# D'Alembert's Formula

## Theorem

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.

## Proof

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 $A + B = 0$.

Therefore:

 $\displaystyle \map u {x,t}$ $=$ $\displaystyle \map f {x + c t} + \map g {x - c t}$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 2 \map \phi {x + c t} + \dfrac 1 {2c} \int_0^{x + c t} \map \psi s \rd s + A + \dfrac 1 2 \map \phi {x - c t} - \dfrac 1 {2c} \int_0^{x - c t} \map \psi s \rd s + B$ 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

$\blacksquare$

## 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 $1746$.