D'Alembert's Formula

Theorem
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:

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:


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

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:

So we have:

Solving the equations give:

Integrating both equations and using Fundamental Theorem of Calculus:

for some constants $A,B$.

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

Therefore: