Wave Mechanics: Initial Conditions and the d'Alembert SolutionWaveMechanicsInitialConditionsAndTheDAlembertSolution2026-09-12 15:26:192026-09-12 15:26:19Topicwave mechanicsone-dimensional wave equationd'Alembert solutioninitial conditionsinitial displacementinitial velocitycharacteristicsdomain of dependencefinite propagation speedtraveling waves% this is the default PhysicsLibrary preamble. as your knowledge
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% define commands here\section*{Wave Mechanics: Initial Conditions and the d'Alembert Solution}
WM16 established the structural form of a sufficiently smooth solution of the one-dimensional constant-speed wave equation,
\begin{equation}
\boxed{u_{tt}=c^2u_{xx}},
\end{equation}
namely
\begin{equation}
\boxed{u(x,t)=F(x-ct)+G(x+ct)}.
\end{equation}
The two arbitrary functions represent independent right- and left-moving components. The remaining question is physical as well as mathematical:
\begin{equation}
\boxed{\text{How do the initial displacement and initial velocity determine }F\text{ and }G?}
\end{equation}
The answer is the classical d'Alembert formula for the initial-value problem on the whole line. If
\begin{equation}
u(x,0)=f(x)
\end{equation}
and
\begin{equation}
u_t(x,0)=g(x),
\end{equation}
then, under the usual smoothness assumptions,
\begin{equation}
\boxed{
u(x,t)=\frac{1}{2}\left[f(x-ct)+f(x+ct)\right]
+\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)\,ds.}
\end{equation}
This formula solves the initial-value problem explicitly and makes finite propagation speed visible in the mathematics \cite{Strauss2008,Feynman47,MITStrangWave}.
\section{Why two initial conditions are needed}
The wave equation contains a second derivative with respect to time. Just as a second-order ordinary differential equation generally requires an initial position and an initial velocity, the wave equation requires two initial functions:
\begin{equation}
\boxed{u(x,0)=f(x)}
\end{equation}
and
\begin{equation}
\boxed{u_t(x,0)=g(x).}
\end{equation}
The function $f(x)$ describes the initial shape of the string or wave field. The function $g(x)$ describes the initial velocity of every material point.
The logical structure is summarized below.
\begin{center}
\includegraphics{WM17_fig01_dalembert_structure.png}
\vspace{0.45em}
\textbf{Figure.}
The wave equation supplies the two traveling families. Initial displacement and initial velocity determine how those families are combined.
\end{center}
\section{Apply the initial displacement}
Begin with the WM16 form
\begin{equation}
u(x,t)=F(x-ct)+G(x+ct).
\end{equation}
At $t=0$,
\begin{equation}
u(x,0)=F(x)+G(x).
\end{equation}
The initial displacement condition therefore gives
\begin{equation}
\boxed{F(x)+G(x)=f(x).}
\end{equation}
This is one relation between the two unknown functions.
\section{Apply the initial velocity}
Differentiate the general solution with respect to time:
\begin{equation}
u_t(x,t)=-cF'(x-ct)+cG'(x+ct).
\end{equation}
At $t=0$,
\begin{equation}
u_t(x,0)=-cF'(x)+cG'(x).
\end{equation}
The initial velocity condition therefore gives
\begin{equation}
\boxed{-cF'(x)+cG'(x)=g(x).}
\end{equation}
Divide by $c$:
\begin{equation}
-F'(x)+G'(x)=\frac{g(x)}{c}.
\end{equation}
Meanwhile, differentiating
\begin{equation}
F(x)+G(x)=f(x)
\end{equation}
with respect to $x$ gives
\begin{equation}
F'(x)+G'(x)=f'(x).
\end{equation}
We now have two algebraic equations for the two unknown derivative functions $F'$ and $G'$.
\section{Solve for the two traveling components}
Add the two equations:
\begin{align}
F'+G'&=f',\\
-F'+G'&=\frac{g}{c}.
