$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ ##[2.3. Homogeneous $1$D Wave equation](id:sect-2.3) ---------------------------- > 1. [Physical examples](#sect-2.3.1) > 2. [General solution](#sect-2.3.2) > 3. [Cauchy problem](#sect-2.3.3) Consider equation \begin{equation} u\_{tt}-c^2u\_{xx}=0. \label{eq-2.3.1} \end{equation} ###[Physical examples](id:sect-2.3.1) **[Remark 1.](id:remark-2.3.1)** As we mentioned in [Section 1.4](../Chapter1/S1.4.html#sect-1.4.1) this equation describes a lot of things. **[Example 1.](id:example-2.3.1)** Consider a string with the points deviating 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 2.](id:example-2.3.2)** This equation also describes compression-rarefication waves in elastic $1$-dimensional media. Then $u(x,t)$ is displacement along $x$. **[Example 3.](id:example-2.3.3)** 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 3.](id:remark-2.3.3)** $c$ has a dimension of the speed. In the example above $c$ is a speed of sound. ###[General solution](id:sect-2.3.2) Let us rewrite formally equation (\ref{eq-2.3.1}) as \begin{equation} (\partial\_t^2 -c^2 \partial\_x^2)u= (\partial\_t-c \partial\_x)(\partial\_t+c \partial\_x)u=0. \label{eq-2.3.2} \end{equation} 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,\label{eq-2.3.3}\\\\ &w\_t+cw\_x=0.\label{eq-2.3.4} \end{align} But from [Section 2.1](./S2.1.html) we know how to solve these equations \begin{align} &v=2c\phi'(x+ct),\label{eq-2.3.5}\\\\ &w=-2c\psi'(x-ct) \label{eq-2.3.6} \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 $w$ 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, \begin{equation} u=\phi(x+ct)+\psi(x-ct) \label{eq-2.3.7} \end{equation} is a *general* solution to (\ref{eq-2.3.1}). 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 3.](id:remark-2.3.3)** Adding constant $C$ to $\phi$ and $-C$ to $\psi$ we get the same solution $u$. However it is the only arbitrarness. ###[Cauchy problem](id:sect-2.3.3) Let us consider IVP (*initial–value problem*, aka *Cauchy problem*) for (\ref{eq-2.3.1}): \begin{align} &u\_{tt}-c^2u\_{xx}=0,\label{eq-2.3.8}\\\\ &u|\_{t=0}=g(x), &&u\_t|\_{t=0}=h(x). \label{eq-2.3.9} \end{align} Plugging (\ref{eq-2.3.7}) into initial conditions we have \begin{align} &\phi(x)+\psi(x)=g(x),\label{eq-2.3.10}\\\\ &c\phi'(x)-c\psi'(x)=h(x) \implies \phi(x)-\psi(x)=\\frac{1}{c}\int^x h(y)\,dy.\label{eq-2.3.11} \end{align} Then \begin{align} &\phi(x)= \frac{1}{2} g(x)+\frac{1}{2c}\int^x h(y)\,dy,\label{eq-2.3.12}\\\\ &\psi(x)= \frac{1}{2} g(x)-\frac{1}{2c}\int^x h(y)\,dy.\label{eq-2.3.13} \end{align} Plugging into (\ref{eq-2.3.7}) and using property of an integral we get *D'Alembert formula* \begin{equation} u(x,t)=\frac{1}{2}\bigl[g(x+ct)+g(x-ct)\bigr]+ \frac{1}{2c}\int\_{x-ct}^{x+ct} h(y)\,dy. \label{eq-2.3.14} \end{equation} **[Remark 4.](id:remark-2.3.4)** Later we generalize it to the case of inhomogeneous equation (with the right-hand expression $f(x,t)$ in (\ref{eq-2.3.8}). ______________ [$\Leftarrow$](./S2.2.html)  [$\Uparrow$](../contents.html)  [$\downarrow$](./S2.3.P.html)  [$\Rightarrow$](./S2.4.html)