D'Alembert's Formula

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Let $u: \R^2 \to \R$ be a twice-differentiable function in two variables.

Let $\phi: \R \to \R$ be a differentiable function in $x$.

Let $\psi: \R \to \R$ be an integrable function in $x$.

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

Then the solution to the partial differential equation:

$u_{tt} = c^2 u_{xx}$

with initial conditions

\(\displaystyle \map u {x, 0}\) \(=\) \(\displaystyle \map \phi x\)
\(\displaystyle \map {u_t} {x, 0}\) \(=\) \(\displaystyle \map \psi x\)

is given by:

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

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


The general solution to the 1-D wave equation:

$u_{tt} = c^2 u_{xx} \quad \text{for } - \infty < x < \infty$

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:

\(\displaystyle \map \phi x\) \(=\) \(\, \displaystyle \map u {x, 0} \, \) \(\, \displaystyle =\, \) \(\displaystyle \map f x + \map g x\)
\(\displaystyle \map \psi x\) \(=\) \(\, \displaystyle \map {u_t} {x, 0} \, \) \(\, \displaystyle =\, \) \(\displaystyle c \map {f'} x - c \map {g'} x\) Chain Rule for Partial Derivatives

So we have:

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

Solving the equations give:

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

Integrating both equations and using Fundamental Theorem of Calculus:

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

for some constants $A,B$.

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


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


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 $1746$.