Wave Mechanics: Traveling-Wave Solutions of the 1D Wave EquationWaveMechanicsTravelingWaveSolutionsOfThe1DWaveEquation2026-09-12 08:01:312026-09-12 08:01:31Topic% this is the default PhysicsLibrary preamble. as your knowledge
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% define commands here\section*{Wave Mechanics: Traveling-Wave Solutions of the 1D Wave Equation}
WM14 derived the one-dimensional string wave equation from Newton's second law:
\begin{equation}
\boxed{u_{tt}=c^2u_{xx}},
\end{equation}
with
\begin{equation}
\boxed{c=\sqrt{\frac{T}{\mu}}}
\end{equation}
for an ideal stretched string.
Earlier articles introduced translating disturbances of the form
\begin{equation}
F(x-ct)
\end{equation}
and
\begin{equation}
G(x+ct).
\end{equation}
WM15 now connects those two parts of the course. The central question is:
\begin{equation}
\boxed{\text{Do arbitrary translating profiles actually satisfy the wave equation?}}
\end{equation}
The answer is yes, provided the profile is sufficiently smooth for the required derivatives to exist. A right-moving profile
\begin{equation}
\boxed{u(x,t)=F(x-ct)}
\end{equation}
and a left-moving profile
\begin{equation}
\boxed{u(x,t)=G(x+ct)}
\end{equation}
both satisfy
\begin{equation}
u_{tt}=c^2u_{xx}.
\end{equation}
This is one of the most important structural facts about the one-dimensional linear wave equation \cite{French1971,Crawford1968,OpenStax162,Feynman47,MIT803L10}.
\section{The question is verification, not yet the full general solution}
There are two different mathematical tasks that should not be confused.
The first is \emph{verification}: start with a proposed function and check whether it satisfies the partial differential equation.
The second is \emph{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.
\section{A translating profile keeps its shape}
Consider
\begin{equation}
u(x,t)=F(x-ct).
\end{equation}
At $t=0$,
\begin{equation}
u(x,0)=F(x).
\end{equation}
At a later time $\Delta t$,
\begin{equation}
u(x,\Delta t)=F(x-c\Delta t).
\end{equation}
The same value of the profile that was originally at $x=x_0$ appears later at
\begin{equation}
x=x_0+c\Delta t.
\end{equation}
Thus the profile translates toward increasing $x$ at speed $c$ without changing shape.
\begin{center}
\includegraphics{WM15_fig01_arbitrary_profile_translation.png}
\vspace{0.45em}
\textbf{Figure.}
An arbitrary profile translated rigidly toward increasing $x$. The function itself need not be sinusoidal; every recognizable feature moves the same distance $c\Delta t$ in the same time interval.
\end{center}
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 \cite{OpenStax162,Feynman47}.
\section{Introduce a traveling coordinate}
Define the single variable
\begin{equation}
\boxed{\xi=x-ct.}
\end{equation}
Then
\begin{equation}
u(x,t)=F(\xi).
\end{equation}
The key derivatives of $\xi$ are
\begin{equation}
\frac{\partial \xi}{\partial x}=1
\end{equation}
and
\begin{equation}
\frac{\partial \xi}{\partial t}=-c.
\end{equation}
The chain rule will now convert derivatives of $u(x,t)$ into ordinary derivatives of $F$ with respect to its single argument $\xi$.
\section{First spatial derivative}
Differentiate $u=F(\xi)$ with respect to $x$ while holding $t$ fixed:
\begin{align}
u_x
&=F'(\xi)\frac{\partial \xi}{\partial x}\\
&=F'(\xi).
\end{align}
Thus
\begin{equation}
\boxed{u_x=F'(\xi).}
\end{equation}
Differentiate once more with respect to $x$:
\begin{align}
u_{xx}
&=F''(\xi)\frac{\partial \xi}{\partial x}\\
&=F''(\xi).
