Wave Mechanics Examples: Initial Conditions and the d’Alembert Solution

This companion article provides exercises for WM17, wave mechanics: Initial Conditions and the d’Alembert Solution. All exercises are stated first so they can be attempted without seeing the answers. Complete worked solutions follow in Part II.

For the one-dimensional constant-speed wave equation on the whole line,

(1)

with initial data

(2)

and

(3)

the d’Alembert formula is

(4)

The first term carries the initial displacement along the two characteristic directions. The integral term carries the effect of the initial velocity over the characteristic interval. This is the classical whole-line initial-value solution under the usual smoothness assumptions [4, 3, 6].

How to use this problem set

Attempt all exercises in Part I before consulting Part II. In each problem, identify the characteristic endpoints

(5)

before substituting into the formula. Keep the whole-line assumption in mind: a finite string also requires endpoint boundary conditions and generally involves reflections or mode expansions.

Part I: Exercises

Exercise 1: Read the d’Alembert formula

Consider

(6)

Answer the following.

1. Which function represents the initial displacement?
2. Which function represents the initial velocity?
3. What are the two characteristic footpoints on the line t = 0?
4. Which part of the formula survives if g(x) = 0?
5. Which part survives if f(x) = 0?

Exercise 2: Released displacement with zero initial velocity

Suppose

(7)

1. Simplify the d’Alembert formula.
2. Explain why the initial profile splits into two copies.
3. What is the amplitude of each copy relative to the original amplitude?
4. What direction does each copy move?
5. Show explicitly that the two copies add back to f(x) at t = 0.

The geometry is illustrated below.

Exercise 3: Initial velocity required for one-way motion

An initial shape f(x) is prescribed.

1. What initial velocity g(x) produces a pure right-moving solution?
2. What initial velocity produces a pure left-moving solution?
3. For

   (8)

   find g(x) for pure right-moving motion.

4. For the same f(x), find g(x) for pure left-moving motion.

Exercise 4: Recover the two traveling components

Define

(9)

where x_∗ is a fixed reference point.

Starting from

(10)

show that one convenient choice of the component functions is

(11)

and

(12)

Explain why adding a constant to F and subtracting the same constant from G does not change u(x,t).

Exercise 5: Direct numerical evaluation

Let

(13)

and let the initial data be

(14)

(15)

Find

(16)

Show separately the displacement contribution and the initial-velocity contribution.

Exercise 6: Velocity-only compact pulse

Suppose the initial displacement is zero,

(17)

and the initial velocity is

(18)

At the observation point x = 0:

1. Find u(0,t) for 0 ≤ t ≤ a∕c.
2. Find u(0,t) for t ≥ a∕c.
3. Evaluate the result for

   (19)

   at t = 0.10 s and t = 0.30 s.

The integral is controlled by the overlap between the characteristic interval and the support of g.

Exercise 7: Sinusoidal initial displacement

Let

(20)

1. Use the d’Alembert formula to find u(x,t).
2. Use a trigonometric identity to simplify the result.
3. Show that the angular frequency is ω = ck.
4. Explain why the final expression is a standing-wave form even though d’Alembert’s solution is built from traveling waves.

Exercise 8: Sinusoidal initial velocity

Let

(21)

1. Evaluate the velocity integral.
2. Show that

   (22)

3. Rewrite the coefficient using ω = ck.
4. Differentiate the result with respect to time and verify u_t(x, 0) = g(x).

Exercise 9: Domain of dependence

At the observation point

(23)

suppose the wave speed is

(24)

1. Find the two characteristic footpoints on the initial line.
2. State the interval of initial data that can influence (x₀,t₀).
3. Can a change in the initial data at x = 2.2 m influence the observation point?
4. Can a change at x = 5.4 m influence it?
5. Can a change at x = 6.0 m influence it?

Exercise 10: Finite propagation from compactly supported data

Suppose both initial functions vanish outside

(25)

The wave speed is

(26)

At time

(27)

find the largest interval in which the solution can possibly be nonzero. Explain how this demonstrates finite propagation speed.

Exercise 11: Check the initial conditions directly

Starting from

(28)

show directly that

(29)

and

(30)

For the velocity check, use the Leibniz rule for differentiating an integral whose limits depend on time.

Exercise 12: Whole-line formula versus a finite string

A string occupies only

(31)

and is fixed at both ends:

(32)

Answer the following.

1. Does the basic whole-line d’Alembert formula automatically enforce these endpoint conditions?
2. What additional physical phenomenon appears when waves reach the endpoints?
3. Name two methods that can be used to solve the finite-string problem after boundary conditions are imposed.
4. Explain why applying the whole-line formula blindly to a finite string can give the wrong physical answer.