\end{align}
This gives
\begin{equation}
2G'=f'+\frac{g}{c},
\end{equation}
so
\begin{equation}
\boxed{G'=\frac{1}{2}f'+\frac{1}{2c}g.}
\end{equation}
Subtracting instead gives
\begin{equation}
2F'=f'-\frac{g}{c},
\end{equation}
so
\begin{equation}
\boxed{F'=\frac{1}{2}f'-\frac{1}{2c}g.}
\end{equation}
Integrating with respect to the argument gives, for a convenient fixed reference point $x_*$,
\begin{equation}
F(x)=\frac{1}{2}f(x)-\frac{1}{2c}\int_{x_*}^{x}g(s)\,ds+C_F
\end{equation}
and
\begin{equation}
G(x)=\frac{1}{2}f(x)+\frac{1}{2c}\int_{x_*}^{x}g(s)\,ds+C_G.
\end{equation}
The displacement condition requires the constants to satisfy
\begin{equation}
C_F+C_G=0.
\end{equation}
Only the sum $F+G$ matters physically, so the arbitrary constant split cancels from the final solution.
\section{Derive the d'Alembert formula}
Substitute $x-ct$ into $F$ and $x+ct$ into $G$:
\begin{align}
u(x,t)
&=\frac{1}{2}f(x-ct)+\frac{1}{2}f(x+ct)\\
&\quad-\frac{1}{2c}\int_{x_*}^{x-ct}g(s)\,ds
+\frac{1}{2c}\int_{x_*}^{x+ct}g(s)\,ds.
\end{align}
Using
\begin{equation}
\int_{x_*}^{x+ct}g(s)\,ds-
\int_{x_*}^{x-ct}g(s)\,ds
=\int_{x-ct}^{x+ct}g(s)\,ds,
\end{equation}
we obtain
\begin{equation}
\boxed{
u(x,t)=\frac{1}{2}\left[f(x-ct)+f(x+ct)\right]
+\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)\,ds.}
\end{equation}
This is the d'Alembert solution of the one-dimensional wave equation initial-value problem on the whole line \cite{Strauss2008,MITStrangWave}.
\section{Interpret the displacement term}
First suppose the initial velocity is zero:
\begin{equation}
g(x)=0.
\end{equation}
Then
\begin{equation}
\boxed{u(x,t)=\frac{1}{2}f(x-ct)+\frac{1}{2}f(x+ct).}
\end{equation}
The initial shape splits into two copies. One moves right, one moves left, and each has half the original amplitude.
At $t=0$ the copies overlap exactly:
\begin{equation}
\frac{1}{2}f(x)+\frac{1}{2}f(x)=f(x).
\end{equation}
The splitting is illustrated below.
\begin{center}
\includegraphics{WM17_fig02_zero_velocity_split.png}
\vspace{0.45em}
\textbf{Figure.}
With zero initial velocity, the initial displacement profile separates into equal right- and left-moving half-amplitude copies.
\end{center}
This result is one of the clearest physical interpretations of the d'Alembert formula.
\section{Interpret the velocity term}
Now suppose the initial displacement is zero:
\begin{equation}
f(x)=0.
\end{equation}
Then
\begin{equation}
\boxed{u(x,t)=\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)\,ds.}
\end{equation}
The displacement at $(x,t)$ depends on the accumulated initial velocity over the interval
\begin{equation}
[x-ct,\,x+ct].
\end{equation}
\begin{center}
\includegraphics{WM17_fig04_initial_velocity_interval.png}
\vspace{0.45em}
\textbf{Figure.}
The initial-velocity contribution is determined by the integral of $g(s)$ over the characteristic interval from $x-ct$ to $x+ct$.
\end{center}
The width of that interval is
\begin{equation}
2ct.
\end{equation}
As time increases, information from a larger portion of the initial line can influence the observation point.
\section{Domain of dependence and finite propagation speed}
For a point $(x_0,t_0)$, the d'Alembert formula uses initial data only between
\begin{equation}
x_0-ct_0
\end{equation}
and
\begin{equation}
x_0+ct_0.
\end{equation}
These endpoints are reached by the two backward characteristic lines.
\begin{center}
\includegraphics{WM17_fig03_domain_of_dependence.png}
\vspace{0.45em}
\textbf{Figure.}
The value at $(x_0,t_0)$ depends only on initial data in the interval $[x_0-ct_0,x_0+ct_0]$. This is the domain of dependence for the one-dimensional constant-speed wave equation.
