Wave Mechanics Examples: Right- and Left-Traveling Solutions

This companion article provides exercises for WM16, wave mechanics: Right- and Left-Traveling Solutions. All exercises are stated first so they can be attempted without seeing the answers. Complete worked solutions follow in Part II.

The central WM16 result is that, for a sufficiently smooth classical solution of the homogeneous one-dimensional constant-speed wave equation,

(1)

one may write

(2)

The two pure traveling families satisfy the first-order transport relations

(3)

and

(4)

The characteristic variables are

(5)

and in these coordinates the wave equation reduces to

(6)

These results are standard descriptions of the one-dimensional wave equation and traveling-wave decomposition [1, 2, 3, 4, 6].

How to use this problem set

For each problem, separate three questions:

• which part of the field moves toward +x and which part moves toward −x;
• what information follows from the second-order wave equation itself; and
• what extra information requires initial or boundary data.

The distinction between a pure one-way wave and a superposition containing both families is especially important.

Part I: Exercises

Exercise 1: Read the two propagation families from snapshots

The figure below shows the same localized profile at an earlier and a later time for each traveling family.

1. Which panel represents F(x − ct)?
2. Which panel represents G(x + ct)?
3. If the peak in the right-moving panel shifts by 2.8 m during 0.70 s, find c.
4. Write the first-order transport equation satisfied by each pure family.

Exercise 2: Derive and use the first-order direction tests

For

(7)

and

(8)

derive the relation between u_t and u_x for each family.

Then suppose a known pure one-way wave has, at one event,

(9)

Determine its direction and speed.

Exercise 3: Factor the wave operator

Expand

(10)

and show that it equals

(11)

for a sufficiently smooth function u(x,t). State precisely which derivative identity makes the two mixed terms cancel.

Exercise 4: Read characteristic footpoints

The figure shows the two characteristic lines through an event P = (x₀,t₀).

Take

(12)

1. Find the initial-line point on the constant-ξ characteristic.
2. Find the initial-line point on the constant-η characteristic.
3. Find the separation between the two footpoints.
4. Which characteristic family has slope dx∕dt = +c?
5. Which has slope dx∕dt = −c?

Exercise 5: Transform the wave equation

Let

(13)

Derive

(14)

and

(15)

Then derive u_xx and u_tt and show that

(16)

What equation must U satisfy when c≠0?

Exercise 6: Integrate the characteristic-coordinate equation

Starting from

(17)

integrate carefully to show that

(18)

Explain why the “constant of integration” after integrating with respect to η may be an arbitrary function of ξ rather than an ordinary numerical constant.

Exercise 7: Identify the two components in a compound field

Consider

(19)

1. Identify a possible F(ξ) and G(η).
2. State the direction of each component.
3. Find the common wave-speed magnitude.
4. State the wave equation satisfied by the total field.
5. Find the position of the center of the Gaussian component as a function of time.

Exercise 8: Point-by-point addition of the two families

At one instant, suppose the two traveling components are

(20)

and

(21)

The corresponding component profiles and their sum are shown below.

1. Evaluate F(0).
2. Evaluate G(0).
3. Find u(0) = F(0) + G(0).
4. Explain why destructive interference at one location does not mean the two traveling components cease to exist.

Exercise 9: Decompose a non-obvious polynomial solution

Show that

(22)

satisfies

(23)

Then use the identity

(24)

to write the solution explicitly as

(25)

Exercise 10: Reconstruct a standing wave from two traveling families

A Standing Wave is

(26)

where x is in meters and t is in seconds.

The figure illustrates the decomposition conceptually.

1. Use a trigonometric identity to write u as the sum of one right-moving and one left-moving sinusoid.
2. What is the amplitude of each traveling component?
3. Find k, ω, and the common wave speed c.
4. State the wave equation satisfied by the standing wave.

Exercise 11: Why the pure transport tests fail for a two-way field

Let

(27)

Derive u_x and u_t, and then show that

(28)

and

(29)

Explain why a general two-way field normally satisfies neither pure first-order transport equation even though it satisfies the second-order wave equation.

Exercise 12: Characteristic reach back to the initial line

At the event

(30)

let

(31)

1. Find x₀ − ct₀.
2. Find x₀ + ct₀.
3. What interval of the initial line lies between those two characteristic footpoints?
4. Why is this construction useful when initial data are eventually used to determine F and G?