Exercise 13: Connect the initial-value solution to string mechanics

An ideal string has

(33)

The initial displacement is

(34)

and the initial velocity is zero.

1. Find the wave speed.
2. Write the complete d’Alembert solution.
3. At t = 0.010 s, where are the centers of the two Gaussian pulses?
4. What is the peak amplitude of each separated pulse?

Exercise 14: Synthesis – design a pure right-moving sinusoid

An ideal string has

(35)

The desired initial displacement is

(36)

The goal is for this entire shape to travel purely to the right.

1. Find the wave speed c.
2. Find the required initial velocity g(x).
3. Write the resulting traveling-wave solution u(x,t).
4. Identify k and ω.
5. Find the wavelength and ordinary frequency.
6. Explain why choosing g(x) = 0 instead would not produce the same motion.

Part II: Complete Worked Solutions

Solution 1: Read the d’Alembert formula

The formula is

(37)

1. The initial displacement is

   (38)

2. The initial velocity is

   (39)

3. The characteristic footpoints are

   (40)

4. If g = 0, only the displaced copies of f remain:

   (41)

5. If f = 0, only the velocity integral remains:

   (42)

Solution 2: Released displacement with zero initial velocity

Setting

(43)

removes the integral term, so

(44)

The first term is a right-moving copy because its argument is x − ct. The second is a left-moving copy because its argument is x + ct.

Each copy has one-half the initial amplitude. At t = 0,

u(x, 0)	= f(x) + f(x)	(45)
	= f(x) .	(46)

Thus the two half-amplitude copies exactly reconstruct the original profile at the initial instant.

Solution 3: Initial velocity required for one-way motion

For a pure right-moving wave,

(47)

Differentiate with respect to time:

(48)

At t = 0,

(49)

Similarly, for a pure left-moving wave,

(50)

so

(51)

Now let

(52)

Then

(53)

Therefore the required right-moving initial velocity is

(54)

whereas the required left-moving initial velocity is

(55)

The sign of the initial velocity therefore selects the propagation direction.

Solution 4: Recover the two traveling components

WM17 gives

(56)

and

(57)

Define

(58)

Integrating gives one convenient choice,

(59)

and

(60)

Their sum is

(61)

as required by the initial displacement.

If a constant C is added to F and subtracted from G, then

(62)

Therefore the physical field u(x,t) is unchanged by this constant redistribution between the two component functions.

Solution 5: Direct numerical evaluation

The observation event is

(63)

with

(64)

The characteristic endpoints are

x − ct	= 1.0 − (2.5)(0.20) = 0.50 m,	(65)
x + ct	= 1.0 + (2.5)(0.20) = 1.50 m.	(66)

The displacement contribution is

[f(0.50) + f(1.50)]	= α[(0.50)2 + (1.50)2]	(67)
	= (0.020)(0.25 + 2.25) m	(68)
	= 0.025 m .	(69)

The velocity contribution is

∫ 0.501.50βsds	= 0.501.50	(70)
	= m	(71)
	= 0.060 m .	(72)

Therefore

(73)

Solution 6: Velocity-only compact pulse

Because f = 0,

(74)

For early times with

(75)

the entire interval [−ct,ct] lies inside the region where g = V ₀. Thus

u(0,t)	= V 0(2ct)	(76)
	= V 0t , 0 ≤ t ≤.	(77)

For later times,

(78)

the integration interval already covers the full support [−a,a]. Hence

u(0,t)	= V 0(2a)	(79)
	= , t ≥.	(80)

With

(81)

we have at t = 0.10 s,

(82)

At t = 0.30 s,

(83)

The plateau occurs because, after t = a∕c, expanding the characteristic interval does not include any new region where g is nonzero.

Solution 7: Sinusoidal initial displacement

With

(84)

the d’Alembert formula becomes

(85)

Use

(86)

With

(87)

we obtain

(88)

The temporal factor has angular frequency

(89)

The result is a standing-wave form because equal right- and left-moving waves of the same amplitude superpose to make a nontranslating pattern with fixed nodes and antinodes.

Solution 8: Sinusoidal initial velocity

Here

(90)

Therefore

(91)

Integrate:

u(x,t)

= .

(92)

Using

(93)

we obtain

(94)

Since

(95)

this can be written as

(96)

Differentiate with respect to time:

(97)

At t = 0,

(98)

Solution 9: Domain of dependence

The characteristic endpoints are

x0 − ct0	= 4.0 − (5.0)(0.30) = 2.5 m,	(99)
x0 + ct0	= 4.0 + (5.0)(0.30) = 5.5 m.	(100)

Therefore the domain of dependence on the initial line is

(101)

A change at x = 2.2 m lies outside this interval, so it cannot influence (x₀,t₀).