\end{center}
Therefore a disturbance in the initial data cannot influence arbitrarily distant points instantaneously. Information propagates at the finite speed $c$.
This causal structure is one of the most important qualitative consequences of the wave equation \cite{Strauss2008}.
\section{Pure one-way motion requires compatible initial data}
The d'Alembert formula also reveals the initial-data condition for a wave to travel in only one direction.
For a pure right-moving wave,
\begin{equation}
u(x,t)=F(x-ct).
\end{equation}
At $t=0$,
\begin{equation}
f(x)=F(x)
\end{equation}
and
\begin{equation}
g(x)=-cF'(x).
\end{equation}
Therefore
\begin{equation}
\boxed{g(x)=-cf'(x)\qquad\text{produces a pure right-moving wave}.}
\end{equation}
Similarly,
\begin{equation}
\boxed{g(x)=+cf'(x)\qquad\text{produces a pure left-moving wave}.}
\end{equation}
This explains why an arbitrary initial shape by itself does not usually travel in only one direction. The initial velocity must be chosen consistently with the desired direction.
\section{Checking the initial conditions directly}
A useful consistency check is to set $t=0$ in the d'Alembert formula:
\begin{align}
u(x,0)
&=\frac{1}{2}[f(x)+f(x)]
+\frac{1}{2c}\int_x^x g(s)\,ds\\
&=f(x).
\end{align}
Thus the displacement condition is satisfied.
To check the velocity, differentiate the formula with respect to time. The displacement part gives
\begin{equation}
-\frac{c}{2}f'(x-ct)+\frac{c}{2}f'(x+ct).
\end{equation}
For the integral term, the Leibniz rule gives
\begin{align}
\frac{\partial}{\partial t}
\left[\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)\,ds\right]
&=\frac{1}{2c}
\left[cg(x+ct)+cg(x-ct)\right]\\
&=\frac{1}{2}[g(x+ct)+g(x-ct)].
\end{align}
At $t=0$, the two $f'$ terms cancel and the velocity term becomes
\begin{equation}
\frac{1}{2}[g(x)+g(x)]=g(x).
\end{equation}
Therefore
\begin{equation}
\boxed{u_t(x,0)=g(x).}
\end{equation}
\section{The role of boundaries}
The formula derived here is most naturally stated for the initial-value problem on the entire real line.
On a finite interval, boundary conditions also matter. Reflections from fixed or free boundaries can be incorporated through reflected extensions, mode expansions, or other boundary-value methods. The PhysicsLibrary treatment of boundary conditions in WM12 explains why endpoint constraints modify which solutions are physically allowed.
Thus
\begin{equation}
\boxed{\text{d'Alembert formula on the whole line} \neq
\text{complete finite-string boundary-value solution}.}
\end{equation}
\section{Regularity assumptions}
The derivation above assumes enough differentiability for the chain rule, mixed derivatives, and the classical wave equation to be meaningful. A common sufficient setting is to take $f$ twice continuously differentiable and $g$ continuously differentiable.
Less regular initial data can also be treated using weaker notions of solution, but that belongs to a more advanced PDE treatment \cite{Strauss2008}.
\section{Worked example 1: released Gaussian displacement}
Suppose
\begin{equation}
f(x)=A\exp\left(-\frac{x^2}{a^2}\right)
\end{equation}
and
\begin{equation}
g(x)=0.
\end{equation}
The d'Alembert formula immediately gives
\begin{align}
u(x,t)
&=\frac{A}{2}\exp\left[-\frac{(x-ct)^2}{a^2}\right]\\
&\quad+\frac{A}{2}\exp\left[-\frac{(x+ct)^2}{a^2}\right].
\end{align}
Therefore the initial Gaussian separates into two Gaussian pulses of amplitude $A/2$ traveling in opposite directions at speed $c$:
\begin{equation}
\boxed{
u(x,t)=\frac{A}{2}e^{-(x-ct)^2/a^2}
+\frac{A}{2}e^{-(x+ct)^2/a^2}.}
\end{equation}
At $t=0$, the two halves add to recover the original amplitude $A$.