Exercise 13: Connect string mechanics to the two traveling families

An ideal string has

(32)

A sinusoidal component has

(33)

1. Find the wave speed c.
2. Find the angular frequency ω.
3. Write a right-moving sinusoid of amplitude 3.0 mm and zero phase constant.
4. Write the corresponding left-moving sinusoid.
5. State the mechanical wave equation in both T,μ form and c form.

Exercise 14: Synthesis - two pulses and their initial data

Let

(34)

and

(35)

Take

(36)

and define

(37)

1. Write the complete function u(x,t) explicitly.
2. State the propagation direction of each pulse.
3. Write the initial displacement u(x, 0).
4. Compute the initial velocity u_t(x, 0).
5. At t = 0.50 s, locate the center of each pulse.
6. State the wave equation satisfied by u and explain why no direct second-derivative calculation is needed to establish it.

Part II: Complete Worked Solutions

Solution 1: Read the two propagation families from snapshots

1. The left panel moves toward increasing x, so it represents

   (38)

2. The right panel moves toward decreasing x, so it represents

   (39)

3. The speed magnitude is displacement divided by elapsed time:

   c	=	(40)
   	= 4.0 m/s .	(41)

4. The pure right-moving family satisfies

   (42)

   while the pure left-moving family satisfies

   (43)

Solution 2: Derive and use the first-order direction tests

For the right-moving family, let

(44)

Then

(45)

and

(46)

Therefore

(47)

or

(48)

For the left-moving family, let

(49)

Then

(50)

and

(51)

so

(52)

or

(53)

For the measured event,

(54)

The two derivatives have opposite signs, which is consistent with the right-moving relation. Solving for c,

c	= −	(55)
	= − m/s	(56)
	= 60 m/s .	(57)

Thus the wave is purely right-moving at speed 60 m/s under the stated one-way-wave assumption.

Solution 3: Factor the wave operator

Expand the two operators in order:

	u	(58)
	=	(59)
	= utt + cuxt − cutx − c2u xx.	(60)

For a sufficiently smooth function, the mixed partial derivatives commute:

(61)

Therefore the middle terms cancel and

(62)

This factorization exposes the two first-order propagation operators hidden inside the second-order wave equation.

Solution 4: Read characteristic footpoints

We have

(63)

First compute

(64)

1. The constant-ξ characteristic reaches the initial line at

   x0 − ct0	= 5.0 − 2.0	(65)
   	= 3.0 m .	(66)

2. The constant-η characteristic reaches the initial line at

   x0 + ct0	= 5.0 + 2.0	(67)
   	= 7.0 m .	(68)

3. Their separation is

   (69)

4. Constant ξ = x − ct gives

   (70)

   so

   (71)

5. Constant η = x + ct gives

   (72)

   so

   (73)

Solution 5: Transform the wave equation

Let

(74)

Because

(75)

the chain rule gives

(76)

Differentiate once more with respect to x:

(77)

Similarly,

(78)

so

(79)

Differentiating again with respect to t gives

(80)

Subtracting c²u _xx,

utt − c2u xx	= c2U ξξ − 2c2U ξη + c2U ηη	(81)
	− c2	(82)
	= −4c2U ξη.	(83)

Thus the wave equation requires

(84)

For c≠0,

(85)

Solution 6: Integrate the characteristic-coordinate equation

Start from

(86)

Write this as

(87)

Therefore U_η does not depend on ξ. It may still depend on η, so write

(88)

Integrating with respect to η gives

(89)

The term F(ξ) appears because integration was performed with respect to η. Anything that depends only on ξ differentiates to zero with respect to η, so it plays the role of an integration constant.

Therefore

(90)

and hence

(91)

Solution 7: Identify the two components in a compound field

The field is

(92)

Define

(93)

Then one valid choice is

(94)

and

(95)

The F component moves toward increasing x and the G component moves toward decreasing x. Both have speed magnitude

(96)

Therefore the total field satisfies

(97)

The Gaussian is centered when its squared argument vanishes:

(98)

Hence its center is

(99)

The decreasing center position confirms leftward propagation.

Solution 8: Point-by-point addition of the two families

At x = 0,

F(0)	= 0.90e−0.9(1)2	(100)
	≈ 0.366 .	(101)

For the second component,

G(0)	= −0.60e−1.15(−0.8)2	(102)
	= −0.60e−0.736	(103)
	≈−0.287 .	(104)

Therefore

u(0)	= F(0) + G(0)	(105)
	≈ 0.366 − 0.287	(106)
	≈ 0.0785 .	(107)

The two components partially cancel at this location, but the linear wave model still contains both traveling contributions. Interference changes their observed sum; it does not destroy the underlying component solutions.

