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2.3. Homogeneous $1$D Wave equation

1. Physical examples
2. General solution
3. Cauchy problem

Consider equation \begin{equation} u_{tt}-c^2u_{xx}=0. \label{eq-2.3.1} \end{equation}

Physical examples

Remark 1. As we mentioned in Section 1.4 this equation describes a lot of things.

Remark 2. $c$ has a dimension of the speed. In the example above $c$ is a speed of sound.

Example 1.

1. This equation describes oscillations of the string (see Section 1.4, Example 1).
2. It also describes 1-dimensional gas oscillations (see Section 1.4, Example 3).
3. Further, this equation with $c=c_{\parallel}$ also describes compression-rarefication waves in the elastic $1$-dimensional media. Then $u(x,t)$ is displacement along $x$.
4. And also this equation with $c=c_{\perp}<c_{\parallel}$ also describes sheer waves in the elastic $1$-dimensional media. Then $u(x,t)$ is displacement in the direction, perpendicular to $x$.

General solution

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 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 (a.k.a. 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*} Observe that the right-hand side of (\ref{eq-2.3.5}) equals to $(\partial_t +c\partial_x)\phi(x+ct)$ and therefore $(\partial_t +c\partial_x)(u-\phi(x+ct))=0$. Then $u-\phi(x+ct)$ must be a function of $x-ct$: $u-\phi(x+ct)=\chi(x-ct)$ and plugging into (\ref{eq-2.3.6}) we conclude that $\chi=\psi$ (up to a constant, but both $\phi$ and $\psi$ are defined up to some constants).

Therefore \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 right respectively with the speed $c$. So $c$ is a propagation speed.

Remark 3. Adding constant $C$ to $\phi$ and $-C$ to $\psi$ we get the same solution $u$. However it is the only arbitrariness.

Cauchy problem

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. Later we generalize it to the case of inhomogeneous equation (with the right-hand expression $f(x,t)$ in (\ref{eq-2.3.8})).

$\Leftarrow$  $\Uparrow$  $\downarrow$  $\Rightarrow$