\section{Worked example 2: zero displacement but sinusoidal initial velocity}
Suppose
\begin{equation}
f(x)=0
\end{equation}
and
\begin{equation}
g(x)=V_0\cos(kx).
\end{equation}
Then
\begin{equation}
u(x,t)=\frac{V_0}{2c}\int_{x-ct}^{x+ct}\cos(ks)\,ds.
\end{equation}
Integrate:
\begin{align}
u(x,t)
&=\frac{V_0}{2ck}
\left[\sin(kx+kct)-\sin(kx-kct)\right].
\end{align}
Using
\begin{equation}
\sin(\alpha+\beta)-\sin(\alpha-\beta)
=2\cos\alpha\sin\beta,
\end{equation}
we obtain
\begin{equation}
\boxed{
u(x,t)=\frac{V_0}{ck}\cos(kx)\sin(kct).}
\end{equation}
Since
\begin{equation}
\omega=ck,
\end{equation}
this is a standing-wave form:
\begin{equation}
\boxed{u(x,t)=\frac{V_0}{\omega}\cos(kx)\sin(\omega t).}
\end{equation}
Different initial data can therefore generate a standing pattern even though the underlying solution is still built from right- and left-moving components.
\section{Worked example 3: choose data for a pure right-moving pulse}
Suppose the desired initial shape is
\begin{equation}
f(x)=A e^{-x^2/a^2}.
\end{equation}
To make it move purely to the right, choose
\begin{equation}
g(x)=-cf'(x).
\end{equation}
Differentiate:
\begin{equation}
f'(x)=-\frac{2Ax}{a^2}e^{-x^2/a^2}.
\end{equation}
Therefore
\begin{equation}
\boxed{g(x)=\frac{2cAx}{a^2}e^{-x^2/a^2}.}
\end{equation}
With this compatible initial velocity, the left-moving component vanishes and
\begin{equation}
\boxed{u(x,t)=A\exp\left[-\frac{(x-ct)^2}{a^2}\right].}
\end{equation}
The entire initial pulse moves right without splitting.
\section{Worked example 4: evaluate the formula numerically}
Let
\begin{equation}
c=2.0\,\text{m/s},
\end{equation}
with initial data
\begin{equation}
f(x)=x^2
\end{equation}
and
\begin{equation}
g(x)=3x.
\end{equation}
Find $u(1.0\,\text{m},0.25\,\text{s})$.
First compute the characteristic endpoints:
\begin{align}
x-ct&=1.0-(2.0)(0.25)=0.50\,\text{m},\\
x+ct&=1.0+(2.0)(0.25)=1.50\,\text{m}.
\end{align}
The displacement contribution is
\begin{align}
\frac{1}{2}[f(0.50)+f(1.50)]
&=\frac{1}{2}[0.50^2+1.50^2]\\
&=\frac{1}{2}[0.25+2.25]\\
&=1.25.
\end{align}
The velocity contribution is
\begin{align}
\frac{1}{2c}\int_{0.50}^{1.50}3s\,ds
&=\frac{1}{4}\left[\frac{3}{2}s^2\right]_{0.50}^{1.50}\\
&=\frac{1}{4}\left[\frac{3}{2}(2.25-0.25)\right]\\
&=0.75.
\end{align}
Therefore
\begin{equation}
\boxed{u(1.0,0.25)=2.00}
\end{equation}
in the displacement units implied by the chosen initial data.
\section{Worked example 5: choose data for a pure left-moving wave}
Suppose
\begin{equation}
f(x)=A\cos(kx).
\end{equation}
To obtain only a left-moving wave, choose
\begin{equation}
g(x)=+cf'(x).
\end{equation}
Since
\begin{equation}
f'(x)=-Ak\sin(kx),
\end{equation}
we require
\begin{equation}
\boxed{g(x)=-cAk\sin(kx).}
\end{equation}
The resulting solution is
\begin{equation}
\boxed{u(x,t)=A\cos[k(x+ct)]},
\end{equation}
or, using $\omega=ck$,
\begin{equation}
\boxed{u(x,t)=A\cos(kx+\omega t).}
\end{equation}
The initial velocity therefore determines which of the two characteristic families survives.
