple damped modal model is

|--------------------------| q¨ + 2γ ˙q + ω2 q = F (t).| -n------n-n----n-n----n-----

(49)

The coefficient

γ n

(50)

has units of inverse time.

For harmonic forcing,

Fn(t) = Fn0 cos(Ωt ),

(51)

the steady-state response can be written

------------------------------- | | -qn(t) =-An-(Ω-)cos[Ωt −-δn(Ω)].|

(52)

11 Damped steady-state amplitude

The amplitude is

|----------------------------------| | Fn0 | |An (Ω) = ∘----2----2-2----------2.| -------------(ω-n −-Ω-)--+-(2γnΩ-)--

(53)

Damping removes the ideal divergence.

The peak is finite.

For weak damping, the amplitude resonance occurs near

|--------∘---2-----2-| -Ωpeak =---ω-n −-2-γn.

(54)

When

γn ≪ ωn,

(55)

Ωpeak ≈ ωn.

(56)

Figure 4. Damping converts the ideal resonance divergence into a finite peak. Smaller damping produces a taller and narrower resonance curve.

12 Phase lag

The steady-state phase lag satisfies

|------------------| |tanδ = -2γn-Ω--.| | n ω2n − Ω2 | -------------------

(57)

A quadrant-aware interpretation is required.

At low frequency,

δ ≈ 0. n

(58)

Near resonance,

δ ≈ π-. n 2

(59)

At high frequency,

δn → π.

(60)

Figure 5. The driven modal response changes from nearly in phase with the drive at low frequency to nearly opposite phase at high frequency, passing through approximately 90^∘ near resonance.

13 Quality factor

For weak damping, define

|----------| | ωn--| Qn = 2γn.| ------------

(61)

A large

Qn

(62)

means:

weak damping,
slow free-decay rate,
narrow resonance,
large resonant amplification.

For weak damping, the full width at half power is approximately

|-----------------| | -ωn | Δ Ω ≈ 2γn = Qn . | -------------------

(63)

14 Example 2: resonance width

Suppose

ωn = 1000 rad∕s

(64)

and

Q = 50. n

(65)

Then

ωn- 2γn = Qn = 20rad ∕s.

(66)

Thus

|------------| |γn = 10 s−1,| -------------

(67)

and the weak-damping half-power bandwidth is

|----------------| |Δ Ω ≈ 20rad∕s. | -----------------

(68)

15 Impulse forcing

Now drive one modal oscillator with an impulse at time

t = t0 :

(69)

|-------------------| Fn (t) = Inδ(t − t0). | ---------------------

(70)

Integrate the equation

¨qn + ω2nqn = Inδ(t − t0)

(71)

across a very small interval surrounding

t0.

(72)

The displacement remains continuous, while the velocity jumps:

|--------------------| q˙n(t+) − q˙n(t−) = In.| ----0--------0--------

(73)

An impulse therefore gives the mode an instantaneous change in generalized velocity.

16 Retarded modal Green function

Define

g (t,t′) n

(74)

(----------)--------------------| | ∂2 2 ′ ′ | | --2-+ ωn gn(t,t ) = δ(t − t ).| --∂t-----------------------------

(75)

The retarded condition requires

|-------------------------| gn(t,t′) = 0 for t < t′.| ---------------------------

(76)

For the undamped oscillator,

|----------------------------------| | ′ ′ sin[ωn-(t −-t′)]| |gn(t,t) = H (t − t) ω . | ---------------------------n-------

(77)

Here

H

(78)

is the Heaviside step function.

Figure 6. The retarded modal Green function vanishes before the impulse time and then produces the free sinusoidal response required by the velocity jump.

17 Why the Green function works

For

t > t′,

(79)

the function

gn

(80)

satisfies the homogeneous equation.

Across

t = t′,

(81)

integrating the differential equation gives

|----|-------------|-----------| | ∂gn| ∂gn | | | ---|| − ----|| = 1. | --∂t--t→(t′)+----∂t--t→-(t′)−------|

(82)

The retarded condition makes the derivative before the impulse zero.

The factor

1∕ωn

(83)

gives exactly the required unit derivative jump.

18 Response to an arbitrary modal force

For zero initial conditions,

|----------------------------| | ∫ ∞ | qn(t) = gn(t,t′)Fn(t′) dt′.| ---------−-∞------------------

(84)

Because the Green function is retarded,

|------------------------------------| | ∫ t sin[ωn(t − t′)] ′ ′ | |qn(t) = -------------Fn (t )dt. | ----------−∞-------ωn----------------

(85)

If forcing begins at

t = 0,

(86)

this becomes

|----------------------------------| | ∫ tsin [ωn (t − t′)] ′ ′ | |qn(t) = -------------Fn (t)dt .| ----------0------ωn----------------

(87)

This is Duhamel’s principle for one modal coordinate.

