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

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

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

## Details

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

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

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

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

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

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

We can integrate the last equation to get

$cF(x)-cG(x)=\int _{-\infty }^{x}h(\xi )\,d\xi +c_{1}.$ Now we can solve this system of equations to get

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

$u(x,t)=F(x+ct)+G(x-ct)$ d'Alembert's formula becomes:

$u(x,t)={\frac {1}{2}}\left[g(x-ct)+g(x+ct)\right]+{\frac {1}{2c}}\int _{x-ct}^{x+ct}h(\xi )\,d\xi .$ ## Generalization for inhomogeneous canonical hyperbolic differential equations

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

$u_{tt}-c^{2}u_{xx}=f(x,t),\,u(x,0)=g(x),\,u_{t}(x,0)=h(x),$ for $-\infty 0,f\in C^{2}(\mathbb {R} ^{2},\mathbb {R} )$ .

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 ($f(x,t)=0$ ), while the inhomogeneous one is much more general, as in $f(x,t)$ 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:

$u(x,t)={\frac {1}{2}}{\bigl (}g(x+ct)+g(x-ct){\bigr )}+{\frac {1}{2c}}\int _{x-ct}^{x+ct}h(s)\,ds+{\frac {1}{2c}}\int _{0}^{t}\int _{x-c(t-\tau )}^{x+c(t-\tau )}f(s,\tau )\,ds\,d\tau .$ If $g(x)=0$ , the first part disappears, if $h(x)=0$ , the second part disappears, and if $f(x)=0$ , the third part disappears from the solution, since integrating the 0-function between any two bounds always results in 0.