\end{align}
Therefore
\begin{equation}
\boxed{u_{xx}=F''(\xi).}
\end{equation}
\section{First temporal derivative}
Now differentiate $u=F(\xi)$ with respect to time while holding $x$ fixed:
\begin{align}
u_t
&=F'(\xi)\frac{\partial \xi}{\partial t}\\
&=-cF'(\xi).
\end{align}
Thus
\begin{equation}
\boxed{u_t=-cF'(\xi).}
\end{equation}
Differentiate again with respect to $t$:
\begin{align}
u_{tt}
&=-cF''(\xi)\frac{\partial \xi}{\partial t}\\
&=(-c)(-c)F''(\xi)\\
&=c^2F''(\xi).
\end{align}
Therefore
\begin{equation}
\boxed{u_{tt}=c^2F''(\xi).}
\end{equation}
Since
\begin{equation}
u_{xx}=F''(\xi),
\end{equation}
we immediately obtain
\begin{equation}
\boxed{u_{tt}=c^2u_{xx}.}
\end{equation}
Hence every sufficiently smooth function of $x-ct$ is a solution of the one-dimensional wave equation.
\begin{center}
\includegraphics{WM15_fig02_chain_rule_verification.png}
\vspace{0.45em}
\textbf{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 $+c^2$.
\end{center}
\section{The left-moving solution}
Now consider
\begin{equation}
u(x,t)=G(x+ct).
\end{equation}
Define
\begin{equation}
\eta=x+ct.
\end{equation}
Then
\begin{equation}
\eta_x=1,
\qquad
\eta_t=c.
\end{equation}
The spatial derivatives are
\begin{equation}
u_x=G'(\eta),
\qquad
u_{xx}=G''(\eta).
\end{equation}
The time derivatives are
\begin{equation}
u_t=cG'(\eta)
\end{equation}
and
\begin{equation}
u_{tt}=c^2G''(\eta).
\end{equation}
Therefore
\begin{equation}
\boxed{u_{tt}=c^2u_{xx}}
\end{equation}
again.
So both propagation directions satisfy the same wave equation:
\begin{equation}
\boxed{F(x-ct)\quad\text{and}\quad G(x+ct).}
\end{equation}
\section{Why the signs correspond to opposite directions}
The sign inside the traveling coordinate determines the propagation direction.
For
\begin{equation}
x-ct=\text{constant},
\end{equation}
solve for $x$:
\begin{equation}
x=ct+\text{constant}.
\end{equation}
As time increases, $x$ increases. The feature moves toward $+x$.
For
\begin{equation}
x+ct=\text{constant},
\end{equation}
we obtain
\begin{equation}
x=-ct+\text{constant}.
\end{equation}
As time increases, $x$ decreases. The feature moves toward $-x$.
\begin{center}
\includegraphics{WM15_fig03_right_left_characteristics.png}
\vspace{0.45em}
\textbf{Figure.}
Lines of constant traveling coordinate in the $x$-$t$ plane. A right-moving feature follows $x-ct=\text{constant}$, while a left-moving feature follows $x+ct=\text{constant}$.
\end{center}
\section{The derivative relation remembers the direction}
For the right-moving profile,
\begin{equation}
u_t=-cu_x.
\end{equation}
For the left-moving profile,
\begin{equation}
u_t=+cu_x.
\end{equation}
Thus
\begin{equation}
\boxed{\text{right-moving: }u_t=-cu_x}
\end{equation}
and
\begin{equation}
\boxed{\text{left-moving: }u_t=+cu_x.}
\end{equation}
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.
\section{Sinusoidal waves are a special case}
Take the right-moving sinusoid
\begin{equation}
u(x,t)=A\cos(kx-\omega t+\phi).
\end{equation}
This can be written as
\begin{equation}
u(x,t)=A\cos\left[k\left(x-\frac{\omega}{k}t\right)+\phi\right].
\end{equation}
Therefore the translation speed is
\begin{equation}
\frac{\omega}{k}.