19 Damped modal Green function

For

2 q¨n + 2γn ˙qn + ω nqn = Fn (t),

(88)

define

|------∘---2----2| ωd,n-=---ω-n −-γn-

(89)

for the underdamped case.

The retarded Green function is

|-----------------------------------------′--| |g(d)(t,t′) = H (t − t′)e−γn(t−t′)sin[ωd,n(t-−-t)]. | n ωd,n | ----------------------------------------------

(90)

The impulse response now decays with time.

20 Green function for the full wave field

Define the retarded Green function of the wave operator by

[-----------]---------------------------------| | ∂2 2 2 ′ ′ ′ ′ | | --2 − c ∇ G (r,t;r,t) = δ(r − r )δ(t − t).| --∂t------------------------------------------

(91)

It must satisfy the same spatial boundary conditions as the physical field and the retarded condition

G = 0

(92)

for

t < t′.

(93)

Expand

G

(94)

in spatial eigenfunctions.

For nonzero-frequency modes,

------------------------------------------------------- | ∑ ′ | |G (r,t;r′,t′) = H (t − t′) Φn-(r)Φn-(r-)sin[ωn(t − t′)].| | n Nn ωn | -------------------------------------------------------|

(95)

Figure 7. The full retarded Green function propagates a localized source to an observation point by summing the response of every spatial eigenmode.

21 Field response to a general source

For zero initial conditions,

ψ(r,t)	= ∫ _−∞^∞dt′∫ _VdV ′	(96)
	× G(r,t; r′,t′)S(r′,t′).	(97)

Retardation automatically restricts the effective time integration to

t′ ≤ t.

(98)

The Green function therefore packages:

the geometry,
the boundary conditions,
the normal-mode spectrum,
causal time evolution.

22 Modal interpretation of source and receiver

In the Green-function expansion, the factor

Φn(r′)

(99)

measures how strongly the source point couples to mode

n.

(100)

The factor

Φn (r)

(101)

measures how strongly that mode appears at the observation point.

Therefore a mode can be suppressed because:

the source lies at a node,
the receiver lies at a node,
both.

This is a useful physical interpretation of modal Green functions.

23 Frequency-domain response

Suppose the forcing varies as

−iΩt e .

(102)

For one undamped modal coordinate,

|------------------| | -^Fn(Ω-)- | |^qn(Ω ) = ω2n − Ω2 .| -------------------|

(103)

The frequency-domain modal Green function is therefore

|------------------| |^gn(Ω ) = ---1----.| ----------ω2n-−-Ω2--|

(104)

The poles occur at

|----------| |Ω = ± ωn. | -----------

(105)

These poles are the normal-mode resonances.

24 Damped frequency-domain response

For

¨qn + 2γn ˙qn + ω2nqn = Fn,

(106)

with

− iΩt e

(107)

time dependence,

|----------------------------| | 1 | |^g(nd)(Ω ) = -2-----2---------.| -----------ωn-−-Ω--−--i2γn-Ω--

(108)

The poles move away from the real frequency axis.

This mathematical shift corresponds physically to finite decay time and finite resonance width.

Figure 8. In the ideal system the modal response has real-frequency poles at ±ω_n. Damping shifts the poles off the real axis and produces finite-width resonances.

25 Frequency-domain spatial Green function

For a time-harmonic source at frequency

Ω,

(109)

the field can be expanded as

∑ ^ψ (r,Ω ) = ^qn(Ω)Φn (r). n

(110)

The spatial modal Green function is

|------------------------------| | ∑ Φ (r)Φ (r′) | G^(r,r′;Ω) = --n---2-n--2-.| | n Nn (ω n − Ω ) | --------------------------------

(111)

With modal damping,

|-----∑------------------′-------| |^Gd = -----Φn-(r)Φn(r-)-----.| | n Nn (ω2n − Ω2 − i2γnΩ ) | ----------------------------------

(112)

Near one isolated resonance, the response is often dominated by a single term.

26 Single-mode approximation near resonance

Suppose

Ω

(113)

is close to one eigenfrequency

ωp

(114)

and well separated from all others.

Then

|-------------------| | Φp (r)Φp (r′) | G^ ≈ ---(--2----2). | -----Np--ω-p −-Ω-----

(115)

With damping,

|--------------------′-------| |^Gd ≈ ---(-Φp(r)Φp(r-)----).| | Np ω2p − Ω2 − i2γpΩ | ------------------------------

(116)

This approximation makes the resonant spatial pattern especially clear.

