# 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 ${\displaystyle u_{tt}(x,t)=c^{2}u_{xx}(x,t)}$ (where subscript indices indicate partial differentiation, using the d'Alembert operator, the PDE becomes: ${\displaystyle \Box u=0}$).

The solution depends on the initial conditions at ${\displaystyle t=0}$: ${\displaystyle u(x,0)}$ and ${\displaystyle u_{t}(x,0)}$. It consists of separate terms for the initial conditions ${\displaystyle u(x,0)}$ and ${\displaystyle u_{t}(x,0)}$:

${\displaystyle u(x,t)={\frac {1}{2}}\left[u(x-ct,0)+u(x+ct,0)\right]+{\frac {1}{2c}}\int _{x-ct}^{x+ct}u_{t}(\xi ,0)\,d\xi .}$

It is named after the mathematician Jean le Rond d'Alembert, who derived it in 1747 as a solution to the problem of a vibrating string.[1]

## Details

The characteristics of the PDE are ${\displaystyle x\pm ct=\mathrm {const} }$ (where ${\displaystyle \pm }$ sign states the two solutions to quadratic equation), so we can use the change of variables ${\displaystyle \mu =x+ct}$ (for the positive solution) and ${\displaystyle \eta =x-ct}$ (for the negative solution) 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 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:[2]

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

## Generalization for inhomogeneous canonical hyperbolic differential equations

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

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

${\displaystyle u(x,t)={\frac {1}{2}}{\bigl (}g(x+ct)+g(x-ct){\bigr )}+{\frac {1}{2c}}\int _{x-ct}^{x+ct}h(s)\,ds+{\frac {1}{2c}}\int _{0}^{t}\int _{x-c(t-\tau )}^{x+c(t-\tau )}f(s,\tau )\,ds\,d\tau .}$

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.