\end{equation}
For this sinusoid to satisfy
\begin{equation}
u_{tt}=c^2u_{xx},
\end{equation}
we need
\begin{equation}
\omega^2=c^2k^2.
\end{equation}
For positive $\omega$ and positive $k$,
\begin{equation}
\boxed{\omega=ck.}
\end{equation}
Equivalently,
\begin{equation}
\boxed{c=\frac{\omega}{k}.}
\end{equation}
This is the same result obtained kinematically in WM08.
\begin{center}
\includegraphics{WM15_fig04_sinusoid_dispersion_relation.png}
\vspace{0.45em}
\textbf{Figure.}
For the ideal nondispersive string equation, $\omega$ is proportional to $k$. The slope of the line is the wave speed $c$, so every sinusoidal component propagates with the same phase speed.
\end{center}
Feynman gives the same criterion for sinusoidal solutions, namely $\omega^2=k^2c^2$, and MIT 8.03 treats traveling-wave solutions of the wave equation explicitly in its traveling-wave lecture \cite{Feynman48,MIT803L10}.
\section{Why arbitrary shapes can propagate unchanged}
The result
\begin{equation}
u(x,t)=F(x-ct)
\end{equation}
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
\begin{equation}
u_{tt}=c^2u_{xx}.
\end{equation}
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.
\section{Smoothness matters}
The chain-rule proof used the second derivatives
\begin{equation}
F''(\xi)
\end{equation}
and
\begin{equation}
G''(\eta).
\end{equation}
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
\begin{equation}
\boxed{\text{arbitrary traveling profile}}
\end{equation}
should therefore be understood as
\begin{equation}
\boxed{\text{arbitrary sufficiently smooth traveling profile}.}
\end{equation}
\section{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:
\begin{enumerate}
\item Compute $u_{xx}$.
\item Compute $u_{tt}$.
\item Substitute both into the PDE.
\item Check whether the equality holds for every $x$ and $t$ in the region of interest.
\item Check whether the proposed solution also satisfies any required initial or boundary conditions.
\end{enumerate}
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.
\section{Worked example 1: verify a Gaussian pulse}
Consider
\begin{equation}
u(x,t)=A\exp\left[-(x-3t)^2\right].
\end{equation}
Does this satisfy
\begin{equation}
u_{tt}=9u_{xx}?
\end{equation}
Define
\begin{equation}
\xi=x-3t.
\end{equation}
Then
\begin{equation}
u=F(\xi)
\end{equation}
with
\begin{equation}
F(\xi)=Ae^{-\xi^2}.
\end{equation}
Because the profile is of the form $F(x-ct)$ with
\begin{equation}
c=3,
\end{equation}
the general result already predicts
\begin{equation}
u_{tt}=3^2u_{xx}=9u_{xx}.
\end{equation}
To verify directly, first compute
\begin{equation}
F'(\xi)=-2\xi Ae^{-\xi^2}.
\end{equation}
Then
\begin{equation}
F''(\xi)=A(4\xi^2-2)e^{-\xi^2}.
\end{equation}
Therefore
\begin{equation}
u_{xx}=A(4\xi^2-2)e^{-\xi^2}
\end{equation}
and
\begin{equation}
u_{tt}=9A(4\xi^2-2)e^{-\xi^2}.
\end{equation}
Hence
\begin{equation}
\boxed{u_{tt}=9u_{xx}.}
\end{equation}
The Gaussian pulse is a classical right-moving solution with speed
\begin{equation}
\boxed{3\,\text{m/s}}
\end{equation}
when $x$ is measured in meters and $t$ in seconds.
\section{Worked example 2: verify a left-moving sinusoid}
Consider
\begin{equation}
u(x,t)=0.020\cos(4x+20t).
\end{equation}
The plus sign between the spatial and temporal terms indicates leftward propagation.
Here
\begin{equation}
k=4\,\text{rad/m}
\end{equation}
and
\begin{equation}
\omega=20\,\text{rad/s}.
\end{equation}
The implied wave speed is
\begin{equation}
c=\frac{\omega}{k}
=\frac{20}{4}
=5\,\text{m/s}.
