Wave Mechanics: Traveling-Wave Solutions of the 1D Wave Equation
WM14 derived the one-dimensional string wave equation from Newton’s second law:
with
for an ideal stretched string.
Earlier articles introduced translating disturbances of the form
and
WM15 now connects those two parts of the course. The central question is:
The answer is yes, provided the profile is sufficiently smooth for the required derivatives to exist. A
right-moving profile
and a left-moving profile
both satisfy
This is one of the most important structural facts about the one-dimensional linear wave equation
[1, 2, 3, 4, 7].
1 The question is verification, not yet the full general solution
There are two different mathematical tasks that should not be confused.
The first is verification: start with a proposed function and check whether it satisfies the partial
differential equation.
The second is solution construction: start with the partial differential equation plus initial or
boundary data and derive the complete solution.
WM15 focuses on the first task. We will prove that the translating forms F(x − ct) and G(x + ct)
are solutions. Later articles will develop how right- and left-moving pieces combine and how initial
conditions determine them.
2 A translating profile keeps its shape
Consider
At t = 0,
At a later time Δt,
The same value of the profile that was originally at x = x0 appears later at
Thus the profile translates toward increasing x at speed c without changing shape.
Figure. An arbitrary profile translated rigidly toward increasing x. The function itself
need not be sinusoidal; every recognizable feature moves the same distance cΔt in the same
time interval.
This kinematic form is used in standard wave treatments. OpenStax writes a right-moving
wave as a function of x − vt, and Feynman explicitly shows that such profiles satisfy
the one-dimensional wave equation when the translation speed equals the wave speed
[3, 4].
3 Introduce a traveling coordinate
Define the single variable
Then
The key derivatives of ξ are
and
The chain rule will now convert derivatives of u(x,t) into ordinary derivatives of F with respect to
its single argument ξ.
4 First spatial derivative
Differentiate u = F(ξ) with respect to x while holding t fixed:
| ux | = F′(ξ) | (17)
|
| = F′(ξ). | (18) |
Thus
Differentiate once more with respect to x:
| uxx | = F′′(ξ) | (20)
|
| = F′′(ξ). | (21) |
Therefore
5 First temporal derivative
Now differentiate u = F(ξ) with respect to time while holding x fixed:
| ut | = F′(ξ) | (23)
|
| = −cF′(ξ). | (24) |
Thus
Differentiate again with respect to t:
| utt | = −cF′′(ξ) | (26)
|
| = (−c)(−c)F′′(ξ) | (27)
|
| = c2F′′(ξ). | (28) |
Therefore
Since
we immediately obtain
Hence every sufficiently smooth function of x − ct is a solution of the one-dimensional wave
equation.
Figure. The chain-rule verification for a right-moving profile. The first time derivative
introduces one factor of −c; the second introduces a second factor, producing +c2.
6 The left-moving solution
Now consider
Define
Then
The spatial derivatives are
The time derivatives are
and
Therefore
again.
So both propagation directions satisfy the same wave equation:
7 Why the signs correspond to opposite directions
The sign inside the traveling coordinate determines the propagation direction.
For
solve for x:
As time increases, x increases. The feature moves toward +x.
For
we obtain
As time increases, x decreases. The feature moves toward −x.
Figure. Lines of constant traveling coordinate in the x-t plane. A right-moving feature
follows x − ct = constant, while a left-moving feature follows x + ct = constant.
8 The derivative relation remembers the direction
For the right-moving profile,
For the left-moving profile,
Thus
and
The second-order wave equation loses this sign information because the first-order sign disappears
when the time derivative is taken twice. That is why the same second-order PDE supports
propagation in both directions.
9 Sinusoidal waves are a special case
Take the right-moving sinusoid
This can be written as
Therefore the translation speed is
For this sinusoid to satisfy
we need
For positive ω and positive k,
Equivalently,
This is the same result obtained kinematically in WM08.
Figure. For the ideal nondispersive string equation, ω is proportional to k. The slope of
the line is the wave speed c, so every sinusoidal component propagates with the same phase
speed.
Feynman gives the same criterion for sinusoidal solutions, namely ω2 = k2c2, and MIT 8.03 treats
traveling-wave solutions of the wave equation explicitly in its traveling-wave lecture
[5, 7].
10 Why arbitrary shapes can propagate unchanged
The result
is much stronger than a statement about sine waves. The function F can represent a smooth
localized pulse, a broad bump, an asymmetric disturbance, or another differentiable
profile.
The equation does not require one particular waveform. It requires that the spatial curvature and
temporal acceleration be related by
Every sufficiently smooth profile that simply translates at the correct speed automatically
maintains that relationship.
This is the mathematical origin of shape-preserving propagation in the ideal nondispersive
model.
11 Smoothness matters
The chain-rule proof used the second derivatives
and
Therefore, for the ordinary classical interpretation of the PDE, the profile must be sufficiently
smooth for these derivatives to exist.
A function with a sharp cusp can still be useful physically as an idealization, but at the cusp the
classical second derivative may fail to exist. More advanced treatments allow weaker notions of
solution, but those belong to a later level of PDE theory.
For WM15, the phrase
should therefore be understood as
12 A practical solution-checking workflow
When given a proposed function u(x,t) and asked whether it satisfies a wave equation, use the
following procedure:
- Compute uxx.
