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


 * $\map u {x, 0} = \map \phi x$
 * $\map {u_t} {x, 0} = \map \psi x$

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:


 * $\map \phi x = \map u {x,0} = \map f x + \map g x$


 * $\map \psi x = \map {u_t} {x,0} = c \map {f'} x - c \map {g'} x$

So we have:


 * $\map {\phi'} x = \map {f'} x + \map {g'} x$


 * $\dfrac {\map \psi x} c = \map {f'} x - \map {g'} x$

Solving the equations give:


 * $\map {f'} x = \dfrac 1 2 \paren {\map {\phi'} x + \dfrac {\map \psi x} c}$


 * $\map {g'} x = \dfrac 1 2 \paren {\map {\phi'} x - \dfrac {\map \psi x} c}$

Integrating both equations and using Fundamental Theorem of Calculus:

$\displaystyle \map f s = \dfrac 1 2 \map \phi s + \dfrac 1 {2c} \int_0^s \map \psi y \rd y + A$

$\displaystyle \map g s = \dfrac 1 2 \map \phi s - \dfrac 1 {2c} \int_0^s \map \psi y \rd y + B$

for some constants $A,B$.

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

Therefore: