D'Alembert's Formula

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Theorem

Let $t$ be time.

Let $x$ be position.

Let $\tuple {t, x} \stackrel u {\longrightarrow} \map u {t, x}: \R^2 \to \R$ be a twice-differentiable function in both variables.

Let $x \stackrel \phi {\longrightarrow} \map \phi x: \R \to \R$ be a differentiable function.

Let $x \stackrel \psi {\longrightarrow} \map \psi x: \R \to \R$ be a Riemann integrable function.

Let $c \in \R_{> 0}$ be a constant.


Then the solution to the partial differential equation:

$\dfrac {\partial^2} {\partial t^2} \map u {x, t} = c^2 \dfrac {\partial^2} {\partial x^2} \map u {x, t}$

with initial conditions:

\(\ds \map u {x, 0}\) \(=\) \(\ds \map \phi x\)
\(\ds \valueat {\dfrac {\partial} {\partial t} \map u {x, t} } {t \mathop = 0}\) \(=\) \(\ds \map \psi x\)

is given by:

$\ds \map u {x, t} = \dfrac 1 2 \paren {\map \phi {x + c t} + \map \phi {x - c t} } + \dfrac 1 {2 c} \int_{x - c t}^{x + c t} \map \psi s \rd s$


The above solution formula is called d'Alembert's formula.


Proof

The general solution to the $1$-D wave equation:

$\dfrac {\partial^2} {\partial t^2} \map u {x, t} = c^2 \dfrac {\partial^2} {\partial x^2} \map u {x, t}$

is given by:

$\map u {x, t} = \map f {x + c t} + \map g {x - c t}$

where $f, g$ are arbitrary twice-differentiable functions.


From initial conditions we have:

\(\ds \map \phi x\) \(=\) \(\, \ds \map u {x, 0} \, \) \(\, \ds = \, \) \(\ds \map f x + \map g x\)
\(\ds \map \psi x\) \(=\) \(\, \ds \valueat {\dfrac \partial {\partial t} \map u {x, t} } {t \mathop = 0} \, \) \(\, \ds = \, \) \(\ds c \map {f'} x - c \map {g'} x\) Chain Rule for Partial Derivatives


So we have:

\(\ds \map {\phi'} x\) \(=\) \(\ds \map {f'} x + \map {g'} x\) Sum Rule for Derivatives
\(\ds \dfrac {\map \psi x} c\) \(=\) \(\ds \map {f'} x - \map {g'} x\)


Solving the equations give:

\(\ds \map {f'} x\) \(=\) \(\ds \dfrac 1 2 \paren {\map {\phi'} x + \dfrac {\map \psi x} c}\)
\(\ds \map {g'} x\) \(=\) \(\ds \dfrac 1 2 \paren {\map {\phi'} x - \dfrac {\map \psi x} c}\)


Integrating both equations and using Fundamental Theorem of Calculus:

\(\ds \map f x\) \(=\) \(\ds \dfrac 1 2 \map \phi x + \dfrac 1 {2 c} \int_0^x \map \psi s \rd s + A\)
\(\ds \map g x\) \(=\) \(\ds \dfrac 1 2 \map \phi x - \dfrac 1 {2 c} \int_0^x \map \psi s \rd s + B\)

for some constants $A, B$.


From $\map \phi x = \map f x + \map g x$, we have $A + B = 0$.

Therefore:

\(\ds \map u {x, t}\) \(=\) \(\ds \map f {x + c t} + \map g {x - c t}\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \map \phi {x + c t} + \dfrac 1 {2 c} \int_0^{x + c t} \map \psi s \rd s + A + \dfrac 1 2 \map \phi {x - c t} - \dfrac 1 {2 c} \int_0^{x - c t} \map \psi s \rd s + B\) substitution
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {\map \phi {x + c t} + \map \phi {x - c t} } + \dfrac 1 {2 c} \int_{x - c t}^{x + c t} \map \psi s \rd s\) simplification

$\blacksquare$


Source of Name

This entry was named for Jean le Rond d'Alembert.


Historical Note

Jean le Rond d'Alembert devised this solution to the $1$-dimensional wave equation in $1746$.


Sources

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