Near resonance, the response looks like the resonant eigenfunction.

27 Example 3: source at a node

Suppose a fixed string is driven harmonically at

x = L ∕2. 0

(117)

The second mode is

( ) Φ2 (x) = sin 2πx- . L

(118)

At the source,

Φ (L∕2 ) = sin π = 0. 2

(119)

Therefore

|----------| |F2(t) = 0.| -----------

(120)

Even if the drive frequency is exactly

ω2,

(121)

the ideal centered point drive does not excite mode 2.

Resonance requires both temporal matching and nonzero spatial coupling.

28 Example 4: off-center forcing

Move the string force to

x0 = L ∕4.

(122)

Then

( π) Φ2(L ∕4) = sin -- = 1. 2

(123)

The same mode now couples maximally in this simple normalization.

Thus source placement can qualitatively change the resonance spectrum observed in an experiment.

29 Uniform forcing and symmetry selection

Suppose a fixed string is driven by a source uniform in space:

S(x,t) = s(t).

(124)

The modal overlap is proportional to

∫ L ( ) sin n-πx dx. 0 L

(125)

This integral vanishes for even

n

(126)

and is nonzero for odd

n.

(127)

Thus a spatially symmetric drive can impose selection rules even without being localized.

30 Green function and initial conditions

The retarded Green function naturally handles forcing with zero initial conditions.

Nonzero initial displacement and velocity may be included by adding the homogeneous solution

|-----------∑-------------------------------------| ψhom (r,t) = Φn (r)[An cos(ωnt) + Bn sin(ωnt )]. | | n | ---------------------------------------------------

(128)

The complete solution is

|------------------| ψ--=-ψhom-+-ψforced.-

(129)

This separation distinguishes stored initial energy from energy injected by the source.

31 Zero modes require separate treatment

For a Neumann problem, a zero mode may satisfy

ω = 0. 0

(130)

The formula

sin(ω0-τ)- ω0

(131)

should then be interpreted by its limit:

|----------------| | sin(ω τ) | |lim --------= τ.| ω-→0---ω----------

(132)

Therefore the zero-mode retarded Green function behaves as

|--------------------------| |g0(t,t′) = H (t − t′)(t − t′).| ----------------------------

(133)

Zero modes must not simply be discarded or divided by zero.

32 Causality

A retarded Green function obeys

|--------------------------| |G (r,t;r′,t′) = 0 t < t′. ---------------------------

(134)

The response cannot occur before the source acts.

In a finite modal expansion this causality appears through the Heaviside step function.

In unbounded wave propagation, the Green function also contains the travel-time structure associated with finite propagation speed.

33 Common mistakes

Expanding the source but not the field in the same eigenfunction basis.
Forgetting to divide the modal source projection by the mode norm.
Assuming every source excites every mode.
Looking only at forcing frequency and ignoring spatial overlap.
Treating a point source at a modal node as strongly coupled to that mode.
Using the nonresonant particular solution at exact resonance.
Interpreting ideal undamped resonant growth as a prediction of infinite real amplitude.
Forgetting the damping term 2γq.
Confusing amplitude resonance with the undamped natural frequency when damping is not weak.
Using tan δ without selecting the correct phase quadrant.
Forgetting that Q = ω∕(2γ) for the damping convention used here.
Omitting the Heaviside factor from a retarded Green function.
Forgetting the factor 1∕ω_n in the undamped impulse response.
Writing the frequency-domain denominator with the wrong damping sign for the chosen e^−iΩt convention.
Dividing by ω_n for a zero mode without taking the proper limit.

34 Practice exercises

Starting from the forced wave equation, derive the independent modal equations.
Derive the modal forcing coefficient for a non-unit orthogonal eigenbasis.
Show that a point force produces a coupling proportional to the mode value at the source point.
Prove that a centered point force on a fixed string excites only odd modes.
Derive the nonresonant harmonic particular solution for one modal coordinate.
Derive the linearly growing particular solution at exact undamped resonance.
Derive the damped steady-state amplitude and phase.
Starting from Q = ω∕(2γ), derive the weak-damping bandwidth relation.
Derive the velocity jump caused by an impulse.
Verify directly that the retarded oscillator Green function has a unit derivative jump.
Use the Green function to derive the convolution solution for arbitrary modal forcing.
Derive the underdamped retarded Green function.
Insert the spatial eigenfunction expansion into the wave-operator Green-function equation and derive the modal sum.
Derive the frequency-domain modal denominator for the e^−iΩt convention.
Explain physically why a resonant source placed at a node does not excite that mode.