$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ ##[2.4. 1D Wave equation reloaded:
characteristic coordinates](id:sect-2.4) ----------------------------------------------------- > 1. [Characteristic coordinates](#sect-2.4.1) > 2. [Application of characteristc coordinates](#sect-2.4.2) > 3. [d'Alembert formula](#sect-2.4.3) ###[Characteristic coordinates](id:sect-2.4.1) We realize that lines $x+ct=\const$ and $x-ct=\const$ play a very special role in our analysis. We call these lines *characteristics*. Let us introduce *characteristic coordinates* \begin{equation} \left\\{\begin{aligned} &\xi=x+ct,\\\\ &\eta=x-ct. \end{aligned}\right. \label{eq-2.4.1} \end{equation} **[Proposition 1](id:prop-4.1)** \begin{equation} u\_{tt}-c^2u\_{xx}=-4c^2u\_{\xi \eta}. \label{eq-2.4.2} \end{equation} *Proof.* From (\ref{eq-2.4.1}) we see that $x=\frac{1}{2}(\xi+\eta)$ and $t=\frac{1}{2c}(\xi-\eta)$ and therefore due to chain rule $v\_\xi = \frac{1}{2}v\_x+\frac{1}{2c}v\_t$ and $v\_\eta = \frac{1}{2}v\_x-\frac{1}{2c}v\_t$ and therefore \begin{equation\*} -4c^2u\_{\xi\eta}= -\frac{1}{4}(c\partial\_x+\partial\_t) (c\partial\_x-\partial\_t)u=u\_{tt}-c^2u\_{xx}. \tag\*{QED} \end{equation\*} Therefore wave equation [(2.3.1)](./S2.3.html#mjx-eqn-eq-2.3.1) [(2.3.1) means (1) from Section 2.3)] becomes in the characteristic coordinates \begin{equation} u\_{\xi\eta}=0 \label{eq-2.4.3} \end{equation} which we rewrite as $(u\_\xi)\_\eta=0\ \implies u\_\xi =\phi'(\xi)$ (really, $u\_\xi$ should not depend on $\eta$ and it is convenient to denote by $\phi(\xi)$ the primitive of $u\_\xi$). Then $(u-\phi(\xi))\_\xi =0 \implies u-\phi(\xi)=\psi(\eta)$ (due to the same arguments) and therefore \begin{equation} u=\phi(\xi)+\psi(\eta) \label{eq-2.4.4} \end{equation} is the *general solution* to (\ref{eq-2.4.3}). ###[Application of characteristc coordinates](id:sect-2.4.2) **[Example 1.](id:example-2.4.1)** Consider *Goursat problem* for (\ref{eq-2.4.3}): \begin{gather\*} u\_{\xi\eta}=0\qquad \text{as } \xi\>0,\eta\>0\\\\[3pt] u|\_{\eta=0}=g(\xi)\qquad \text{as }\xi\>0 ,\\\\[3pt] u|\_{\xi=0}=h(\eta)\qquad \text{as } \eta\>0 \end{gather\*} where $g$ and $h$ must satisfy *compatibility condition* $g(0)=h(0)$ (really $g(0)=u(0,0)=h(0)$). Then one can see easily that $u(\xi,\eta)=g(\xi)+h(\eta)-g(0)$ solves Goursat problem. Plugging (\ref{eq-2.4.1}) into (\ref{eq-2.4.4}) we get for a general solution [(2.3.1)](./S2.3.html#mjx-eqn-eq-2.3.1) \begin{equation} u=\phi (x+ct)+\psi(x-ct) \label{eq-2.4.5} \end{equation} which is exactly [(2.3.7)](./L3.html#mjx-eqn-eq-2.3.7). ###[d'Alembert formula](id:sect-2.4.3) So far we achieved nothing new. Consider now IVP: \begin{align} &u\_{tt}-c^2u\_{xx}=f(x,t),\label{eq-2.4.6} \\\\[3pt] &u|\_{t=0}=g(x),\label{eq-2.4.7}\\\\[3pt] &u\_t|\_{t=0}=h(x). \label{eq-2.4.8} \end{align} It is convenient for us to assume that $g=h=0$. Later we will get rid off this assumption. Rewriting (\ref{eq-2.4.6}) as \begin{equation\*} \tilde{u}\_{\xi\eta}= -\frac{1}{4c^2}\tilde{f}(\xi,\eta ) \end{equation\*} (where $\tilde{u}$ etc means that we use characteristic coordinates) we get after integration \begin{equation\*} \tilde{u}\_{\xi}= -\frac{1}{4c^2}\int^\eta \tilde{f}(\xi,\eta' ) \,d\eta'= -\frac{1}{4c^2}\int\_\xi ^\eta \tilde{f}(\xi,\eta') \,d\eta' {\color{red}{+ \phi'(\xi)}} \end{equation\*} with an indefinite integral in the middle. Note that $t=0$ means exactly that $\xi=\eta$ but then $u\_\xi=0$ there. Really, $u\_\xi$ is a linear combination of $u\_t$ and $u\_x$ but both of them are $0$ as $t=0$. Therefore $\phi'(\xi)=0$ and \begin{equation\*} \tilde{u}\_{\xi}= \frac{1}{4c^2}\int^\xi\_\eta \tilde{f}(\xi,\eta' )\,d\eta' \end{equation\*} where we flipped limits and changed sign. Integrating with respect to $\xi$ we arrive to \begin{equation\*} \tilde{u}= \frac{1}{4c^2} \int^\xi \Bigl[\int\_{\xi'} ^\eta \tilde{f}(\xi',\eta' ) \,d\eta'\\Bigr] \,d\xi' = \frac{1}{4c^2} \int\_\eta ^\xi \\Bigl[\int\_\eta ^{\xi'} \tilde{f}(\xi',\eta') \,d\eta'\Bigr]\,d\xi' {\color{red}{+ \psi(\eta)}} \end{equation\*} and $\psi(\eta)$ also must vanish because $u=0$ as $t=0$ (i.e. $\xi=\eta$). So \begin{equation} \tilde{u}(\xi,\eta)= \frac{1}{4c^2} \int\_\eta ^\xi \Bigl[\int\_\eta ^{\xi'} \tilde{f}(\xi',\eta' )\\,d\eta'\Bigr]\\,d\xi' . \label{eq-2.4.9} \end{equation} We got a solution as a *double integral* but we want to write it down as *2-dimensional integral* \begin{equation} \tilde{u}(\xi,\eta)= \frac{1}{4c^2} \iint\_{\tilde{\Delta}(\xi,\eta)} \tilde{f}(\xi',\eta' )\\,d\eta’d\xi' . \label{eq-2.4.10} \end{equation} But what is $\tilde{\Delta}$? Consider $\xi\>\eta$. Then $\xi'$ should run from $\eta$ to $\xi$ and for fixed $\xi'$, $\eta\<\xi'\<\xi$ eta should run from $\eta $ to $\xi'$. So, we get a triangle bounded by $\xi'=\eta'$, $\xi'=\xi$ and $\eta'=\eta$: ->![fig-2.4.1](./F2.4-1.svg)<- But in coordinates $(x,t)$ this domain $\Delta(x,t)$ is bounded by $t=0$ and two characteristics: ->![fig-2.4.2](./F2.4-2.svg)<- So, we get \begin{equation} u(x,t)= \frac{1}{2c} \iint\_{\Delta (x,t)} f(x',t' )\,dx'd t' . \label{eq-2.4.11} \end{equation} because we need to replace $d\xi'd\eta' $ by $|J|dx'dt'$ with *Jacobian* $J$. **Exercise.** Calculate $J$ and justify factor $2c$. [$\Leftarrow$](./S2.3.html)  [$\Uparrow$](../contents.html)  [$\downarrow$](./S2.4.P.html)  [$\Rightarrow$](./S2.5.html)