\section{Worked example 6: connect d'Alembert's solution to string mechanics}
A string has
\begin{equation}
T=144\,\text{N}
\end{equation}
and
\begin{equation}
\mu=0.016\,\text{kg/m}.
\end{equation}
Its wave speed is
\begin{align}
c&=\sqrt{\frac{T}{\mu}}\\
&=\sqrt{\frac{144}{0.016}}\\
&=\sqrt{9000}\\
&\approx \boxed{94.9\,\text{m/s}}.
\end{align}
Suppose the initial displacement is
\begin{equation}
f(x)=0.010\cos(4x)\,\text{m}
\end{equation}
and the initial velocity is zero.
Then
\begin{align}
u(x,t)
&=0.005\cos[4(x-ct)]\\
&\quad+0.005\cos[4(x+ct)].
\end{align}
Using the cosine sum identity,
\begin{equation}
u(x,t)=0.010\cos(4x)\cos(4ct).
\end{equation}
The angular frequency is
\begin{align}
\omega&=ck\\
&=(94.9)(4)\\
&\approx \boxed{380\,\text{rad/s}}.
\end{align}
Thus
\begin{equation}
\boxed{u(x,t)\approx0.010\cos(4x)\cos(380t)\,\text{m}.}
\end{equation}
The same result can be viewed either as two counter-propagating traveling waves or as a standing-wave pattern generated by the specified initial data.
\section{Common mistakes}
\begin{itemize}
\item \textbf{Mistake:} using only the initial displacement. A second-order-in-time wave equation also requires the initial velocity.
\item \textbf{Mistake:} forgetting the factor $1/2$ multiplying the two displaced copies of $f$.
\item \textbf{Mistake:} reversing the limits of the velocity integral. The correct interval is from $x-ct$ to $x+ct$.
\item \textbf{Mistake:} forgetting the factor $1/(2c)$ in front of the velocity integral.
\item \textbf{Mistake:} assuming zero initial velocity produces one traveling copy of $f$. It produces equal left- and right-moving half-amplitude copies.
\item \textbf{Mistake:} using the whole-line formula without considering finite-domain boundary conditions.
\item \textbf{Mistake:} assuming arbitrary $f$ and $g$ produce a pure one-way wave. Pure right- or left-moving motion requires the compatibility conditions $g=\mp cf'$.
\end{itemize}
\section{What WM17 establishes}
The structural solution from WM16,
\begin{equation}
u(x,t)=F(x-ct)+G(x+ct),
\end{equation}
becomes a complete initial-value solution once the functions
\begin{equation}
f(x)=u(x,0)
\end{equation}
and
\begin{equation}
g(x)=u_t(x,0)
\end{equation}
are specified:
\begin{equation}
\boxed{
u(x,t)=\frac{1}{2}[f(x-ct)+f(x+ct)]
+\frac{1}{2c}\int_{x-ct}^{x+ct}g(s)\,ds.}
\end{equation}
The formula exposes three major physical ideas at once:
\begin{itemize}
\item waves propagate along two characteristic directions,
\item information travels at finite speed $c$, and
\item both initial displacement and initial velocity are required to determine the subsequent motion.
\end{itemize}
This completes the basic initial-value solution of the one-dimensional ideal wave equation and prepares the way for later treatments of energy transport, interfaces, modal expansions, Fourier methods, and dispersive wave systems.
\section*{References}
\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{Feynman47}
Richard P. Feynman, Robert B. Leighton, and Matthew Sands,
\emph{The Feynman Lectures on Physics, Volume I},
Chapter 47, ``Sound. The Wave Equation.''
\bibitem{Strauss2008}
Walter A. Strauss,
\emph{Partial Differential Equations: An Introduction},
Second Edition,
John Wiley \& Sons, 2008.
\bibitem{MIT803L10}
Massachusetts Institute of Technology,
\emph{8.03SC Physics III: Vibrations and Waves},
Lecture 10, ``Traveling Waves,''
MIT OpenCourseWare.
\bibitem{MITStrangWave}
Gilbert Strang and Cleve Moler,
\emph{Learn Differential Equations: Up Close},
``Wave Equation,''
MIT OpenCourseWare, 2015.
\end{thebibliography}