\end{equation}
Differentiate twice:
\begin{equation}
u_{xx}=-16u
\end{equation}
and
\begin{equation}
u_{tt}=-400u.
\end{equation}
Because
\begin{equation}
25u_{xx}=25(-16u)=-400u,
\end{equation}
we have
\begin{equation}
\boxed{u_{tt}=25u_{xx}.}
\end{equation}
Thus the function is a solution of the wave equation with
\begin{equation}
\boxed{c=5\,\text{m/s}.}
\end{equation}
\section{Worked example 3: reject a mismatched sinusoid}
Suppose the governing equation is
\begin{equation}
u_{tt}=25u_{xx}.
\end{equation}
Test the candidate
\begin{equation}
u(x,t)=A\cos(4x-12t).
\end{equation}
The candidate has
\begin{equation}
k=4
\end{equation}
and
\begin{equation}
\omega=12.
\end{equation}
Its translation speed is
\begin{equation}
\frac{\omega}{k}=3.
\end{equation}
But the PDE requires
\begin{equation}
c=\sqrt{25}=5.
\end{equation}
Direct differentiation confirms the mismatch:
\begin{equation}
u_{xx}=-16u
\end{equation}
while
\begin{equation}
u_{tt}=-144u.
\end{equation}
The PDE right-hand side is
\begin{equation}
25u_{xx}=-400u,
\end{equation}
which is not equal to $u_{tt}$ except at isolated points where $u=0$.
Therefore
\begin{equation}
\boxed{A\cos(4x-12t)\text{ is not a solution of }u_{tt}=25u_{xx}.}
\end{equation}
A PDE must hold throughout the domain, not merely at selected points.
\section{Worked example 4: propagate an initial pulse}
Suppose
\begin{equation}
u_{tt}=144u_{xx}
\end{equation}
and a right-moving pulse initially has shape
\begin{equation}
u(x,0)=F(x).
\end{equation}
The wave speed is
\begin{equation}
c=\sqrt{144}=12\,\text{m/s}.
\end{equation}
A right-moving solution that preserves the initial profile is
\begin{equation}
\boxed{u(x,t)=F(x-12t).}
\end{equation}
After
\begin{equation}
t=0.25\,\text{s},
\end{equation}
every identifiable feature has shifted by
\begin{align}
\Delta x
&=ct\\
&=(12\,\text{m/s})(0.25\,\text{s})\\
&=3.0\,\text{m}.
\end{align}
So the profile has moved
\begin{equation}
\boxed{3.0\,\text{m}}
\end{equation}
toward increasing $x$.
\section{Worked example 5: a cusp and the meaning of classical solution}
Consider the translating profile
\begin{equation}
u(x,t)=|x-ct|.
\end{equation}
Away from
\begin{equation}
x=ct,
\end{equation}
the function is piecewise linear and its second derivatives vanish. However, at
\begin{equation}
x=ct,
\end{equation}
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
\begin{equation}
\boxed{|x-ct|\text{ is not a classical twice-differentiable solution everywhere}.}
\end{equation}
More advanced PDE theory can treat nonsmooth waves using generalized notions of solution, but that is beyond the present scope.
\section{Worked example 6: superpose right- and left-moving solutions}
Let
\begin{equation}
u_1(x,t)=F(x-ct)
\end{equation}
and
\begin{equation}
u_2(x,t)=G(x+ct).
\end{equation}
Each separately satisfies
\begin{equation}
u_{tt}=c^2u_{xx}.
\end{equation}
Because the wave equation is linear, define
\begin{equation}
u=u_1+u_2.
\end{equation}
Then
\begin{align}
u_{tt}
&=(u_1)_{tt}+(u_2)_{tt}\\
&=c^2(u_1)_{xx}+c^2(u_2)_{xx}\\
&=c^2\left[(u_1)_{xx}+(u_2)_{xx}\right]\\
&=c^2u_{xx}.
