2.4. 1D Wave equation reloaded

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

2.4. 1D Wave equation reloaded:
characteristic coordinates

  1. Characteristic coordinates
  2. Application of characteristc coordinates
  3. d'Alembert formula

Characteristic coordinates

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 \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) [(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

Example 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) \begin{equation} u=\phi (x+ct)+\psi(x-ct) \label{eq-2.4.5} \end{equation} which is exactly (2.3.7).

d'Alembert formula

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 $\eta'=\eta$:


But in coordinates $(x,t)$ this domain $\Delta(x,t)$ is bounded by $t=0$ and two characteristics:


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 $2c$.

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