$\newcommand{\const}{\mathrm{const}}$

## Homogeneous 1D Wave equation

Consider equation $$u_{tt}-c^2u_{xx}=0. \label{WE}$$

#### Physical examples

Remark. This equation describes a lot of things.

Example. Consider a string with the points defiating from the original position (along $x$) in the orthogonal direction ($y$); so the string is described by $y=u(x,t)$ at the moment $t$ (so $u$ is a displacement along $y$). In this case $c^2= T/\rho$ where $T$ is a tension of the string and $\rho$ is a linear density of it.

Example. This equation also describes compression-rarefication waves in elastic 1-dimensional media. Then $u(x,t)$ is displacement along $x$.

Example. Consider a pipe filled by an ideal gas. Then $c^2= p(\rho)/\rho$ where $\rho$ is a density and $p(\rho)$ is a pressure (for an ideal gas at the given temperature such ratio is constant and due to Mendeleev-Clapeyron equation it is proportional to absolute temperature $T$ which is assumed to be a constant). Then f.e. $u$ may denote a density $\rho(x,t)$ at point $x$ at time $t$.

Remark. $c$ has a dimension of the speed. In the example above $c$ is a speed of sound.

#### General solution

Let us rewrite formally equation (\ref{WE}) as $$(\partial_t^2 -c^2 \partial_x^2)= (\partial_t-c \partial_x)(\partial_t+c \partial_x)u=0. \label{WE2}$$ Denoting $v=(\partial_t+c \partial_x)u= u_t+cu_x$ and $w=(\partial_t-c \partial_x)u= u_t-cu_x$ we have \begin{align} &v_t-cv_x=0,\\ &w_t+cw_x=0. \end{align} But we know how to solve these equations \begin{align} &v=2c\phi'(x+ct),\\ &w=-2c\psi'(x-ct) \end{align} where $\phi'$ and $\psi'$ are arbitrary functions. We find convenient to have factors $2c$ and $-2c$ and to denote by $\phi$ and $\psi$ their primitives (aka indefinite integrals). Recalling definitions of $v$ and $v$ we have \begin{align*} &u_t+cu_x=2c\phi'(x+ct),\\ &u_t-cu_x=-2c\psi'(x-ct). \end{align*} Then \begin{align*} c^{-1}&u_t=\phi'(x+ct)-\psi'(x-ct),\\ &u_x=\phi'(x+ct)+\psi'(x-ct). \end{align*} The second equation implies that $u=\phi(x+ct)+\psi(x-ct)+\Phi(t)$ and plugging to the first equation we get $\Phi'=0$, thus $\Phi=\const$).

So, $$u=\phi(x+ct)+\psi(x-ct) \label{GS}$$ is a general solution to (\ref{WE}). This solution is a superposition of two waves $u_1=\phi(x+ct)$ and $u_2=\psi(x-ct)$ running to the left and to the rightrespectively with the speed $c$. So $c$ is a propagation speed.

Remark. Adding constant $C$ to $\phi$ and $-C$ to $\psi$ we get the same $u$. However it is the only arbitrarness.

#### Cauchy problem

Let us consider IVP for (\ref{WE}): \begin{align} &u_{tt}-c^2u_{xx}=0, &u|_{t=0}=f(x),\\ &u_t|_{t=0}=g(x). \end{align} Plugging (\ref{GS}) into initial conditions we have \begin{align} &\phi(x)+\psi(x)=f(x),\\ &c\phi'(x)-c\psi'(x)=g(x)\implies \phi(x)-\psi(x)=\frac{1}{c}\int^x g(y)\,dy. \end{align} Then \begin{align} &\phi(x)= \frac{1}{2} f(x)+\frac{1}{2c}\int^x g(y)\,dy,\\ &\psi(x)= \frac{1}{2} f(x)-\frac{1}{2c}\int^x g(y)\,dy. \end{align} Plugging into (\ref{GS}) and using property of an integral we get D'Alembert formula $$u(x,t)=\frac{1}{2}\bigl[ f(x+ct)+f(x-ct)\bigr]+\frac{1}{2c}\int_{x-ct}^{x+ct} g(y)\,dy. \label{DAF}$$