D'Alembert's formula

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

for ${\displaystyle -\mathrm{\infty }<x<\mathrm{\infty },t>0}$

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 ${\displaystyle x\pm ct=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}}$ (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 ${\displaystyle c{F}^{\prime }(x)-c{G}^{\prime }(x)=h(x)}$.

We can integrate the last equation to get $${\displaystyle cF(x)-cG(x)={\int }_{-\mathrm{\infty }}^{x}h(\xi )d\xi +c_1.}$$

Now we can solve this system of equations to get $${\displaystyle F(x)=\frac{-1}{2c}\left(-cg(x)-\left({\int }_{-\mathrm{\infty }}^{x}h(\xi )d\xi +c_1\right)\right)}$$ $${\displaystyle G(x)=\frac{-1}{2c}\left(-cg(x)+\left({\int }_{-\mathrm{\infty }}^{x}h(\xi )d\xi +c_1\right)\right).}$$

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

d'Alembert's formula becomes: $${\displaystyle u(x,t)=\frac{1}{2}\left[g(x-ct)+g(x+ct)\right]+\frac{1}{2c}{\int }_{x-ct}^{x+ct}h(\xi )d\xi .}$$