- Compute utt.
- Substitute both into the PDE.
- Check whether the equality holds for every x and t in the region of interest.
- Check whether the proposed solution also satisfies any required initial or boundary
conditions.
The final step matters. Satisfying the differential equation in the interior is necessary, but a
particular physical problem also includes its initial and boundary data.
13 Worked example 1: verify a Gaussian pulse
Consider
Does this satisfy
Define
Then
with
Because the profile is of the form F(x − ct) with
the general result already predicts
To verify directly, first compute
Then
Therefore
and
Hence
The Gaussian pulse is a classical right-moving solution with speed
when x is measured in meters and t in seconds.
14 Worked example 2: verify a left-moving sinusoid
Consider
The plus sign between the spatial and temporal terms indicates leftward propagation.
Here
and
The implied wave speed is
Differentiate twice:
and
Because
we have
Thus the function is a solution of the wave equation with
15 Worked example 3: reject a mismatched sinusoid
Suppose the governing equation is
Test the candidate
The candidate has
and
Its translation speed is
But the PDE requires
Direct differentiation confirms the mismatch:
while
The PDE right-hand side is
which is not equal to utt except at isolated points where u = 0.
Therefore
A PDE must hold throughout the domain, not merely at selected points.
16 Worked example 4: propagate an initial pulse
Suppose
and a right-moving pulse initially has shape
The wave speed is
A right-moving solution that preserves the initial profile is
After
every identifiable feature has shifted by
| Δx | = ct | (98)
|
| = (12 m/s)(0.25 s) | (99)
|
| = 3.0 m. | (100) |
So the profile has moved
toward increasing x.
17 Worked example 5: a cusp and the meaning of classical solution
Consider the translating profile
Away from
the function is piecewise linear and its second derivatives vanish. However, at
the profile has a cusp. The first derivative changes discontinuously, so the ordinary second
derivative is not defined there.
Therefore the chain-rule proof used in WM15 cannot establish a classical solution at the cusp
itself.
The correct conclusion at this level is
More advanced PDE theory can treat nonsmooth waves using generalized notions of solution, but
that is beyond the present scope.
18 Worked example 6: superpose right- and left-moving solutions
Let
and
Each separately satisfies
Because the wave equation is linear, define
Then
| utt | = (u1)tt + (u2)tt | (110)
|
| = c2(u
1)xx + c2(u
2)xx | (111)
|
| = c2![[(u1)xx + (u2)xx]](https://images.physicslibrary.org/cache/objects/1177/make4ht/WaveMechanicsTravelingWaveSolutionsOfThe1DWaveEquation101x.png) | (112)
|
| = c2u
xx. | (113) |
Therefore
is also a solution whenever the two profiles are sufficiently smooth.
This result previews the general structure developed in the next stage of the course. Feynman gives
the same two-direction form as the general one-dimensional wave solution, while WM15 uses it
here only as a consequence of the already-established linearity and the two verified traveling-wave
families [6].
19 Common mistakes
- Mistake: assuming F(x−ct) works for any numerical value of c. The translation speed
must match the coefficient in the governing PDE.
- Mistake: losing the minus sign in ut for a right-moving profile. The minus sign comes
from differentiating x − ct with respect to time.
- Mistake: carrying that minus sign into utt. The second time derivative introduces
another factor of −c, so the product is +c2.
- Mistake: deciding propagation direction from the sign of c alone. In this series, c
denotes a positive speed magnitude; direction is encoded by x ∓ ct.
- Mistake: checking the PDE at one point and declaring success. A solution must satisfy
the PDE throughout the relevant domain.
- Mistake: forgetting initial and boundary conditions. Satisfying the PDE does not
automatically solve a particular physical problem.
- Mistake: treating a nonsmooth cusp exactly like a smooth pulse. Classical second
derivatives may fail at the cusp.
20 What WM15 establishes
WM14 showed why an ideal stretched string obeys
WM15 has now shown that this equation supports arbitrary sufficiently smooth translating
profiles:
for propagation toward +x, and
for propagation toward −x.
For sinusoidal waves, the same condition becomes
or
The conceptual chain is now complete:
The next articles can therefore focus on combining right- and left-moving pieces and determining
them from initial data.
References
[1] A. P. French, Vibrations and Waves, M.I.T. Introductory Physics Series, W. W.
Norton & Company, 1971.
[2] Frank S. Crawford, Jr., Waves, Berkeley Physics Course, Volume 3, McGraw-Hill,
1968.
[3] William Moebs, Samuel J. Ling, and Jeff Sanny, University Physics, Volume 1,
OpenStax, 2016, Section 16.2, “Mathematics of Waves.”
[4] Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures
on Physics, Volume I, Chapter 47, “Sound. The wave equation,” especially Section 47–4,
“Solutions of the wave equation.”
[5] Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures
on Physics, Volume I, Chapter 48, “Beats,” especially the discussion of sinusoidal
traveling waves and ω2 = k2c2.
[6] Richard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman
Lectures on Physics, Volume I, Chapter 49, “Modes,” including the two-direction form
F(x − ct) + G(x + ct).
[7] Massachusetts Institute of Technology, 8.03SC Physics III: Vibrations and Waves,
Lecture 10, “Traveling Waves,” Fall 2016, MIT OpenCourseWare.