A change at x = 5.4 m lies inside the interval, so it can influence the observation point.

A change at x = 6.0 m lies outside the interval, so it cannot influence the observation point.

Thus

(102)

Solution 10: Finite propagation from compactly supported data

Initially, both f and g vanish outside

(103)

In time t, information can propagate outward by at most the distance

(104)

Therefore the largest possible support at that time is

(105)

Hence

(106)

Outside this interval, neither characteristic reaching (x,t) can intersect nonzero initial data. The disturbance therefore cannot appear arbitrarily far away instantly. This is finite propagation at speed c.

Solution 11: Check the initial conditions directly

Start with

(107)

Set t = 0:

u(x, 0)	= [f(x) + f(x)] + ∫ xxg(s) ds	(108)
	= f(x).	(109)

Thus

(110)

Now differentiate with respect to time. The first two terms give

(111)

For the integral term, the Leibniz rule gives

	=	(112)
	= [g(x + ct) + g(x − ct)].	(113)

Therefore

ut(x,t)	= −f′(x − ct) + f′(x + ct)	(114)
	+ [g(x + ct) + g(x − ct)].	(115)

At t = 0, the two f′ terms cancel:

(116)

Hence

(117)

Solution 12: Whole-line formula versus a finite string

1. No. The basic whole-line d’Alembert formula is constructed for an unbounded spatial domain. It does not automatically impose

   (118)

2. When a wave reaches a finite endpoint, it reflects. The boundary condition determines the reflected behavior.
3. Two standard methods are reflected or extended initial data and normal-mode/Fourier-series expansion.
4. Applying the whole-line formula blindly would allow the wave to pass beyond the physical endpoints instead of reflecting and satisfying the endpoint constraints. The resulting field could therefore violate the actual boundary conditions.

Thus the whole-line formula solves an initial-value problem, while a finite string requires an initial-boundary-value problem.

Solution 13: Connect the initial-value solution to string mechanics

For an ideal string,

(119)

With

(120)

we obtain

c	=	(121)
	=	(122)
	≈ 63.2 m/s .	(123)

Because g = 0, the solution is

u(x,t)	= 0.004 exp	(124)
	+ 0.004 exp m.	(125)

At

(126)

the pulse centers are at

(127)

Therefore

(128)

Each separated pulse has one-half the initial peak amplitude:

(129)

Solution 14: Synthesis – design a pure right-moving sinusoid

The wave speed is

c	=	(130)
	=	(131)
	=	(132)
	= 40 m/s .	(133)

The initial displacement is

(134)

Differentiate:

(135)

For pure right-moving motion,

(136)

Therefore

g(x)	= −40[−0.030 sin(5x)]	(137)
	= 1.20 sin(5x) m/s .	(138)

The complete right-moving solution is

(139)

Expanding the phase,

(140)

Hence

(141)

The wavelength is

λ	=	(142)
	=	(143)
	≈ 1.26 m .	(144)

The ordinary frequency is

fwave	=	(145)
	=	(146)
	≈ 31.8 Hz .	(147)

If instead g(x) = 0, the d’Alembert formula would produce

(148)

which is an equal two-way split, not a single right-moving waveform. The nonzero initial velocity is what removes the unwanted left-moving component.

Common mistakes

• Mistake: forgetting that both f and g are required for a second-order-in-time wave equation.
• Mistake: omitting the factor 1∕2 on the two displaced copies of f.
• Mistake: reversing the limits of the velocity integral.
• Mistake: omitting the factor 1∕(2c) in front of the velocity integral.
• Mistake: assuming g = 0 creates a single traveling copy. It creates two equal half-amplitude copies moving in opposite directions.
• Mistake: confusing the domain of dependence with the entire initial line. Only [x − ct,x + ct] can affect (x,t).
• Mistake: using the whole-line d’Alembert formula on a finite string without enforcing boundary conditions.
• Mistake: using the wrong sign in g = ∓cf′ for one-way motion.

What WM17E1 reinforces

The d’Alembert formula converts physical initial data directly into a wave field:

(149)

The exercises reinforce four structural ideas:

• initial displacement and initial velocity play different roles,
• the wave equation contains independent right- and left-moving information,
• information propagates only within characteristic cones at speed c, and
• pure one-way motion requires compatible initial displacement and velocity.

These ideas complete the basic whole-line initial-value problem for the one-dimensional ideal wave equation and prepare the way for more advanced topics such as energy transport, Fourier decompositions, modal expansions, interfaces, and dispersive waves.

References

References