\end{align}
Therefore
\begin{equation}
\boxed{u(x,t)=F(x-ct)+G(x+ct)}
\end{equation}
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 \cite{Feynman49}.
\section{Common mistakes}
\begin{itemize}
\item \textbf{Mistake:} assuming $F(x-ct)$ works for any numerical value of $c$. The translation speed must match the coefficient in the governing PDE.
\item \textbf{Mistake:} losing the minus sign in $u_t$ for a right-moving profile. The minus sign comes from differentiating $x-ct$ with respect to time.
\item \textbf{Mistake:} carrying that minus sign into $u_{tt}$. The second time derivative introduces another factor of $-c$, so the product is $+c^2$.
\item \textbf{Mistake:} deciding propagation direction from the sign of $c$ alone. In this series, $c$ denotes a positive speed magnitude; direction is encoded by $x\mp ct$.
\item \textbf{Mistake:} checking the PDE at one point and declaring success. A solution must satisfy the PDE throughout the relevant domain.
\item \textbf{Mistake:} forgetting initial and boundary conditions. Satisfying the PDE does not automatically solve a particular physical problem.
\item \textbf{Mistake:} treating a nonsmooth cusp exactly like a smooth pulse. Classical second derivatives may fail at the cusp.
\end{itemize}
\section{What WM15 establishes}
WM14 showed why an ideal stretched string obeys
\begin{equation}
u_{tt}=c^2u_{xx}.
\end{equation}
WM15 has now shown that this equation supports arbitrary sufficiently smooth translating profiles:
\begin{equation}
\boxed{F(x-ct)}
\end{equation}
for propagation toward $+x$, and
\begin{equation}
\boxed{G(x+ct)}
\end{equation}
for propagation toward $-x$.
For sinusoidal waves, the same condition becomes
\begin{equation}
\boxed{\omega=ck}
\end{equation}
or
\begin{equation}
\boxed{c=\frac{\omega}{k}.}
\end{equation}
The conceptual chain is now complete:
\begin{equation}
\boxed{
\text{Newton's law}
\longrightarrow
\text{wave equation}
\longrightarrow
\text{traveling-wave solutions}
}.
\end{equation}
The next articles can therefore focus on combining right- and left-moving pieces and determining them from initial data.
\begin{thebibliography}{9}
\bibitem{French1971}
A.~P. French,
\emph{Vibrations and Waves},
M.I.T. Introductory Physics Series,
W. W. Norton \& Company, 1971.
\bibitem{Crawford1968}
Frank S. Crawford, Jr.,
\emph{Waves},
Berkeley Physics Course, Volume 3,
McGraw-Hill, 1968.
\bibitem{OpenStax162}
William Moebs, Samuel J. Ling, and Jeff Sanny,
\emph{University Physics, Volume 1},
OpenStax, 2016,
Section 16.2, ``Mathematics of Waves.''
\bibitem{Feynman47}
Richard P. Feynman, Robert B. Leighton, and Matthew Sands,
\emph{The Feynman Lectures on Physics, Volume I},
Chapter 47, ``Sound. The wave equation,'' especially Section 47--4,
``Solutions of the wave equation.''
\bibitem{Feynman48}
Richard P. Feynman, Robert B. Leighton, and Matthew Sands,
\emph{The Feynman Lectures on Physics, Volume I},
Chapter 48, ``Beats,'' especially the discussion of sinusoidal traveling waves
and $\omega^2=k^2c^2$.
\bibitem{Feynman49}
Richard P. Feynman, Robert B. Leighton, and Matthew Sands,
\emph{The Feynman Lectures on Physics, Volume I},
Chapter 49, ``Modes,'' including the two-direction form
$F(x-ct)+G(x+ct)$.
\bibitem{MIT803L10}
Massachusetts Institute of Technology,
\emph{8.03SC Physics III: Vibrations and Waves},
Lecture 10, ``Traveling Waves,'' Fall 2016,
MIT OpenCourseWare.
\end{thebibliography}