Solution 9: Decompose a non-obvious polynomial solution

For

(108)

we have

(109)

and

(110)

Thus

(111)

so the wave equation is satisfied.

Now use

(112)

Dividing by 2,

(113)

Therefore choose

(114)

and

(115)

Hence

(116)

This again shows that the two-family representation is not limited to pulses and sinusoids.

Solution 10: Reconstruct a standing wave from two traveling families

The standing wave is

(117)

Use

(118)

Then

u	= 4.0 mm cos(2x − 20t)	(119)
	+ 4.0 mm cos(2x + 20t).	(120)

Thus

(121)

and

(122)

Each component amplitude is therefore

(123)

The Wavenumber and angular frequency are

(124)

Hence

(125)

The wave equation is therefore

(126)

Solution 11: Why the pure transport tests fail for a two-way field

Let

(127)

For

(128)

we obtain

(129)

and

(130)

Therefore

ut + cux	= + c	(131)
	= 2cG′(η) .	(132)

Likewise,

ut − cux	= − c	(133)
	= −2cF′(ξ) .	(134)

Thus the right-moving transport equation u_t + cu_x = 0 holds for the total field only where the left-moving derivative contribution vanishes. Similarly, u_t − cu_x = 0 holds only where the right-moving derivative contribution vanishes.

A general two-way field therefore satisfies the second-order wave equation but normally satisfies neither pure one-way first-order transport equation.

Solution 12: Characteristic reach back to the initial line

At

(135)

we have

(136)

Therefore

(137)

and

(138)

The interval between the footpoints is

(139)

The two characteristics show where the two traveling families reaching the event P trace back to the initial line. In WM17, initial displacement and velocity data will be used to determine the specific functions F and G, and these characteristic locations will enter naturally into the resulting formula.

Solution 13: Connect string mechanics to the two traveling families

For an ideal string,

(140)

Therefore

c	=	(141)
	=	(142)
	≈ 79.1 m/s .	(143)

With

(144)

the nondispersive relation gives

ω	= ck	(145)
	= (79.1)(5.0)	(146)
	≈ 395 rad/s .	(147)

A right-moving sinusoid with amplitude 3.0 mm is

(148)

The corresponding left-moving sinusoid is

(149)

The mechanical form of the string equation is

(150)

or equivalently

(151)

Here

(152)

Solution 14: Synthesis - two pulses and their initial data

The two profile functions are

(153)

and

(154)

with

(155)

1. Substitute s = x − 4t into F and s = x + 4t into G:

   (156)

2. The first pulse depends on x − 4t, so it moves toward +x. The second depends on x + 4t, so it moves toward −x.
3. At t = 0,

   (157)

4. For the general two-family form,

   (158)

   Here

   (159)

   and

   (160)

   Therefore at t = 0,

   (161)

5. The first pulse is centered where

   (162)

   so at t = 0.50 s,

   (163)

   The second pulse is centered where

   (164)

   giving

   (165)

6. Each component is a sufficiently smooth function of one characteristic variable, and the wave equation is linear. Therefore the sum satisfies

   (166)

   No fresh second-derivative calculation is required because the WM15 verification theorem and linearity already establish the result.

Common mistakes

• Mistake: reading the sign inside x ∓ ct as the direction itself. Direction comes from holding the whole argument constant.
• Mistake: applying u_t = −cu_x or u_t = cu_x to a field that contains both traveling families.
• Mistake: forgetting that factorization uses differential operators and requires commuting mixed partial derivatives for the classical derivation.
• Mistake: treating the integration function F(ξ) as an ordinary numerical constant when integrating with respect to η.
• Mistake: assuming F and G have been determined merely because the PDE has been solved structurally. Initial or boundary data are still required.
• Mistake: assuming a standing wave is unrelated to traveling waves. It can be built from equal counter-propagating components.

What WM16E1 reinforces

The one-dimensional constant-speed wave equation contains two independent characteristic propagation families. In characteristic coordinates,

(167)

the second-order equation reduces to

(168)

whose sufficiently smooth solutions have the form

(169)

WM17 will use initial displacement and initial velocity to determine the specific functions F and G and obtain the d’Alembert initial-value formula.

References

References