# d'Alembert's formula

In mathematics, and specifically partial differential equations (PDEs), d'Alembert's formula is the general solution to the one-dimensional wave equation:

${\displaystyle u_{tt}-c^{2}u_{xx}=0,\,u(x,0)=g(x),\,u_{t}(x,0)=h(x),}$

for ${\displaystyle -\infty 0}$. It is named after the mathematician Jean le Rond d'Alembert.[1]

## Details

The characteristics of the PDE are ${\displaystyle x\pm ct=\mathrm {const} \,}$, so we can use the change of variables ${\displaystyle \mu =x+ct,\eta =x-ct\,}$ to transform the PDE to ${\displaystyle u_{\mu \eta }=0\,}$. The general solution of this PDE is ${\displaystyle u(\mu ,\eta )=F(\mu )+G(\eta )\,}$ where ${\displaystyle F\,}$ and ${\displaystyle G\,}$ are ${\displaystyle C^{1}\,}$ functions. Back in ${\displaystyle x,t\,}$ coordinates,

${\displaystyle u(x,t)=F(x+ct)+G(x-ct)\,}$
${\displaystyle u\,}$ is ${\displaystyle C^{2}\,}$ if ${\displaystyle F\,}$ and ${\displaystyle G\,}$ are ${\displaystyle C^{2}\,}$.

This solution ${\displaystyle u\,}$ can be interpreted as two waves with constant velocity ${\displaystyle c\,}$ moving in opposite directions along the x-axis.

Now let us consider this solution with the Cauchy data ${\displaystyle u(x,0)=g(x),u_{t}(x,0)=h(x)\,}$.

Using ${\displaystyle u(x,0)=g(x)\,}$ we get ${\displaystyle F(x)+G(x)=g(x)\,}$.

Using ${\displaystyle u_{t}(x,0)=h(x)\,}$ we get ${\displaystyle cF'(x)-cG'(x)=h(x)\,}$.

We can integrate the last equation to get

${\displaystyle cF(x)-cG(x)=\int _{-\infty }^{x}h(\xi )\,d\xi +c_{1}.\,}$

Now we can solve this system of equations to get

${\displaystyle F(x)={\frac {-1}{2c}}\left(-cg(x)-\left(\int _{-\infty }^{x}h(\xi )\,d\xi +c_{1}\right)\right)\,}$
${\displaystyle G(x)={\frac {-1}{2c}}\left(-cg(x)+\left(\int _{-\infty }^{x}h(\xi )d\xi +c_{1}\right)\right).\,}$

Now, using

${\displaystyle u(x,t)=F(x+ct)+G(x-ct)\,}$

d'Alembert's formula becomes:

${\displaystyle u(x,t)={\frac {1}{2}}\left[g(x-ct)+g(x+ct)\right]+{\frac {1}{2c}}\int _{x-ct}^{x+ct}h(\xi )\,d\xi .}$

## Generalization for inhomogeneous canonical hyperbolic differential equations

The general form of an inhomogeneous canonical hyperbolic type differential equation takes the form of:

${\displaystyle u_{tt}-u_{xx}=f(x,t),\,u(x,0)=g(x),\,u_{t}(x,0)=h(x),}$

for ${\displaystyle -\infty 0,f\in C^{2}(\mathbb {R} ^{2},\mathbb {R} )}$.

All second order differential equations with constant coefficients can be transformed into their respective canonic forms. This equation is one of these three cases: Elliptic partial differential equation, Parabolic partial differential equation and Hyperbolic partial differential equation.

The only difference between a homogeneous and an inhomogeneous (partial) differential equation is that in the homogeneous form we only allow 0 to stand on the right side ( ${\displaystyle f(x,t)=0}$ ), while the inhomogeneous one is much more general, as in ${\displaystyle f(x,t)}$ could be any function as long as it's continuous and can be continuously differentiated twice.

The solution of the above equation is given by the formula:

${\displaystyle u(x,t)={\frac {1}{2}}{\Bigl (}g(x+t)+g(x-t){\biggr )}+{\frac {1}{2}}{\biggl (}\int \limits _{x-t}^{x+t}h(s)ds{\biggr )}+{\frac {1}{2}}{\biggl (}\int \limits _{0}^{t}\int \limits _{x-t+\tau }^{x+t-\tau }f(s,\tau )dsd\tau {\Bigr )}}$.

If ${\displaystyle g(x)=0}$, the first part disappears, if ${\displaystyle h(x)=0}$, the second part disappears, and if ${\displaystyle f(x)=0}$, the third part disappears from the solution, since integrating the 0-function between any two bounds always results in 0.

This means, that the homogeneous equation ( ${\displaystyle f(x,t)=0}$ ) gives back our original formula for the case of ${\displaystyle c=1}$.