Solution to the Cauchy Problem for the Wave Equation
Description
$$ \begin{cases} u_{tt} = c^2 u_{xx} \\ u(0,x) = f(x) \\ u_{t}(0,x) = g(x) \end{cases} $$
This equation is a case where both the density $\rho (x) > 0$ and stiffness $\kappa (x) > 0$ are constants in the following wave equation, referring to $\displaystyle c : = {{\kappa} \over {\rho}}$ as the wave speed.
Here, $t$ is time, $x$ is position, and $u(t,x)$ represents the waveform at time $t$. $t$ is time, $x$ is position, and $u(t,x)$ represents the waveform at position $x$ at time $t$. $f$ and $g$ are initial conditions, where, notably, $f$ represents the waveform at time $t=0$.
A Cauchy problem refers to a wave equation with given initial values but without boundary conditions. The solution in this case is represented in the form of a simple formula, referred to as d’Alembert’s formula.
Solution
Step 1. $\Box u : = (\partial_{t}^{2} - c^2 \partial_{x}^{2} ) u = u_{tt} - c^2 u_{xx} = 0$
Defining the linear operator $\Box$ as above,
$$ \Box = (\partial_{t}^{2} - c^2 \partial_{x}^{2} ) = (\partial_{t} + c \partial_{x} ) (\partial_{t} - c \partial_{x} ) $$
Means that any element $u$ of $\ker ( \Box )$ becomes a solution to the given equation.
Where $u \in \ker ( \Box )$ can be represented as $u(t,x) = p (x - ct) + q(x + ct)$.
Step 2.
By the initial condition, we have $f(x) = u(0,x) = p(x) + q(x)$. Therefore,
$$ p(x - ct) = {{f(x +ct) + f(x - ct) } \over {2}} $$
Step 3.
By the initial condition, we have $f ' (x) = p '(x) + q’(x)$. Also, we have $g(x ) = u_{t} (0,x) = -c p '(x) + c q’ (x)$.
$$ \begin{align*} &\begin{cases} p '(x) = \dfrac{1}{2} f '(x) + \dfrac{1}{2}g(x) \\ q’(x) = \dfrac{1}{2} f '(x) + \dfrac{1}{2} g(x) \end{cases} \\ \implies& \begin{cases} p(x) = \dfrac{1}{2} f(x) + \dfrac{1}{2} \displaystyle \int_{0}^{x} g(z) dz + a \\ q(x) = \dfrac{1}{2} f(x) + \dfrac{1}{2} \displaystyle \int_{0}^{x} g(z) dz - a \end{cases} \\ \implies& q(x + ct) = \dfrac{1}{2c} \displaystyle \int_{x - ct}^{x + ct} g(z) dz \end{align*} $$
Step 4. d’Alembert’s Formula
Summarizing,
$$ u(t,x) = {{f(x +ct) + f(x - ct) } \over {2}} + {{1} \over {2c}} \int_{x - ct}^{x + ct} g(z) dz $$
is obtained.
■